id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-018401 | Real Analysis: Uniform Continuity (Variant B) | 10 | Task: Let $f:[-782,782]\to\mathbb{R}$ be $f(x)=(3)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-782,782]$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Lipschitz on Bounded Domain",
"approach": "On a bounded interval, $|x^2-y^2|=|x-y||x+y|$ and $|x+y|$ is uniformly bounded, giving a global Lipschitz constant.",
"steps": [
"Step 1: For $x,y\\in[-782,782]$, $|f(x)-f(y)|=|3|\\,|x^2-y^2|=|3|\\,|x-y||x+y|$.",
"Step 2: Bound $|x... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Yes}$.\nHeine–Cantor implies uniform continuity abstractly from compactness. The Lipschitz bound constructs an explicit modulus $\\delta(\\varepsilon)=\\varepsilon/(2M)$; both agree on the same con... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018402 | Real Analysis: Uniform Continuity (Core) | 10 | Give a fully justified solution: Let $f:(0,247)\to\mathbb{R}$ be $f(x)=\sqrt{5x}$. Prove that $f$ is uniformly continuous on $(0,247)$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Direct Inequality",
"approach": "Bound $|\\sqrt{x}-\\sqrt{y}|$ in terms of $|x-y|$ uniformly using rationalization.",
"steps": [
"Step 1: For $x,y>0$, $|\\sqrt{x}-\\sqrt{y}|=\\frac{|x-y|}{\\sqrt{x}+\\sqrt{y}}$.",
"Step 2: Since $\\sqrt{x}+\\sqrt{y}\\ge \\sqrt{|x-y|}$, we ge... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Yes}$.\nThe direct inequality provides an explicit modulus $\\delta=\\varepsilon^2$. Heine–Cantor guarantees the existence of some modulus on the compact extension; both give the same conc... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. |
math-018403 | Real Analysis: Uniform Continuity (Variant C) | 10 | Problem: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(8x)$. Prove that $f$ is uniformly continuous on $\mathbb{R}$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Lipschitz via Mean Value Theorem",
"approach": "Use MVT to get a global Lipschitz bound $|\\sin x-\\sin y|\\le |x-y|$.",
"steps": [
"Step 1: Apply MVT to $g(t)=\\sin(ct)$: there exists $\\xi$ between $x$ and $y$ with $g(x)-g(y)=g'(\\xi)(x-y)$.",
"Step 2: Since $g'(t)=(8)\\c... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Yes}$.\nThe MVT proof gives an explicit global Lipschitz modulus, which immediately implies the Cauchy-sequence property. Both are equivalent characterizations of uniform continuity.",
"robustnes... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018404 | Real Analysis: Uniform Continuity (Variant B) | 10 | Track quantifiers carefully: Let $f:(0,1513)\to\mathbb{R}$ be $f(x)=\frac{1}{(-29)x}$. Is $f$ uniformly continuous on $(0,1513)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both c... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018405 | Real Analysis: Uniform Continuity (Core) | 10 | Complete the analysis: Let $f:(0,445)\to\mathbb{R}$ be $f(x)=\frac{1}{(14)x}$. Is $f$ uniformly continuous on $(0,445)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{445}{n}$ and $y_n=\\frac{445}{2n}$ in $(0,445)$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018406 | Real Analysis: Uniform Continuity (Core) | 10 | Show all reasoning: Let $f:(0,1634)\to\mathbb{R}$ be $f(x)=\ln(22x)$. Is $f$ uniformly continuous on $(0,1634)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{1634}{n}$ and $y_n=\\frac{1634}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\l... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018407 | Real Analysis: Uniform Continuity (Variant A) | 10 | Track units/moduli carefully: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(-9x)$. Prove that $f$ is uniformly continuous on $\mathbb{R}$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Lipschitz via Mean Value Theorem",
"approach": "Use MVT to get a global Lipschitz bound $|\\sin x-\\sin y|\\le |x-y|$.",
"steps": [
"Step 1: Apply MVT to $g(t)=\\sin(ct)$: there exists $\\xi$ between $x$ and $y$ with $g(x)-g(y)=g'(\\xi)(x-y)$.",
"Step 2: Since $g'(t)=(-9)\\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Yes}$.\nThe MVT proof gives an explicit global Lipschitz modulus, which immediately implies the Cauchy-sequence property. Both are equivalent characterizations of uniform continuity.",
"robustness_analysis": "If the pr... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018408 | Real Analysis: Uniform Continuity (Core) | 10 | Provide both a computational and a conceptual explanation: Let $f:(0,1073)\to\mathbb{R}$ be $f(x)=\ln(15x)$. Is $f$ uniformly continuous on $(0,1073)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{1073}{n}$ and $y_n=\\frac{1073}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\l... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robus... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018409 | Real Analysis: Uniform Continuity (Variant C) | 10 | Work carefully and justify each inference: Let $f:(0,874)\to\mathbb{R}$ be $f(x)=\ln(41x)$. Is $f$ uniformly continuous on $(0,874)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{874}{n}$ and $y_n=\\frac{874}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\ln(... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": "Sensi... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018410 | Real Analysis: Uniform Continuity (Core) | 10 | Give a fully justified solution: Let $f:(0,495)\to\mathbb{R}$ be $f(x)=\ln(39x)$. Is $f$ uniformly continuous on $(0,495)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": "Gener... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018411 | Real Analysis: Uniform Continuity (Variant B) | 10 | Prompt: Let $f:(0,747)\to\mathbb{R}$ be $f(x)=\ln(32x)$. Is $f$ uniformly continuous on $(0,747)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018412 | Real Analysis: Uniform Continuity (Variant A) | 10 | Track quantifiers carefully: Let $f:(0,657)\to\mathbb{R}$ be $f(x)=\frac{1}{(24)x}$. Is $f$ uniformly continuous on $(0,657)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{657}{n}$ and $y_n=\\frac{657}{2n}$ in $(0,657)$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018413 | Real Analysis: Uniform Continuity (Variant C) | 10 | Give reasoning, not just computation: Let $f:(0,1411)\to\mathbb{R}$ be $f(x)=\frac{1}{(-4)x}$. Is $f$ uniformly continuous on $(0,1411)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}}.",
... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018414 | Real Analysis: Uniform Continuity (Core) | 10 | Make each step logically reversible (or explain if not): Let $f:(0,1040)\to\mathbb{R}$ be $f(x)=\frac{1}{(26)x}$. Is $f$ uniformly continuous on $(0,1040)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{1040}{n}$ and $y_n=\\frac{1040}{2n}$ in $(0,1040)$.",
"Step 2: Then $|x_n-y_n|=\\frac{... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}}.",
... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018415 | Real Analysis: Uniform Continuity (Variant B) | 10 | Challenge: Let $f:(0,1056)\to\mathbb{R}$ be $f(x)=\ln(32x)$. Is $f$ uniformly continuous on $(0,1056)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": "Robus... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018416 | Real Analysis: Uniform Continuity (Core) | 10 | Warm-up: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(2)x^2$ with $c\ne 0$. Is $f$ uniformly continuous on $\mathbb{R}$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Mean Value Theorem Blow-Up",
"approach": "Use MVT to show that on unbounded domains, the derivative growth makes a uniform modulus impossible.",
"steps": [
"Step 1: Suppose uniform continuity holds. Take $\\varepsilon=1$ and let $\\delta>0$ correspond.",
"Step 2: Choose $x$... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nThe sequence method explicitly exhibits $|x_n-y_n|\\to 0$ while $|f(x_n)-f(y_n)|\\not\\to 0$. The MVT method formalizes the same idea via unbounded derivative growth. Both conclude non-uniform continuity.",
"rob... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018417 | Real Analysis: Uniform Continuity (Variant B) | 10 | Warm-up: Let $f:(0,379)\to\mathbb{R}$ be $f(x)=\ln(26x)$. Is $f$ uniformly continuous on $(0,379)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{379}{n}$ and $y_n=\\frac{379}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\ln(... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018418 | Real Analysis: Uniform Continuity (Variant B) | 10 | Checkpoint: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(-10)x^2$ with $c\ne 0$. Is $f$ uniformly continuous on $\mathbb{R}$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Mean Value Theorem Blow-Up",
"approach": "Use MVT to show that on unbounded domains, the derivative growth makes a uniform modulus impossible.",
"steps": [
"Step 1: Suppose uniform continuity holds. Take $\\varepsilon=1$ and let $\\delta>0$ correspond.",
"Step 2: Choose $x$... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nThe sequence method explicitly exhibits $|x_n-y_n|\\to 0$ while $|f(x_n)-f(y_n)|\\not\\to 0$. The MVT method formalizes the same idea via unbounded derivative growth. Both conclude ... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018419 | Real Analysis: Uniform Continuity (Variant A) | 10 | Answer with a short justification: Let $f:(0,671)\to\mathbb{R}$ be $f(x)=\frac{1}{(7)x}$. Is $f$ uniformly continuous on $(0,671)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{671}{n}$ and $y_n=\\frac{671}{2n}$ in $(0,671)$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}}.",
... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018420 | Real Analysis: Uniform Continuity (Variant A) | 10 | Give reasoning, not just computation: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(-19x)$. Prove that $f$ is uniformly continuous on $\mathbb{R}$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Lipschitz via Mean Value Theorem",
"approach": "Use MVT to get a global Lipschitz bound $|\\sin x-\\sin y|\\le |x-y|$.",
"steps": [
"Step 1: Apply MVT to $g(t)=\\sin(ct)$: there exists $\\xi$ between $x$ and $y$ with $g(x)-g(y)=g'(\\xi)(x-y)$.",
"Step 2: Since $g'(t)=(-19)\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Yes}$.\nThe MVT proof gives an explicit global Lipschitz modulus, which immediately implies the Cauchy-sequence property. Both are equivalent characterizations of uniform continuity.",
"robustness_analysis": "Sensitivi... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018421 | Real Analysis: Uniform Continuity (Variant A) | 10 | Provide both a computational and a conceptual explanation: Let $f:(0,173)\to\mathbb{R}$ be $f(x)=\sqrt{54x}$. Prove that $f$ is uniformly continuous on $(0,173)$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Direct Inequality",
"approach": "Bound $|\\sqrt{x}-\\sqrt{y}|$ in terms of $|x-y|$ uniformly using rationalization.",
"steps": [
"Step 1: For $x,y>0$, $|\\sqrt{x}-\\sqrt{y}|=\\frac{|x-y|}{\\sqrt{x}+\\sqrt{y}}$.",
"Step 2: Since $\\sqrt{x}+\\sqrt{y}\\ge \\sqrt{|x-y|}$, we ge... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Yes}$.\nThe direct inequality provides an explicit modulus $\\delta=\\varepsilon^2$. Heine–Cantor guarantees the existence of some modulus on the compact extension; both give the same conclusion.",... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018422 | Real Analysis: Uniform Continuity (Variant B) | 10 | Derive the result step-by-step: Let $f:(0,667)\to\mathbb{R}$ be $f(x)=\sqrt{9x}$. Prove that $f$ is uniformly continuous on $(0,667)$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Compact Extension + Heine–Cantor",
"approach": "Extend continuously to a compact interval and invoke Heine–Cantor, then restrict back.",
"steps": [
"Step 1: Define $g:[0,667]\\to\\mathbb{R}$ by $g(x)=\\sqrt{x}$; this is continuous on $[0,667]$.",
"Step 2: The interval $[0,6... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Yes}$.\nThe direct inequality provides an explicit modulus $\\delta=\\varepsilon^2$. Heine–Cantor guarantees the existence of some modulus on the compact extension; both give the same conc... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018423 | Real Analysis: Uniform Continuity (Variant B) | 10 | Proceed methodically: Let $f:(0,1510)\to\mathbb{R}$ be $f(x)=\sqrt{56x}$. Prove that $f$ is uniformly continuous on $(0,1510)$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Compact Extension + Heine–Cantor",
"approach": "Extend continuously to a compact interval and invoke Heine–Cantor, then restrict back.",
"steps": [
"Step 1: Define $g:[0,1510]\\to\\mathbb{R}$ by $g(x)=\\sqrt{x}$; this is continuous on $[0,1510]$.",
"Step 2: The interval $[0... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Yes}$.\nThe direct inequality provides an explicit modulus $\\delta=\\varepsilon^2$. Heine–Cantor guarantees the existence of some modulus on the compact extension; both give the same conc... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. |
math-018424 | Real Analysis: Uniform Continuity (Variant B) | 10 | Problem: Let $f:(0,980)\to\mathbb{R}$ be $f(x)=\frac{1}{(3)x}$. Is $f$ uniformly continuous on $(0,980)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{980}{n}$ and $y_n=\\frac{980}{2n}$ in $(0,980)$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}}.",
... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018425 | Real Analysis: Uniform Continuity (Variant C) | 10 | Give reasoning, not just computation: Let $f:(0,212)\to\mathbb{R}$ be $f(x)=\ln(32x)$. Is $f$ uniformly continuous on $(0,212)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robus... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018426 | Real Analysis: Uniform Continuity (Variant C) | 10 | Provide both a computational and a conceptual explanation: Let $f:(0,1790)\to\mathbb{R}$ be $f(x)=\ln(39x)$. Is $f$ uniformly continuous on $(0,1790)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": "Robus... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018427 | Real Analysis: Uniform Continuity (Variant C) | 10 | Answer using clear logical steps: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(17x)$. Prove that $f$ is uniformly continuous on $\mathbb{R}$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Characterization",
