id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-018701 | Real Analysis: Uniform Continuity (Core) | 10 | Answer using clear logical steps: Let $f:(0,1418)\to\mathbb{R}$ be $f(x)=\frac{1}{(-14)x}$. Is $f$ uniformly continuous on $(0,1418)$? 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-018702 | Real Analysis: Uniform Continuity (Variant A) | 10 | Work this out carefully: Let $f:(0,1804)\to\mathbb{R}$ be $f(x)=\ln(28x)$. Is $f$ uniformly continuous on $(0,1804)$? 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 $x_n=\\frac{1804}{n}$ and $y_n=\\frac{1804}{2n}$.",
"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 ... | 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-018703 | Real Analysis: Uniform Continuity (Core) | 10 | Carefully track domains: Let $f:(0,1858)\to\mathbb{R}$ be $f(x)=\ln(8x)$. Is $f$ uniformly continuous on $(0,1858)$? 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 $x_n=\\frac{1858}{n}$ and $y_n=\\frac{1858}{2n}$.",
"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": "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 ... | 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-018704 | Real Analysis: Uniform Continuity (Variant A) | 10 | Find the exact value: Let $f:(0,1288)\to\mathbb{R}$ be $f(x)=\ln(32x)$. Is $f$ uniformly continuous on $(0,1288)$? 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 $x_n=\\frac{1288}{n}$ and $y_n=\\frac{1288}{2n}$.",
"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": "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 ... | 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-018705 | Real Analysis: Uniform Continuity (Variant B) | 10 | Answer using clear logical steps: Let $f:(0,46)\to\mathbb{R}$ be $f(x)=\ln(5x)$. Is $f$ uniformly continuous on $(0,46)$? 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 $x_n=\\frac{46}{n}$ and $y_n=\\frac{46}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\ln(c ... | {
"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-018706 | Real Analysis: Uniform Continuity (Variant A) | 10 | Start by stating any domain restrictions: Let $f:(0,911)\to\mathbb{R}$ be $f(x)=\frac{1}{(13)x}$. Is $f$ uniformly continuous on $(0,911)$? 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 ... | 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-018707 | Real Analysis: Uniform Continuity (Variant A) | 10 | Warm-up: Let $f:(0,1220)\to\mathbb{R}$ be $f(x)=\sqrt{55x}$. Prove that $f$ is uniformly continuous on $(0,1220)$.
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 $g:[0,1220]\\to\\mathbb{R}$ by $g(x)=\\sqrt{x}$; this is continuous on $[0,1220]$.",
"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-018708 | Real Analysis: Uniform Continuity (Variant A) | 10 | State any required conditions first: Let $f:(0,293)\to\mathbb{R}$ be $f(x)=\ln(9x)$. Is $f$ uniformly continuous on $(0,293)$? 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 $x_n=\\frac{293}{n}$ and $y_n=\\frac{293}{2n}$.",
"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": "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. |
math-018709 | Real Analysis: Uniform Continuity (Core) | 10 | Explain each transformation: Let $f:(0,421)\to\mathbb{R}$ be $f(x)=\frac{1}{(24)x}$. Is $f$ uniformly continuous on $(0,421)$? 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 $x_n=\\frac{421}{n}$ and $y_n=\\frac{421}{2n}$ in $(0,421)$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{... | {
"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-018710 | Real Analysis: Uniform Continuity (Variant C) | 10 | Compute the requested quantity: Let $f:(0,645)\to\mathbb{R}$ be $f(x)=\ln(43x)$. Is $f$ uniformly continuous on $(0,645)$? 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 $x_n=\\frac{645}{n}$ and $y_n=\\frac{645}{2n}$.",
"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": "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 ... | 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-018711 | Real Analysis: Uniform Continuity (Variant A) | 10 | Task: Let $f:(0,181)\to\mathbb{R}$ be $f(x)=\frac{1}{(2)x}$. Is $f$ uniformly continuous on $(0,181)$? 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 $x_n=\\frac{181}{n}$ and $y_n=\\frac{181}{2n}$ in $(0,181)$.",
"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-018712 | Real Analysis: Uniform Continuity (Variant C) | 10 | Do not skip justification steps: Let $f:[-1678,1678]\to\mathbb{R}$ be $f(x)=(-25)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-1678,1678]$.