"approach": "A function is uniformly continuous iff it sends Cauchy sequences to Cauchy sequences; use the global inequality $|\\sin x-\\sin y|\\le |x-y|$.",
"steps": [
"Step 1: For all real $x,y$, $|\\sin x-\\sin y|\\le |x-y|$ (e.g., by MVT).",... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Yes}$.\nThe MVT proof gives an explicit global Lipschitz modulus, which immediately implies the Cauchy-sequence property. Both are equivalent characterizations of uniform continuity.",
"robustnes... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018428 | Real Analysis: Uniform Continuity (Variant C) | 10 | Solve and sanity-check: Let $f:(0,328)\to\mathbb{R}$ be $f(x)=\sqrt{56x}$. Prove that $f$ is uniformly continuous on $(0,328)$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Compact Extension + Heine–Cantor",
"approach": "Extend continuously to a compact interval and invoke Heine–Cantor, then restrict back.",
"steps": [
"Step 1: Define $g:[0,328]\\to\\mathbb{R}$ by $g(x)=\\sqrt{x}$; this is continuous on $[0,328]$.",
"Step 2: The interval $[0,3... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Yes}$.\nThe direct inequality provides an explicit modulus $\\delta=\\varepsilon^2$. Heine–Cantor guarantees the existence of some modulus on the compact extension; both give the same conclusion.",
"robustness_an... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018429 | Real Analysis: Uniform Continuity (Variant C) | 10 | Carefully track domains: Let $f:(0,1054)\to\mathbb{R}$ be $f(x)=\frac{1}{(8)x}$. Is $f$ uniformly continuous on $(0,1054)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both c... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018430 | Real Analysis: Uniform Continuity (Variant A) | 10 | Exercise: Let $f:(0,1740)\to\mathbb{R}$ be $f(x)=\ln(26x)$. Is $f$ uniformly continuous on $(0,1740)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018431 | Real Analysis: Uniform Continuity (Core) | 10 | Prompt: Let $f:(0,1673)\to\mathbb{R}$ be $f(x)=\ln(17x)$. Is $f$ uniformly continuous on $(0,1673)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{1673}{n}$ and $y_n=\\frac{1673}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\l... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": ... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018432 | Real Analysis: Uniform Continuity (Variant C) | 10 | Complete the analysis: Let $f:(0,776)\to\mathbb{R}$ be $f(x)=\ln(48x)$. Is $f$ uniformly continuous on $(0,776)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robus... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018433 | Real Analysis: Uniform Continuity (Variant A) | 10 | Carefully track domains: Let $f:[-654,654]\to\mathbb{R}$ be $f(x)=(-7)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-654,654]$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Lipschitz on Bounded Domain",
"approach": "On a bounded interval, $|x^2-y^2|=|x-y||x+y|$ and $|x+y|$ is uniformly bounded, giving a global Lipschitz constant.",
"steps": [
"Step 1: For $x,y\\in[-654,654]$, $|f(x)-f(y)|=|-7|\\,|x^2-y^2|=|-7|\\,|x-y||x+y|$.",
"Step 2: Bound $... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Yes}$.\nHeine–Cantor implies uniform continuity abstractly from compactness. The Lipschitz bound constructs an explicit modulus $\\delta(\\varepsilon)=\\varepsilon/(2M)$; both agree on the same con... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018434 | Real Analysis: Uniform Continuity (Variant C) | 10 | Prompt: Let $f:(0,756)\to\mathbb{R}$ be $f(x)=\frac{1}{(-23)x}$. Is $f$ uniformly continuous on $(0,756)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{756}{n}$ and $y_n=\\frac{756}{2n}$ in $(0,756)$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018435 | Real Analysis: Uniform Continuity (Variant A) | 10 | Compute the requested quantity: Let $f:(0,1736)\to\mathbb{R}$ be $f(x)=\sqrt{17x}$. Prove that $f$ is uniformly continuous on $(0,1736)$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Compact Extension + Heine–Cantor",
"approach": "Extend continuously to a compact interval and invoke Heine–Cantor, then restrict back.",
"steps": [
"Step 1: Define $g:[0,1736]\\to\\mathbb{R}$ by $g(x)=\\sqrt{x}$; this is continuous on $[0,1736]$.",
"Step 2: The interval $[0... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Yes}$.\nThe direct inequality provides an explicit modulus $\\delta=\\varepsilon^2$. Heine–Cantor guarantees the existence of some modulus on the compact extension; both give the same conclusion.",
"robustness_analysis... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018436 | Real Analysis: Uniform Continuity (Core) | 10 | Write the solution set clearly: Let $f:[-1857,1857]\to\mathbb{R}$ be $f(x)=(27)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-1857,1857]$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Heine–Cantor",
"approach": "Continuous functions on compact metric spaces are uniformly continuous.",
"steps": [
"Step 1: The interval $[-1857,1857]$ is compact in $\\mathbb{R}$.",
"Step 2: The function $x\\mapsto x^2$ is continuous on $\\mathbb{R}$, hence on the compact se... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Yes}$.\nHeine–Cantor implies uniform continuity abstractly from compactness. The Lipschitz bound constructs an explicit modulus $\\delta(\\varepsilon)=\\varepsilon/(2M)$; both agree on the same conclusion.",
"rob... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. |
math-018437 | Real Analysis: Uniform Continuity (Core) | 10 | Answer using clear logical steps: Let $f:(0,945)\to\mathbb{R}$ be $f(x)=\frac{1}{(4)x}$. Is $f$ uniformly continuous on $(0,945)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}}.",
... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018438 | Real Analysis: Uniform Continuity (Variant A) | 10 | Give a fully justified solution: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(21)x^2$ with $c\ne 0$. Is $f$ uniformly continuous on $\mathbb{R}$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Mean Value Theorem Blow-Up",
"approach": "Use MVT to show that on unbounded domains, the derivative growth makes a uniform modulus impossible.",
"steps": [
"Step 1: Suppose uniform continuity holds. Take $\\varepsilon=1$ and let $\\delta>0$ correspond.",
"Step 2: Choose $x$... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nThe sequence method explicitly exhibits $|x_n-y_n|\\to 0$ while $|f(x_n)-f(y_n)|\\not\\to 0$. The MVT method formalizes the same idea via unbounded derivative growth. Both conclude non-uniform continuity.",
"rob... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018439 | Real Analysis: Uniform Continuity (Variant B) | 10 | Answer with a short justification: Let $f:(0,825)\to\mathbb{R}$ be $f(x)=\frac{1}{(-16)x}$. Is $f$ uniformly continuous on $(0,825)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{825}{n}$ and $y_n=\\frac{825}{2n}$ in $(0,825)$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018440 | Real Analysis: Uniform Continuity (Variant C) | 10 | Complete the analysis: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(-24)x^2$ with $c\ne 0$. Is $f$ uniformly continuous on $\mathbb{R}$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Explicit Sequence Counterexample",
"approach": "Pick sequences approaching each other at infinity but whose squares stay separated.",
"steps": [
"Step 1: Let $x_n=n$ and $y_n=n+\\frac{1}{n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{1}{n}\\to 0$.",
"Final step: But $|f(x_n)-... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nThe sequence method explicitly exhibits $|x_n-y_n|\\to 0$ while $|f(x_n)-f(y_n)|\\not\\to 0$. The MVT method formalizes the same idea via unbounded derivative growth. Both conclude non-uniform continuity.",
... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018441 | Real Analysis: Uniform Continuity (Variant B) | 10 | Give a theorem-based solution: Let $f:(0,1346)\to\mathbb{R}$ be $f(x)=\sqrt{48x}$. Prove that $f$ is uniformly continuous on $(0,1346)$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Compact Extension + Heine–Cantor",
"approach": "Extend continuously to a compact interval and invoke Heine–Cantor, then restrict back.",
"steps": [
"Step 1: Define $g:[0,1346]\\to\\mathbb{R}$ by $g(x)=\\sqrt{x}$; this is continuous on $[0,1346]$.",
"Step 2: The interval $[0... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Yes}$.\nThe direct inequality provides an explicit modulus $\\delta=\\varepsilon^2$. Heine–Cantor guarantees the existence of some modulus on the compact extension; both give the same conc... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. |
math-018442 | Real Analysis: Uniform Continuity (Core) | 10 | Do not skip justification steps: Let $f:(0,231)\to\mathbb{R}$ be $f(x)=\sqrt{34x}$. Prove that $f$ is uniformly continuous on $(0,231)$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Direct Inequality",
"approach": "Bound $|\\sqrt{x}-\\sqrt{y}|$ in terms of $|x-y|$ uniformly using rationalization.",
"steps": [
"Step 1: For $x,y>0$, $|\\sqrt{x}-\\sqrt{y}|=\\frac{|x-y|}{\\sqrt{x}+\\sqrt{y}}$.",
"Step 2: Since $\\sqrt{x}+\\sqrt{y}\\ge \\sqrt{|x-y|}$, we ge... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Yes}$.\nThe direct inequality provides an explicit modulus $\\delta=\\varepsilon^2$. Heine–Cantor guarantees the existence of some modulus on the compact extension; both give the same conclusion.",... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018443 | Real Analysis: Uniform Continuity (Core) | 10 | Track quantifiers carefully: Let $f:(0,223)\to\mathbb{R}$ be $f(x)=\sqrt{48x}$. Prove that $f$ is uniformly continuous on $(0,223)$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Direct Inequality",
"approach": "Bound $|\\sqrt{x}-\\sqrt{y}|$ in terms of $|x-y|$ uniformly using rationalization.",
"steps": [
"Step 1: For $x,y>0$, $|\\sqrt{x}-\\sqrt{y}|=\\frac{|x-y|}{\\sqrt{x}+\\sqrt{y}}$.",
"Step 2: Since $\\sqrt{x}+\\sqrt{y}\\ge \\sqrt{|x-y|}$, we ge... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Yes}$.\nThe direct inequality provides an explicit modulus $\\delta=\\varepsilon^2$. Heine–Cantor guarantees the existence of some modulus on the compact extension; both give the same conclusion.",... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018444 | Real Analysis: Uniform Continuity (Core) | 10 | Track quantifiers carefully: Let $f:(0,1257)\to\mathbb{R}$ be $f(x)=\ln(53x)$. Is $f$ uniformly continuous on $(0,1257)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": ... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018445 | Real Analysis: Uniform Continuity (Core) | 10 | Task: Let $f:(0,919)\to\mathbb{R}$ be $f(x)=\ln(43x)$. Is $f$ uniformly continuous on $(0,919)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{919}{n}$ and $y_n=\\frac{919}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\ln(... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": ... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018446 | Real Analysis: Uniform Continuity (Variant A) | 10 | Give reasoning, not just computation: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(27)x^2$ with $c\ne 0$. Is $f$ uniformly continuous on $\mathbb{R}$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Explicit Sequence Counterexample",
"approach": "Pick sequences approaching each other at infinity but whose squares stay separated.",
"steps": [
"Step 1: Let $x_n=n$ and $y_n=n+\\frac{1}{n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{1}{n}\\to 0$.",
"Final step: But $|f(x_n)-... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nThe sequence method explicitly exhibits $|x_n-y_n|\\to 0$ while $|f(x_n)-f(y_n)|\\not\\to 0$. The MVT method formalizes the same idea via unbounded derivative growth. Both conclude non-uniform continuity.",
"rob... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018447 | Real Analysis: Uniform Continuity (Variant C) | 10 | Track quantifiers carefully: Let $f:[-192,192]\to\mathbb{R}$ be $f(x)=(-19)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-192,192]$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Lipschitz on Bounded Domain",
"approach": "On a bounded interval, $|x^2-y^2|=|x-y||x+y|$ and $|x+y|$ is uniformly bounded, giving a global Lipschitz constant.",
"steps": [
"Step 1: For $x,y\\in[-192,192]$, $|f(x)-f(y)|=|-19|\\,|x^2-y^2|=|-19|\\,|x-y||x+y|$.",
"Step 2: Bound... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Yes}$.\nHeine–Cantor implies uniform continuity abstractly from compactness. The Lipschitz bound constructs an explicit modulus $\\delta(\\varepsilon)=\\varepsilon/(2M)$; both agree on the same conclusion.",
"robustnes... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018448 | Real Analysis: Uniform Continuity (Variant C) | 10 | Proceed methodically: Let $f:[-1104,1104]\to\mathbb{R}$ be $f(x)=(-11)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-1104,1104]$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Lipschitz on Bounded Domain",
"approach": "On a bounded interval, $|x^2-y^2|=|x-y||x+y|$ and $|x+y|$ is uniformly bounded, giving a global Lipschitz constant.",
"steps": [
"Step 1: For $x,y\\in[-1104,1104]$, $|f(x)-f(y)|=|-11|\\,|x^2-y^2|=|-11|\\,|x-y||x+y|$.",
"Step 2: Bou... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Yes}$.\nHeine–Cantor implies uniform continuity abstractly from compactness. The Lipschitz bound constructs an explicit modulus $\\delta(\\varepsilon)=\\varepsilon/(2M)$; both agree on the same con... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. |