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 $[-1678,1678]$ 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": "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 ... | 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-018713 | Real Analysis: Uniform Continuity (Variant A) | 10 | Solve and justify each step: Let $f:(0,1638)\to\mathbb{R}$ be $f(x)=\ln(35x)$. Is $f$ uniformly continuous on $(0,1638)$? 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. (Here the result is $\boxed{\text{No}$.) |
math-018714 | Real Analysis: Uniform Continuity (Core) | 10 | Exercise: Let $f:(0,686)\to\mathbb{R}$ be $f(x)=\ln(54x)$. Is $f$ uniformly continuous on $(0,686)$? 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 $x_n=\\frac{686}{n}$ and $y_n=\\frac{686}{2n}$.",
"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 ... | 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-018715 | Real Analysis: Uniform Continuity (Core) | 10 | Explain why your operations are valid: Let $f:(0,36)\to\mathbb{R}$ be $f(x)=\sqrt{28x}$. Prove that $f$ is uniformly continuous on $(0,36)$.
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 $g:[0,36]\\to\\mathbb{R}$ by $g(x)=\\sqrt{x}$; this is continuous on $[0,36]$.",
"Step 2: The interval $[0,36]... | {
"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 ... | 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. |
math-018716 | Real Analysis: Uniform Continuity (Variant C) | 10 | Explain why your operations are valid: Let $f:(0,984)\to\mathbb{R}$ be $f(x)=\sqrt{60x}$. Prove that $f$ is uniformly continuous on $(0,984)$.
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 ... | 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-018717 | Real Analysis: Uniform Continuity (Variant A) | 10 | Complete the analysis: Let $f:(0,1984)\to\mathbb{R}$ be $f(x)=\sqrt{53x}$. Prove that $f$ is uniformly continuous on $(0,1984)$.
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-018718 | Real Analysis: Uniform Continuity (Variant C) | 10 | Explain each transformation: Let $f:(0,280)\to\mathbb{R}$ be $f(x)=\frac{1}{(26)x}$. Is $f$ uniformly continuous on $(0,280)$? 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 ... | 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-018719 | Real Analysis: Uniform Continuity (Core) | 10 | Explain what is being counted/optimized: Let $f:(0,359)\to\mathbb{R}$ be $f(x)=\ln(4x)$. Is $f$ uniformly continuous on $(0,359)$? 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-018720 | Real Analysis: Uniform Continuity (Variant C) | 10 | Work this out carefully: Let $f:(0,1657)\to\mathbb{R}$ be $f(x)=\ln(50x)$. Is $f$ uniformly continuous on $(0,1657)$? 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 $x_n=\\frac{1657}{n}$ and $y_n=\\frac{1657}{2n}$.",
"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": "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-018721 | Real Analysis: Uniform Continuity (Variant A) | 10 | Solve (and briefly cross-validate): Let $f:(0,1825)\to\mathbb{R}$ be $f(x)=\ln(2x)$. Is $f$ uniformly continuous on $(0,1825)$? 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. (Here the result is $\boxed{\text{No}$.) |
math-018722 | Real Analysis: Uniform Continuity (Variant B) | 10 | Compute the requested quantity: Let $f:(0,1861)\to\mathbb{R}$ be $f(x)=\ln(30x)$. Is $f$ uniformly continuous on $(0,1861)$? 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-018723 | Real Analysis: Uniform Continuity (Variant A) | 10 | Compute the requested quantity: Let $f:[-372,372]\to\mathbb{R}$ be $f(x)=(-29)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-372,372]$.
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 $[-372,372]$ 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 ... | 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-018724 | Real Analysis: Uniform Continuity (Variant B) | 10 | Problem: Let $f:[-995,995]\to\mathbb{R}$ be $f(x)=(-23)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-995,995]$.