math-018449 | Real Analysis: Uniform Continuity (Core) | 10 | Checkpoint: Let $f:(0,1493)\to\mathbb{R}$ be $f(x)=\ln(18x)$. Is $f$ uniformly continuous on $(0,1493)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": "If th... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018450 | Real Analysis: Uniform Continuity (Core) | 10 | Try to avoid pattern-matching; explain why: Let $f:(0,1189)\to\mathbb{R}$ be $f(x)=\frac{1}{(6)x}$. Is $f$ uniformly continuous on $(0,1189)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018451 | Real Analysis: Uniform Continuity (Variant A) | 10 | Write the solution set clearly: Let $f:(0,1172)\to\mathbb{R}$ be $f(x)=\frac{1}{(-8)x}$. Is $f$ uniformly continuous on $(0,1172)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both c... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018452 | Real Analysis: Uniform Continuity (Variant C) | 10 | Write the solution set clearly: Let $f:(0,135)\to\mathbb{R}$ be $f(x)=\frac{1}{(26)x}$. Is $f$ uniformly continuous on $(0,135)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{135}{n}$ and $y_n=\\frac{135}{2n}$ in $(0,135)$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018453 | Real Analysis: Uniform Continuity (Variant B) | 10 | Where appropriate, name the theorem you use: Let $f:[-89,89]\to\mathbb{R}$ be $f(x)=(23)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-89,89]$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Heine–Cantor",
"approach": "Continuous functions on compact metric spaces are uniformly continuous.",
"steps": [
"Step 1: The interval $[-89,89]$ is compact in $\\mathbb{R}$.",
"Step 2: The function $x\\mapsto x^2$ is continuous on $\\mathbb{R}$, hence on the compact set.",... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Yes}$.\nHeine–Cantor implies uniform continuity abstractly from compactness. The Lipschitz bound constructs an explicit modulus $\\delta(\\varepsilon)=\\varepsilon/(2M)$; both agree on the same con... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. |
math-018454 | Real Analysis: Uniform Continuity (Variant A) | 10 | Complete the analysis: Let $f:(0,568)\to\mathbb{R}$ be $f(x)=\ln(24x)$. Is $f$ uniformly continuous on $(0,568)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{568}{n}$ and $y_n=\\frac{568}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\ln(... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": ... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018455 | Real Analysis: Uniform Continuity (Variant B) | 10 | Do not skip justification steps: Let $f:(0,1915)\to\mathbb{R}$ be $f(x)=\sqrt{11x}$. Prove that $f$ is uniformly continuous on $(0,1915)$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Direct Inequality",
"approach": "Bound $|\\sqrt{x}-\\sqrt{y}|$ in terms of $|x-y|$ uniformly using rationalization.",
"steps": [
"Step 1: For $x,y>0$, $|\\sqrt{x}-\\sqrt{y}|=\\frac{|x-y|}{\\sqrt{x}+\\sqrt{y}}$.",
"Step 2: Since $\\sqrt{x}+\\sqrt{y}\\ge \\sqrt{|x-y|}$, we ge... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Yes}$.\nThe direct inequality provides an explicit modulus $\\delta=\\varepsilon^2$. Heine–Cantor guarantees the existence of some modulus on the compact extension; both give the same conclusion.",
"robustness_an... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. |
math-018456 | Real Analysis: Uniform Continuity (Variant B) | 10 | State any required conditions first: Let $f:(0,1062)\to\mathbb{R}$ be $f(x)=\frac{1}{(5)x}$. Is $f$ uniformly continuous on $(0,1062)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{1062}{n}$ and $y_n=\\frac{1062}{2n}$ in $(0,1062)$.",
"Step 2: Then $|x_n-y_n|=\\frac{... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018457 | Real Analysis: Uniform Continuity (Variant B) | 10 | Start by stating any domain restrictions: Let $f:(0,234)\to\mathbb{R}$ be $f(x)=\sqrt{6x}$. Prove that $f$ is uniformly continuous on $(0,234)$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Compact Extension + Heine–Cantor",
"approach": "Extend continuously to a compact interval and invoke Heine–Cantor, then restrict back.",
"steps": [
"Step 1: Define $g:[0,234]\\to\\mathbb{R}$ by $g(x)=\\sqrt{x}$; this is continuous on $[0,234]$.",
"Step 2: The interval $[0,2... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Yes}$.\nThe direct inequality provides an explicit modulus $\\delta=\\varepsilon^2$. Heine–Cantor guarantees the existence of some modulus on the compact extension; both give the same conclusion.",... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. |
math-018458 | Real Analysis: Uniform Continuity (Variant C) | 10 | Question: Let $f:(0,367)\to\mathbb{R}$ be $f(x)=\frac{1}{(29)x}$. Is $f$ uniformly continuous on $(0,367)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{367}{n}$ and $y_n=\\frac{367}{2n}$ in $(0,367)$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both c... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018459 | Real Analysis: Uniform Continuity (Variant A) | 10 | Proceed methodically: Let $f:(0,913)\to\mathbb{R}$ be $f(x)=\sqrt{32x}$. Prove that $f$ is uniformly continuous on $(0,913)$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Compact Extension + Heine–Cantor",
"approach": "Extend continuously to a compact interval and invoke Heine–Cantor, then restrict back.",
"steps": [
"Step 1: Define $g:[0,913]\\to\\mathbb{R}$ by $g(x)=\\sqrt{x}$; this is continuous on $[0,913]$.",
"Step 2: The interval $[0,9... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Yes}$.\nThe direct inequality provides an explicit modulus $\\delta=\\varepsilon^2$. Heine–Cantor guarantees the existence of some modulus on the compact extension; both give the same conc... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018460 | Real Analysis: Uniform Continuity (Variant A) | 10 | Challenge: Let $f:(0,860)\to\mathbb{R}$ be $f(x)=\ln(50x)$. Is $f$ uniformly continuous on $(0,860)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robus... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018461 | Real Analysis: Uniform Continuity (Variant C) | 10 | Work carefully and justify each inference: Let $f:(0,979)\to\mathbb{R}$ be $f(x)=\sqrt{48x}$. Prove that $f$ is uniformly continuous on $(0,979)$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Compact Extension + Heine–Cantor",
"approach": "Extend continuously to a compact interval and invoke Heine–Cantor, then restrict back.",
"steps": [
"Step 1: Define $g:[0,979]\\to\\mathbb{R}$ by $g(x)=\\sqrt{x}$; this is continuous on $[0,979]$.",
"Step 2: The interval $[0,9... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Yes}$.\nThe direct inequality provides an explicit modulus $\\delta=\\varepsilon^2$. Heine–Cantor guarantees the existence of some modulus on the compact extension; both give the same conclusion.",
"robustness_analysis... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. |