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 $x,y\\in[-995,995]$, $|f(x)-f(y)|=|-23|\\,|x^2-y^2|=|-23|\\,|x-y||x+y|$.",
"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 ... | 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-018725 | Real Analysis: Uniform Continuity (Variant C) | 10 | Problem: Let $f:(0,713)\to\mathbb{R}$ be $f(x)=\ln(11x)$. Is $f$ uniformly continuous on $(0,713)$? 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 $x_n=\\frac{713}{n}$ and $y_n=\\frac{713}{2n}$.",
"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 ... | 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-018726 | Real Analysis: Uniform Continuity (Variant A) | 10 | Solve and justify each step: Let $f:(0,1149)\to\mathbb{R}$ be $f(x)=\ln(13x)$. Is $f$ uniformly continuous on $(0,1149)$? 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 $x_n=\\frac{1149}{n}$ and $y_n=\\frac{1149}{2n}$.",
"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 ... | 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-018727 | Real Analysis: Uniform Continuity (Variant A) | 10 | Start by stating any domain restrictions: Let $f:(0,1491)\to\mathbb{R}$ be $f(x)=\frac{1}{(-6)x}$. Is $f$ uniformly continuous on $(0,1491)$? 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 $x_n=\\frac{1491}{n}$ and $y_n=\\frac{1491}{2n}$ in $(0,1491)$.",
"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 ... | 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-018728 | Real Analysis: Uniform Continuity (Variant A) | 10 | Solve with verification: Let $f:(0,836)\to\mathbb{R}$ be $f(x)=\frac{1}{(3)x}$. Is $f$ uniformly continuous on $(0,836)$? 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 ... | 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-018729 | Real Analysis: Uniform Continuity (Variant A) | 10 | Proceed methodically: Let $f:(0,1345)\to\mathbb{R}$ be $f(x)=\frac{1}{(13)x}$. Is $f$ uniformly continuous on $(0,1345)$? 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 $x_n=\\frac{1345}{n}$ and $y_n=\\frac{1345}{2n}$ in $(0,1345)$.",
"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 ... | 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-018730 | Real Analysis: Uniform Continuity (Variant B) | 10 | Write the solution set clearly: Let $f:(0,831)\to\mathbb{R}$ be $f(x)=\sqrt{23x}$. Prove that $f$ is uniformly continuous on $(0,831)$.
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": "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-018731 | Real Analysis: Uniform Continuity (Variant C) | 10 | Determine the requested value: Let $f:(0,746)\to\mathbb{R}$ be $f(x)=\ln(4x)$. Is $f$ uniformly continuous on $(0,746)$? 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 $x_n=\\frac{746}{n}$ and $y_n=\\frac{746}{2n}$.",
"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 ... | 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-018732 | Real Analysis: Uniform Continuity (Variant B) | 10 | Prompt: Let $f:(0,314)\to\mathbb{R}$ be $f(x)=\sqrt{24x}$. Prove that $f$ is uniformly continuous on $(0,314)$.
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. |
math-018733 | Real Analysis: Uniform Continuity (Variant A) | 10 | Checkpoint: Let $f:(0,1666)\to\mathbb{R}$ be $f(x)=\ln(29x)$. Is $f$ uniformly continuous on $(0,1666)$? 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 ... | 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-018734 | Real Analysis: Uniform Continuity (Variant C) | 10 | Keep the final answer in boxed form: Let $f:(0,1517)\to\mathbb{R}$ be $f(x)=\frac{1}{(-3)x}$. Is $f$ uniformly continuous on $(0,1517)$? 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 ... | 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-018735 | Real Analysis: Uniform Continuity (Variant B) | 10 | Provide a rigorous solution: Let $f:(0,1727)\to\mathbb{R}$ be $f(x)=\ln(24x)$. Is $f$ uniformly continuous on $(0,1727)$? 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 $x_n=\\frac{1727}{n}$ and $y_n=\\frac{1727}{2n}$.",