math-018462 | Real Analysis: Uniform Continuity (Variant C) | 10 | Determine the requested value: Let $f:(0,1260)\to\mathbb{R}$ be $f(x)=\ln(11x)$. Is $f$ uniformly continuous on $(0,1260)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{1260}{n}$ and $y_n=\\frac{1260}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\l... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": ... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018463 | Real Analysis: Uniform Continuity (Variant A) | 10 | Answer using clear logical steps: Let $f:(0,1966)\to\mathbb{R}$ be $f(x)=\frac{1}{(-5)x}$. Is $f$ uniformly continuous on $(0,1966)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018464 | Real Analysis: Uniform Continuity (Core) | 10 | Answer with a short justification: Let $f:(0,1853)\to\mathbb{R}$ be $f(x)=\ln(25x)$. Is $f$ uniformly continuous on $(0,1853)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": "Sensi... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018465 | Real Analysis: Uniform Continuity (Variant C) | 10 | Provide both a computational and a conceptual explanation: Let $f:(0,761)\to\mathbb{R}$ be $f(x)=\frac{1}{(-15)x}$. Is $f$ uniformly continuous on $(0,761)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018466 | Real Analysis: Uniform Continuity (Core) | 10 | Track quantifiers carefully: Let $f:(0,857)\to\mathbb{R}$ be $f(x)=\frac{1}{(13)x}$. Is $f$ uniformly continuous on $(0,857)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{857}{n}$ and $y_n=\\frac{857}{2n}$ in $(0,857)$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}}.",
... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018467 | Real Analysis: Uniform Continuity (Variant C) | 10 | Give an answer and a quick verification: Let $f:(0,335)\to\mathbb{R}$ be $f(x)=\sqrt{57x}$. Prove that $f$ is uniformly continuous on $(0,335)$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Direct Inequality",
"approach": "Bound $|\\sqrt{x}-\\sqrt{y}|$ in terms of $|x-y|$ uniformly using rationalization.",
"steps": [
"Step 1: For $x,y>0$, $|\\sqrt{x}-\\sqrt{y}|=\\frac{|x-y|}{\\sqrt{x}+\\sqrt{y}}$.",
"Step 2: Since $\\sqrt{x}+\\sqrt{y}\\ge \\sqrt{|x-y|}$, we ge... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Yes}$.\nThe direct inequality provides an explicit modulus $\\delta=\\varepsilon^2$. Heine–Cantor guarantees the existence of some modulus on the compact extension; both give the same conclusion.",
"robustness_an... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018468 | Real Analysis: Uniform Continuity (Variant C) | 10 | Work carefully and justify each inference: Let $f:(0,1133)\to\mathbb{R}$ be $f(x)=\ln(57x)$. Is $f$ uniformly continuous on $(0,1133)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{1133}{n}$ and $y_n=\\frac{1133}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\l... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018469 | Real Analysis: Uniform Continuity (Variant C) | 10 | Solve and include a self-check: Let $f:(0,1609)\to\mathbb{R}$ be $f(x)=\ln(44x)$. Is $f$ uniformly continuous on $(0,1609)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{1609}{n}$ and $y_n=\\frac{1609}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\l... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018470 | Real Analysis: Uniform Continuity (Variant C) | 10 | Find the exact value: Let $f:(0,52)\to\mathbb{R}$ be $f(x)=\sqrt{15x}$. Prove that $f$ is uniformly continuous on $(0,52)$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Compact Extension + Heine–Cantor",
"approach": "Extend continuously to a compact interval and invoke Heine–Cantor, then restrict back.",
"steps": [
"Step 1: Define $g:[0,52]\\to\\mathbb{R}$ by $g(x)=\\sqrt{x}$; this is continuous on $[0,52]$.",
"Step 2: The interval $[0,52]... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Yes}$.\nThe direct inequality provides an explicit modulus $\\delta=\\varepsilon^2$. Heine–Cantor guarantees the existence of some modulus on the compact extension; both give the same conclusion.",
"robustness_analysis... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. |
math-018471 | Real Analysis: Uniform Continuity (Variant C) | 10 | Provide both a computational and a conceptual explanation: Let $f:(0,1010)\to\mathbb{R}$ be $f(x)=\frac{1}{(4)x}$. Is $f$ uniformly continuous on $(0,1010)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both c... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018472 | Real Analysis: Uniform Continuity (Variant C) | 10 | Indicate where a theorem is used: Let $f:(0,610)\to\mathbb{R}$ be $f(x)=\ln(24x)$. Is $f$ uniformly continuous on $(0,610)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": ... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018473 | Real Analysis: Uniform Continuity (Core) | 10 | Solve and justify each step: Let $f:(0,122)\to\mathbb{R}$ be $f(x)=\frac{1}{(-6)x}$. Is $f$ uniformly continuous on $(0,122)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018474 | Real Analysis: Uniform Continuity (Variant C) | 10 | Exercise: Let $f:(0,223)\to\mathbb{R}$ be $f(x)=\sqrt{25x}$. Prove that $f$ is uniformly continuous on $(0,223)$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Direct Inequality",
"approach": "Bound $|\\sqrt{x}-\\sqrt{y}|$ in terms of $|x-y|$ uniformly using rationalization.",
"steps": [
"Step 1: For $x,y>0$, $|\\sqrt{x}-\\sqrt{y}|=\\frac{|x-y|}{\\sqrt{x}+\\sqrt{y}}$.",
"Step 2: Since $\\sqrt{x}+\\sqrt{y}\\ge \\sqrt{|x-y|}$, we ge... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Yes}$.\nThe direct inequality provides an explicit modulus $\\delta=\\varepsilon^2$. Heine–Cantor guarantees the existence of some modulus on the compact extension; both give the same conc... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018475 | Real Analysis: Uniform Continuity (Variant C) | 10 | Show all reasoning: Let $f:(0,330)\to\mathbb{R}$ be $f(x)=\ln(31x)$. Is $f$ uniformly continuous on $(0,330)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018476 | Real Analysis: Uniform Continuity (Variant B) | 10 | Challenge: Let $f:(0,166)\to\mathbb{R}$ be $f(x)=\sqrt{6x}$. Prove that $f$ is uniformly continuous on $(0,166)$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Compact Extension + Heine–Cantor",
"approach": "Extend continuously to a compact interval and invoke Heine–Cantor, then restrict back.",
"steps": [
"Step 1: Define $g:[0,166]\\to\\mathbb{R}$ by $g(x)=\\sqrt{x}$; this is continuous on $[0,166]$.",
"Step 2: The interval $[0,1... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Yes}$.\nThe direct inequality provides an explicit modulus $\\delta=\\varepsilon^2$. Heine–Cantor guarantees the existence of some modulus on the compact extension; both give the same conc... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018477 | Real Analysis: Uniform Continuity (Variant A) | 10 | Exercise: Let $f:(0,37)\to\mathbb{R}$ be $f(x)=\frac{1}{(-27)x}$. Is $f$ uniformly continuous on $(0,37)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018478 | Real Analysis: Uniform Continuity (Core) | 10 | State any required conditions first: Let $f:(0,1205)\to\mathbb{R}$ be $f(x)=\sqrt{20x}$. Prove that $f$ is uniformly continuous on $(0,1205)$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Direct Inequality",
"approach": "Bound $|\\sqrt{x}-\\sqrt{y}|$ in terms of $|x-y|$ uniformly using rationalization.",