"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 ... | 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-018736 | Real Analysis: Uniform Continuity (Variant A) | 10 | Task: Let $f:(0,1875)\to\mathbb{R}$ be $f(x)=\frac{1}{(-14)x}$. Is $f$ uniformly continuous on $(0,1875)$? 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 ... | 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-018737 | Real Analysis: Uniform Continuity (Core) | 10 | Answer with a short justification: Let $f:(0,1673)\to\mathbb{R}$ be $f(x)=\frac{1}{(-2)x}$. Is $f$ uniformly continuous on $(0,1673)$? 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. |
math-018738 | Real Analysis: Uniform Continuity (Variant A) | 10 | Explain why your operations are valid: Let $f:(0,608)\to\mathbb{R}$ be $f(x)=\ln(19x)$. Is $f$ uniformly continuous on $(0,608)$? 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 ... | 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-018739 | Real Analysis: Uniform Continuity (Variant A) | 10 | Explain what is being counted/optimized: Let $f:(0,1236)\to\mathbb{R}$ be $f(x)=\ln(25x)$. Is $f$ uniformly continuous on $(0,1236)$? 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-018740 | Real Analysis: Uniform Continuity (Variant C) | 10 | Work this out carefully: Let $f:(0,126)\to\mathbb{R}$ be $f(x)=\frac{1}{(-26)x}$. Is $f$ uniformly continuous on $(0,126)$? 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-018741 | Real Analysis: Uniform Continuity (Variant A) | 10 | Solve and then verify: Let $f:(0,1969)\to\mathbb{R}$ be $f(x)=\ln(37x)$. Is $f$ uniformly continuous on $(0,1969)$? 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 ... | 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-018742 | Real Analysis: Uniform Continuity (Variant B) | 10 | Checkpoint: Let $f:(0,1343)\to\mathbb{R}$ be $f(x)=\frac{1}{(3)x}$. Is $f$ uniformly continuous on $(0,1343)$? 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 $x_n=\\frac{1343}{n}$ and $y_n=\\frac{1343}{2n}$ in $(0,1343)$.",
"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 ... | 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-018743 | Real Analysis: Uniform Continuity (Variant C) | 10 | Work carefully and justify each inference: Let $f:(0,1591)\to\mathbb{R}$ be $f(x)=\ln(3x)$. Is $f$ uniformly continuous on $(0,1591)$? 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 $x_n=\\frac{1591}{n}$ and $y_n=\\frac{1591}{2n}$.",
"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 ... | 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-018744 | Real Analysis: Uniform Continuity (Variant C) | 10 | Answer with a short justification: Let $f:(0,420)\to\mathbb{R}$ be $f(x)=\frac{1}{(8)x}$. Is $f$ uniformly continuous on $(0,420)$? 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-018745 | Real Analysis: Uniform Continuity (Variant B) | 10 | Provide both a computational and a conceptual explanation: Let $f:(0,1636)\to\mathbb{R}$ be $f(x)=\ln(42x)$. Is $f$ uniformly continuous on $(0,1636)$? 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. (Here the result is $\boxed{\text{No}$.) |
math-018746 | Real Analysis: Uniform Continuity (Variant C) | 10 | Try to avoid pattern-matching; explain why: Let $f:[-1375,1375]\to\mathbb{R}$ be $f(x)=(-9)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-1375,1375]$.
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 $[-1375,1375]$ 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": "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. |
math-018747 | Real Analysis: Uniform Continuity (Core) | 10 | Prompt: Let $f:[-387,387]\to\mathbb{R}$ be $f(x)=(6)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-387,387]$.
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 $x,y\\in[-387,387]$, $|f(x)-f(y)|=|6|\\,|x^2-y^2|=|6|\\,|x-y||x+y|$.",
"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 ... | 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-018748 | Real Analysis: Uniform Continuity (Variant A) | 10 | Answer with a short justification: Let $f:[-1229,1229]\to\mathbb{R}$ be $f(x)=(-27)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-1229,1229]$.
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 $x,y\\in[-1229,1229]$, $|f(x)-f(y)|=|-27|\\,|x^2-y^2|=|-27|\\,|x-y||x+y|$.",
"Step 2: Bou... | {
"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 ... | 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-018749 | Real Analysis: Uniform Continuity (Core) | 10 | Solve and include a self-check: Let $f:[-974,974]\to\mathbb{R}$ be $f(x)=(29)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-974,974]$.