"steps": [
"Step 1: For $x,y>0$, $|\\sqrt{x}-\\sqrt{y}|=\\frac{|x-y|}{\\sqrt{x}+\\sqrt{y}}$.",
"Step 2: Since $\\sqrt{x}+\\sqrt{y}\\ge \\sqrt{|x-y|}$, we ge... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Yes}$.\nThe direct inequality provides an explicit modulus $\\delta=\\varepsilon^2$. Heine–Cantor guarantees the existence of some modulus on the compact extension; both give the same conclusion.",... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018479 | Real Analysis: Uniform Continuity (Variant B) | 10 | Problem: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=\sin(18x)$. Prove that $f$ is uniformly continuous on $\mathbb{R}$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Lipschitz via Mean Value Theorem",
"approach": "Use MVT to get a global Lipschitz bound $|\\sin x-\\sin y|\\le |x-y|$.",
"steps": [
"Step 1: Apply MVT to $g(t)=\\sin(ct)$: there exists $\\xi$ between $x$ and $y$ with $g(x)-g(y)=g'(\\xi)(x-y)$.",
"Step 2: Since $g'(t)=(18)\\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Yes}$.\nThe MVT proof gives an explicit global Lipschitz modulus, which immediately implies the Cauchy-sequence property. Both are equivalent characterizations of uniform continuity.",
"robustness_analysis": "Rob... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018480 | Real Analysis: Uniform Continuity (Variant B) | 10 | Be explicit about assumptions: Let $f:[-1911,1911]\to\mathbb{R}$ be $f(x)=(11)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-1911,1911]$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Heine–Cantor",
"approach": "Continuous functions on compact metric spaces are uniformly continuous.",
"steps": [
"Step 1: The interval $[-1911,1911]$ is compact in $\\mathbb{R}$.",
"Step 2: The function $x\\mapsto x^2$ is continuous on $\\mathbb{R}$, hence on the compact se... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Yes}$.\nHeine–Cantor implies uniform continuity abstractly from compactness. The Lipschitz bound constructs an explicit modulus $\\delta(\\varepsilon)=\\varepsilon/(2M)$; both agree on the same conclusion.",
"rob... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018481 | Real Analysis: Uniform Continuity (Variant B) | 10 | Complete the analysis: Let $f:(0,1938)\to\mathbb{R}$ be $f(x)=\ln(55x)$. Is $f$ uniformly continuous on $(0,1938)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{1938}{n}$ and $y_n=\\frac{1938}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\l... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": "Sensi... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018482 | Real Analysis: Uniform Continuity (Core) | 10 | Be explicit about assumptions: Let $f:(0,997)\to\mathbb{R}$ be $f(x)=\frac{1}{(-5)x}$. Is $f$ uniformly continuous on $(0,997)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{997}{n}$ and $y_n=\\frac{997}{2n}$ in $(0,997)$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both c... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018483 | Real Analysis: Uniform Continuity (Variant C) | 10 | Solve (and briefly cross-validate): Let $f:(0,1677)\to\mathbb{R}$ be $f(x)=\sqrt{34x}$. Prove that $f$ is uniformly continuous on $(0,1677)$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Compact Extension + Heine–Cantor",
"approach": "Extend continuously to a compact interval and invoke Heine–Cantor, then restrict back.",
"steps": [
"Step 1: Define $g:[0,1677]\\to\\mathbb{R}$ by $g(x)=\\sqrt{x}$; this is continuous on $[0,1677]$.",
"Step 2: The interval $[0... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Yes}$.\nThe direct inequality provides an explicit modulus $\\delta=\\varepsilon^2$. Heine–Cantor guarantees the existence of some modulus on the compact extension; both give the same conclusion.",... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018484 | Real Analysis: Uniform Continuity (Core) | 10 | Checkpoint: Let $f:(0,1313)\to\mathbb{R}$ be $f(x)=\frac{1}{(-30)x}$. Is $f$ uniformly continuous on $(0,1313)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{1313}{n}$ and $y_n=\\frac{1313}{2n}$ in $(0,1313)$.",
"Step 2: Then $|x_n-y_n|=\\frac{... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018485 | Real Analysis: Uniform Continuity (Variant A) | 10 | Work this out carefully: Let $f:(0,676)\to\mathbb{R}$ be $f(x)=\frac{1}{(5)x}$. Is $f$ uniformly continuous on $(0,676)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both c... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018486 | Real Analysis: Uniform Continuity (Variant C) | 10 | Answer with a short justification: Let $f:(0,1253)\to\mathbb{R}$ be $f(x)=\frac{1}{(17)x}$. Is $f$ uniformly continuous on $(0,1253)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{1253}{n}$ and $y_n=\\frac{1253}{2n}$ in $(0,1253)$.",
"Step 2: Then $|x_n-y_n|=\\frac{... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018487 | Real Analysis: Uniform Continuity (Variant A) | 10 | Work carefully and justify each inference: Let $f:(0,1396)\to\mathbb{R}$ be $f(x)=\frac{1}{(21)x}$. Is $f$ uniformly continuous on $(0,1396)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018488 | Real Analysis: Uniform Continuity (Variant A) | 10 | Explain why your operations are valid: Let $f:[-601,601]\to\mathbb{R}$ be $f(x)=(-1)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-601,601]$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Heine–Cantor",
"approach": "Continuous functions on compact metric spaces are uniformly continuous.",
"steps": [
"Step 1: The interval $[-601,601]$ is compact in $\\mathbb{R}$.",
"Step 2: The function $x\\mapsto x^2$ is continuous on $\\mathbb{R}$, hence on the compact set.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Yes}$.\nHeine–Cantor implies uniform continuity abstractly from compactness. The Lipschitz bound constructs an explicit modulus $\\delta(\\varepsilon)=\\varepsilon/(2M)$; both agree on the same conclusion.",
"robustnes... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018489 | Real Analysis: Uniform Continuity (Variant C) | 10 | Problem: Let $f:[-462,462]\to\mathbb{R}$ be $f(x)=(-26)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-462,462]$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Lipschitz on Bounded Domain",
"approach": "On a bounded interval, $|x^2-y^2|=|x-y||x+y|$ and $|x+y|$ is uniformly bounded, giving a global Lipschitz constant.",
"steps": [
"Step 1: For $x,y\\in[-462,462]$, $|f(x)-f(y)|=|-26|\\,|x^2-y^2|=|-26|\\,|x-y||x+y|$.",
"Step 2: Bound... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Yes}$.\nHeine–Cantor implies uniform continuity abstractly from compactness. The Lipschitz bound constructs an explicit modulus $\\delta(\\varepsilon)=\\varepsilon/(2M)$; both agree on the same con... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018490 | Real Analysis: Uniform Continuity (Variant A) | 10 | Exercise: Let $f:(0,730)\to\mathbb{R}$ be $f(x)=\sqrt{14x}$. Prove that $f$ is uniformly continuous on $(0,730)$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Compact Extension + Heine–Cantor",