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 $x,y\\in[-974,974]$, $|f(x)-f(y)|=|29|\\,|x^2-y^2|=|29|\\,|x-y||x+y|$.",
"Step 2: Bound $... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. 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... | [
{
"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-018750 | Real Analysis: Uniform Continuity (Core) | 10 | Determine the requested value: Let $f:(0,1202)\to\mathbb{R}$ be $f(x)=\frac{1}{(-19)x}$. Is $f$ uniformly continuous on $(0,1202)$? 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 $x_n=\\frac{1202}{n}$ and $y_n=\\frac{1202}{2n}$ in $(0,1202)$.",
"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-018751 | Real Analysis: Uniform Continuity (Variant A) | 10 | Track units/moduli carefully: Let $f:(0,1545)\to\mathbb{R}$ be $f(x)=\ln(6x)$. Is $f$ uniformly continuous on $(0,1545)$? 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 $x_n=\\frac{1545}{n}$ and $y_n=\\frac{1545}{2n}$.",
"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 ... | 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-018752 | Real Analysis: Uniform Continuity (Variant B) | 10 | Solve and include a self-check: Let $f:(0,397)\to\mathbb{R}$ be $f(x)=\frac{1}{(-3)x}$. Is $f$ uniformly continuous on $(0,397)$? 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-018753 | Real Analysis: Uniform Continuity (Variant A) | 10 | Track quantifiers carefully: Let $f:(0,1642)\to\mathbb{R}$ be $f(x)=\ln(60x)$. Is $f$ uniformly continuous on $(0,1642)$? 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-018754 | Real Analysis: Uniform Continuity (Core) | 10 | Give a theorem-based solution: Let $f:(0,1512)\to\mathbb{R}$ be $f(x)=\ln(21x)$. Is $f$ uniformly continuous on $(0,1512)$? 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-018755 | Real Analysis: Uniform Continuity (Core) | 10 | Prompt: Let $f:[-309,309]\to\mathbb{R}$ be $f(x)=(-27)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-309,309]$.
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 $x,y\\in[-309,309]$, $|f(x)-f(y)|=|-27|\\,|x^2-y^2|=|-27|\\,|x-y||x+y|$.",
"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 ... | 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. |
math-018756 | Real Analysis: Uniform Continuity (Variant C) | 10 | Work carefully and justify each inference: Let $f:(0,956)\to\mathbb{R}$ be $f(x)=\ln(22x)$. Is $f$ uniformly continuous on $(0,956)$? 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. |
math-018757 | Real Analysis: Uniform Continuity (Core) | 10 | Explain each transformation: Let $f:[-1971,1971]\to\mathbb{R}$ be $f(x)=(-4)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-1971,1971]$.
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 $x,y\\in[-1971,1971]$, $|f(x)-f(y)|=|-4|\\,|x^2-y^2|=|-4|\\,|x-y||x+y|$.",
"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 ... | 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-018758 | Real Analysis: Uniform Continuity (Variant B) | 10 | Derive the result step-by-step: Let $f:(0,996)\to\mathbb{R}$ be $f(x)=\ln(18x)$. Is $f$ uniformly continuous on $(0,996)$? 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 ... | 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-018759 | Real Analysis: Uniform Continuity (Variant A) | 10 | Give a theorem-based solution: Let $f:(0,114)\to\mathbb{R}$ be $f(x)=\ln(34x)$. Is $f$ uniformly continuous on $(0,114)$? 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-018760 | Real Analysis: Uniform Continuity (Variant C) | 10 | Prompt: Let $f:(0,1626)\to\mathbb{R}$ be $f(x)=\sqrt{49x}$. Prove that $f$ is uniformly continuous on $(0,1626)$.
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 ... | 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-018761 | Real Analysis: Uniform Continuity (Variant B) | 10 | Show all reasoning: Let $f:(0,837)\to\mathbb{R}$ be $f(x)=\ln(15x)$. Is $f$ uniformly continuous on $(0,837)$? 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 $x_n=\\frac{837}{n}$ and $y_n=\\frac{837}{2n}$.",
"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": "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. |
math-018762 | Real Analysis: Uniform Continuity (Variant A) | 10 | Prompt: Let $f:(0,714)\to\mathbb{R}$ be $f(x)=\sqrt{60x}$. Prove that $f$ is uniformly continuous on $(0,714)$.
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 ... | 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-018763 | Real Analysis: Uniform Continuity (Variant C) | 10 | Use two approaches if possible: Let $f:[-1379,1379]\to\mathbb{R}$ be $f(x)=(28)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-1379,1379]$.
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 $[-1379,1379]$ 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": "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 ... | 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-018764 | Real Analysis: Uniform Continuity (Variant B) | 10 | Track units/moduli carefully: Let $f:(0,461)\to\mathbb{R}$ be $f(x)=\sqrt{12x}$. Prove that $f$ is uniformly continuous on $(0,461)$.