"approach": "Extend continuously to a compact interval and invoke Heine–Cantor, then restrict back.",
"steps": [
"Step 1: Define $g:[0,730]\\to\\mathbb{R}$ by $g(x)=\\sqrt{x}$; this is continuous on $[0,730]$.",
"Step 2: The interval $[0,7... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Yes}$.\nThe direct inequality provides an explicit modulus $\\delta=\\varepsilon^2$. Heine–Cantor guarantees the existence of some modulus on the compact extension; both give the same conclusion.",... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018491 | Real Analysis: Uniform Continuity (Variant A) | 10 | Keep the final answer in boxed form: Let $f:(0,210)\to\mathbb{R}$ be $f(x)=\frac{1}{(-22)x}$. Is $f$ uniformly continuous on $(0,210)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}}.",
... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Takeaway: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018492 | Real Analysis: Uniform Continuity (Variant C) | 10 | Make each step logically reversible (or explain if not): Let $f:(0,1185)\to\mathbb{R}$ be $f(x)=\frac{1}{(16)x}$. Is $f$ uniformly continuous on $(0,1185)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{1185}{n}$ and $y_n=\\frac{1185}{2n}$ in $(0,1185)$.",
"Step 2: Then $|x_n-y_n|=\\frac{... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}}.",
... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018493 | Real Analysis: Uniform Continuity (Variant C) | 10 | Work carefully and justify each inference: Let $f:(0,1462)\to\mathbb{R}$ be $f(x)=\frac{1}{(-6)x}$. Is $f$ uniformly continuous on $(0,1462)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Cauchy-Sequence Criterion",
"approach": "To disprove uniform continuity, construct sequences with $|x_n-y_n|\\to 0$ but $|f(x_n)-f(y_n)|\\not\\to 0$.",
"steps": [
"Step 1: Let $x_n=\\frac{1462}{n}$ and $y_n=\\frac{1462}{2n}$ in $(0,1462)$.",
"Step 2: Then $|x_n-y_n|=\\frac{... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both c... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018494 | Real Analysis: Uniform Continuity (Variant C) | 10 | Prompt: Let $f:(0,1151)\to\mathbb{R}$ be $f(x)=\sqrt{7x}$. Prove that $f$ is uniformly continuous on $(0,1151)$.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Direct Inequality",
"approach": "Bound $|\\sqrt{x}-\\sqrt{y}|$ in terms of $|x-y|$ uniformly using rationalization.",
"steps": [
"Step 1: For $x,y>0$, $|\\sqrt{x}-\\sqrt{y}|=\\frac{|x-y|}{\\sqrt{x}+\\sqrt{y}}$.",
"Step 2: Since $\\sqrt{x}+\\sqrt{y}\\ge \\sqrt{|x-y|}$, we ge... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Yes}$.\nThe direct inequality provides an explicit modulus $\\delta=\\varepsilon^2$. Heine–Cantor guarantees the existence of some modulus on the compact extension; both give the same conclusion.",... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is Yes for the stated domain. (Here the result is $\boxed{\text{Yes}$.) |
math-018495 | Real Analysis: Uniform Continuity (Core) | 10 | Do not skip justification steps: Let $f:(0,1214)\to\mathbb{R}$ be $f(x)=\ln(52x)$. Is $f$ uniformly continuous on $(0,1214)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": ... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018496 | Real Analysis: Uniform Continuity (Variant A) | 10 | Task: Let $f:(0,1294)\to\mathbb{R}$ be $f(x)=\frac{1}{(20)x}$. Is $f$ uniformly continuous on $(0,1294)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}}.",
... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Key idea: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018497 | Real Analysis: Uniform Continuity (Variant C) | 10 | Find the exact value: Let $f:(0,693)\to\mathbb{R}$ be $f(x)=\frac{1}{(13)x}$. Is $f$ uniformly continuous on $(0,693)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Quantifier/Epsilon–Delta Contradiction",
"approach": "Assume a global $\\delta(\\varepsilon)$ exists and show it fails near the singularity at 0.",
"steps": [
"Step 1: Suppose $f$ is uniformly continuous and take $\\varepsilon=1$.",
"Step 2: Let $\\delta>0$ be given by unif... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{No}$.\nBoth methods exploit the same obstruction: arbitrarily close inputs near 0 can have arbitrarily far apart outputs for $1/x$, so no single global modulus of continuity exists. Both conclude \\boxed{\\text{No}}.",
... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
math-018498 | Real Analysis: Uniform Continuity (Core) | 10 | Work carefully and justify each inference: Let $f:(0,1952)\to\mathbb{R}$ be $f(x)=\ln(19x)$. Is $f$ uniformly continuous on $(0,1952)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Epsilon–Delta with Scale Trick",
"approach": "Assume uniform continuity; choose $\\varepsilon<|\\ln 2|$ and show $x$ and $2x$ violate it for sufficiently small $x$.",
"steps": [
"Step 1: Assume uniform continuity holds and choose $\\varepsilon=|\\ln 2|/2$.",
"Step 2: Let $\... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Core principle: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018499 | Real Analysis: Uniform Continuity (Variant B) | 10 | Indicate where a theorem is used: Let $f:(0,1860)\to\mathbb{R}$ be $f(x)=\ln(21x)$. Is $f$ uniformly continuous on $(0,1860)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{1860}{n}$ and $y_n=\\frac{1860}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\l... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": ... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. |
math-018500 | Real Analysis: Uniform Continuity (Variant C) | 10 | Carefully track domains: Let $f:(0,1905)\to\mathbb{R}$ be $f(x)=\ln(17x)$. Is $f$ uniformly continuous on $(0,1905)$? Give a proof.
Include a brief verification/cross-check at the end. | [
{
"method_name": "Close Inputs, Fixed Output Gap",
"approach": "Use a sequence approaching 0 where scaling by 2 produces a constant log-gap.",
"steps": [
"Step 1: Let $x_n=\\frac{1905}{n}$ and $y_n=\\frac{1905}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\l... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{No}$.\nBoth arguments exhibit a uniform obstruction near 0: inputs with ratio 2 can be arbitrarily close, while the log difference stays $\\ln 2$. Thus both conclude \\boxed{\\text{No}}.",
"robustness_analysis": ... | [
{
"error_description": "Assumed continuity implies uniform continuity without checking compactness or a global bound.",
"why_plausible": "Heine–Cantor is often remembered without its hypotheses.",
"why_wrong": "Continuity implies uniform continuity only on compact sets; on non-compact domains there are ... | Remember: Uniform continuity is global and depends on the domain: compactness/Lipschitz bounds guarantee it, while singularities or unbounded derivative growth often destroy it. Here the answer is No for the stated domain. (Here the result is $\boxed{\text{No}$.) |
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