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 $g:[0,461]\\to\\mathbb{R}$ by $g(x)=\\sqrt{x}$; this is continuous on $[0,461]$.",
"Step 2: The interval $[0,4... | {
"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. |
math-018765 | Real Analysis: Uniform Continuity (Variant B) | 10 | Complete the analysis: Let $f:(0,461)\to\mathbb{R}$ be $f(x)=\frac{1}{(5)x}$. Is $f$ uniformly continuous on $(0,461)$? 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 $x_n=\\frac{461}{n}$ and $y_n=\\frac{461}{2n}$ in $(0,461)$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{... | {
"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. |
math-018766 | Real Analysis: Uniform Continuity (Variant A) | 10 | Complete the analysis: Let $f:(0,695)\to\mathbb{R}$ be $f(x)=\ln(23x)$. Is $f$ uniformly continuous on $(0,695)$? 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-018767 | Real Analysis: Uniform Continuity (Variant C) | 10 | Keep the final answer in boxed form: Let $f:[-845,845]\to\mathbb{R}$ be $f(x)=(30)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-845,845]$.
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 $x,y\\in[-845,845]$, $|f(x)-f(y)|=|30|\\,|x^2-y^2|=|30|\\,|x-y||x+y|$.",
"Step 2: Bound $... | {
"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 ... | 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. |
math-018768 | Real Analysis: Uniform Continuity (Variant A) | 10 | Solve and then verify: Let $f:[-1196,1196]\to\mathbb{R}$ be $f(x)=(29)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-1196,1196]$.
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 $x,y\\in[-1196,1196]$, $|f(x)-f(y)|=|29|\\,|x^2-y^2|=|29|\\,|x-y||x+y|$.",
"Step 2: Bound... | {
"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-018769 | Real Analysis: Uniform Continuity (Variant C) | 10 | Work this out carefully: Let $f:(0,1341)\to\mathbb{R}$ be $f(x)=\ln(2x)$. Is $f$ uniformly continuous on $(0,1341)$? 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 $x_n=\\frac{1341}{n}$ and $y_n=\\frac{1341}{2n}$.",
"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 ... | 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-018770 | Real Analysis: Uniform Continuity (Variant B) | 10 | Checkpoint: Let $f:[-1275,1275]\to\mathbb{R}$ be $f(x)=(20)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-1275,1275]$.
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 $x,y\\in[-1275,1275]$, $|f(x)-f(y)|=|20|\\,|x^2-y^2|=|20|\\,|x-y||x+y|$.",
"Step 2: Bound... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. 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... | [
{
"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-018771 | Real Analysis: Uniform Continuity (Variant A) | 10 | Warm-up: Let $f:(0,1804)\to\mathbb{R}$ be $f(x)=\frac{1}{(16)x}$. Is $f$ uniformly continuous on $(0,1804)$? 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 $x_n=\\frac{1804}{n}$ and $y_n=\\frac{1804}{2n}$ in $(0,1804)$.",
"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 ... | 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-018772 | Real Analysis: Uniform Continuity (Variant C) | 10 | Find the exact value: Let $f:(0,604)\to\mathbb{R}$ be $f(x)=\ln(45x)$. Is $f$ uniformly continuous on $(0,604)$? 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. |
math-018773 | Real Analysis: Uniform Continuity (Variant A) | 10 | Complete the analysis: Let $f:(0,1781)\to\mathbb{R}$ be $f(x)=\ln(22x)$. Is $f$ uniformly continuous on $(0,1781)$? 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 $x_n=\\frac{1781}{n}$ and $y_n=\\frac{1781}{2n}$.",
"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. |
math-018774 | Real Analysis: Uniform Continuity (Variant A) | 10 | Do not skip justification steps: Let $f:(0,1465)\to\mathbb{R}$ be $f(x)=\ln(55x)$. Is $f$ uniformly continuous on $(0,1465)$? 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 $x_n=\\frac{1465}{n}$ and $y_n=\\frac{1465}{2n}$.",
"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": "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. |
math-018775 | Real Analysis: Uniform Continuity (Variant A) | 10 | Provide both a computational and a conceptual explanation: Let $f:(0,219)\to\mathbb{R}$ be $f(x)=\ln(59x)$. Is $f$ uniformly continuous on $(0,219)$? 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-018776 | Real Analysis: Uniform Continuity (Core) | 10 | Solve and include a self-check: Let $f:(0,377)\to\mathbb{R}$ be $f(x)=\ln(11x)$. Is $f$ uniformly continuous on $(0,377)$? 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 $x_n=\\frac{377}{n}$ and $y_n=\\frac{377}{2n}$.",
"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. (Here the result is $\boxed{\text{No}$.) |
math-018777 | Real Analysis: Uniform Continuity (Variant B) | 10 | Make each step logically reversible (or explain if not): Let $f:(0,1013)\to\mathbb{R}$ be $f(x)=\ln(25x)$. Is $f$ uniformly continuous on $(0,1013)$? 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 ... | 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-018778 | Real Analysis: Uniform Continuity (Variant A) | 10 | Complete the analysis: Let $f:(0,1443)\to\mathbb{R}$ be $f(x)=\frac{1}{(-9)x}$. Is $f$ uniformly continuous on $(0,1443)$? 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. |
math-018779 | Real Analysis: Uniform Continuity (Variant B) | 10 | Problem: Let $f:(0,266)\to\mathbb{R}$ be $f(x)=\sqrt{39x}$. Prove that $f$ is uniformly continuous on $(0,266)$.
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": "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-018780 | Real Analysis: Uniform Continuity (Variant B) | 10 | Try to avoid pattern-matching; explain why: Let $f:(0,743)\to\mathbb{R}$ be $f(x)=\ln(39x)$. Is $f$ uniformly continuous on $(0,743)$? 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 ... | 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-018781 | Real Analysis: Uniform Continuity (Variant C) | 10 | Answer using clear logical steps: Let $f:(0,1262)\to\mathbb{R}$ be $f(x)=\frac{1}{(-16)x}$. Is $f$ uniformly continuous on $(0,1262)$? 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 $x_n=\\frac{1262}{n}$ and $y_n=\\frac{1262}{2n}$ in $(0,1262)$.",
"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 ... | 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-018782 | Real Analysis: Uniform Continuity (Variant A) | 10 | Answer using clear logical steps: Let $f:[-1927,1927]\to\mathbb{R}$ be $f(x)=(-30)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-1927,1927]$.
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 $[-1927,1927]$ 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": "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-018783 | Real Analysis: Uniform Continuity (Variant C) | 10 | Solve and justify each step: Let $f:(0,1564)\to\mathbb{R}$ be $f(x)=\ln(26x)$. Is $f$ uniformly continuous on $(0,1564)$? 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 $x_n=\\frac{1564}{n}$ and $y_n=\\frac{1564}{2n}$.",
"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": "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. |
math-018784 | Real Analysis: Uniform Continuity (Variant B) | 10 | Proceed methodically: Let $f:(0,1238)\to\mathbb{R}$ be $f(x)=\ln(30x)$. Is $f$ uniformly continuous on $(0,1238)$? 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 ... | 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-018785 | Real Analysis: Uniform Continuity (Core) | 10 | Work this out carefully: Let $f:(0,459)\to\mathbb{R}$ be $f(x)=\frac{1}{(3)x}$. Is $f$ uniformly continuous on $(0,459)$? 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 $x_n=\\frac{459}{n}$ and $y_n=\\frac{459}{2n}$ in $(0,459)$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{... | {
"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 ... | 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-018786 | Real Analysis: Uniform Continuity (Core) | 10 | Track quantifiers carefully: Let $f:(0,1272)\to\mathbb{R}$ be $f(x)=\frac{1}{(-9)x}$. Is $f$ uniformly continuous on $(0,1272)$? 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 $x_n=\\frac{1272}{n}$ and $y_n=\\frac{1272}{2n}$ in $(0,1272)$.",
"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-018787 | Real Analysis: Uniform Continuity (Variant B) | 10 | Track quantifiers carefully: Let $f:(0,1441)\to\mathbb{R}$ be $f(x)=\ln(47x)$. Is $f$ uniformly continuous on $(0,1441)$? 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-018788 | Real Analysis: Uniform Continuity (Core) | 10 | Give a fully justified solution: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(-16)x^2$ 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 ... | 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-018789 | Real Analysis: Uniform Continuity (Variant C) | 10 | Where appropriate, name the theorem you use: Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=(-30)x^2$ 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 ... | 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-018790 | Real Analysis: Uniform Continuity (Core) | 10 | Provide both a computational and a conceptual explanation: Let $f:(0,1328)\to\mathbb{R}$ be $f(x)=\sqrt{36x}$. Prove that $f$ is uniformly continuous on $(0,1328)$.
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. |
math-018791 | Real Analysis: Uniform Continuity (Variant C) | 10 | Provide both a computational and a conceptual explanation: Let $f:(0,35)\to\mathbb{R}$ be $f(x)=\frac{1}{(-9)x}$. Is $f$ uniformly continuous on $(0,35)$? 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 $x_n=\\frac{35}{n}$ and $y_n=\\frac{35}{2n}$ in $(0,35)$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}... | {
"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-018792 | Real Analysis: Uniform Continuity (Variant A) | 10 | Complete the analysis: Let $f:[-1953,1953]\to\mathbb{R}$ be $f(x)=(-1)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-1953,1953]$.
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 $[-1953,1953]$ 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": "Consistency verification shows both paths yield the identical boxed result. 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... | [
{
"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-018793 | Real Analysis: Uniform Continuity (Variant A) | 10 | Checkpoint: Let $f:(0,632)\to\mathbb{R}$ be $f(x)=\ln(6x)$. Is $f$ uniformly continuous on $(0,632)$? 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 $x_n=\\frac{632}{n}$ and $y_n=\\frac{632}{2n}$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{2n}\\to 0$.",
"Step 3: But $|\\ln(... | {
"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. (Here the result is $\boxed{\text{No}$.) |
math-018794 | Real Analysis: Uniform Continuity (Variant C) | 10 | Work this out carefully: Let $f:(0,635)\to\mathbb{R}$ be $f(x)=\sqrt{13x}$. Prove that $f$ is uniformly continuous on $(0,635)$.
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-018795 | Real Analysis: Uniform Continuity (Variant B) | 10 | Compute the requested quantity: Let $f:[-770,770]\to\mathbb{R}$ be $f(x)=(-21)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-770,770]$.
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 $x,y\\in[-770,770]$, $|f(x)-f(y)|=|-21|\\,|x^2-y^2|=|-21|\\,|x-y||x+y|$.",
"Step 2: Bound... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. 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... | [
{
"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-018796 | Real Analysis: Uniform Continuity (Variant C) | 10 | Prompt: Let $f:[-741,741]\to\mathbb{R}$ be $f(x)=(-14)x^2$ with $c\ne 0$. Prove that $f$ is uniformly continuous on $[-741,741]$.
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 $x,y\\in[-741,741]$, $|f(x)-f(y)|=|-14|\\,|x^2-y^2|=|-14|\\,|x-y||x+y|$.",
"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 ... | 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-018797 | Real Analysis: Uniform Continuity (Variant C) | 10 | Work this out carefully: Let $f:(0,224)\to\mathbb{R}$ be $f(x)=\ln(5x)$. Is $f$ uniformly continuous on $(0,224)$? 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 ... | 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-018798 | Real Analysis: Uniform Continuity (Core) | 10 | Challenge: Let $f:(0,1983)\to\mathbb{R}$ be $f(x)=\sqrt{20x}$. Prove that $f$ is uniformly continuous on $(0,1983)$.
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": "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-018799 | Real Analysis: Uniform Continuity (Variant C) | 10 | Explain what is being counted/optimized: Let $f:(0,504)\to\mathbb{R}$ be $f(x)=\ln(43x)$. Is $f$ uniformly continuous on $(0,504)$? 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 $x_n=\\frac{504}{n}$ and $y_n=\\frac{504}{2n}$.",
"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 ... | 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-018800 | Real Analysis: Uniform Continuity (Core) | 10 | Solve and sanity-check: Let $f:(0,131)\to\mathbb{R}$ be $f(x)=\frac{1}{(5)x}$. Is $f$ uniformly continuous on $(0,131)$? 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 $x_n=\\frac{131}{n}$ and $y_n=\\frac{131}{2n}$ in $(0,131)$.",
"Step 2: Then $|x_n-y_n|=\\frac{A}{... | {
"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 $\boxed{\text{No}$.) |
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