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values | question stringlengths 17 24.5k | options stringlengths 2 4.26k | correct_option stringclasses 6
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|---|---|---|---|---|---|---|---|---|---|---|---|
1l54ug8v8 | maths | functions | composite-functions | <p>Let f(x) and g(x) be two real polynomials of degree 2 and 1 respectively. If $$f(g(x)) = 8{x^2} - 2x$$ and $$g(f(x)) = 4{x^2} + 6x + 1$$, then the value of $$f(2) + g(2)$$ is _________.</p> | [] | null | 18 | $f(g(x))=8 x^{2}-2 x$
$$
g(f(x))=4 x^{2}+6 x+1
$$
<br/><br/>
let $f(x)=c x^{2}+d x+e$
<br/><br/>
$g(x)=a x+b$
<br/><br/>
$f(g(x))=c(a x+b)^{2}+d(a x+b)+e \equiv 8 x^{2}-2 x$
<br/><br/>
$g(f(x))=a\left(c x^{2}+d x+e\right)+b \equiv 4 x^{2}+6 x+1$
<br/><br/>
$\therefore \quad a c=4 \quad a d=6 \quad a e+b=1$
<br/><br/>
... | integer | jee-main-2022-online-29th-june-evening-shift | 6,208 |
1l56rupn7 | maths | functions | composite-functions | <p>Let S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Define f : S $$\to$$ S as</p>
<p>$$f(n) = \left\{ {\matrix{
{2n} & , & {if\,n = 1,2,3,4,5} \cr
{2n - 11} & , & {if\,n = 6,7,8,9,10} \cr
} } \right.$$.</p>
<p>Let g : S $$\to$$ S be a function such that $$fog(n) = \left\{ {\matrix{
{n + 1} & ... | [] | null | 190 | <p>$$\because$$ $$f(n) = \left\{ {\matrix{
{2n,} & {n = 1,2,3,4,5} \cr
{2n - 11,} & {n = 6,7,8,9,10} \cr} } \right.$$</p>
<p>$$\therefore$$ f(1) = 2, f(2) = 4, ......, f(5) = 10</p>
<p>and f(6) = 1, f(7) = 3, f(8) = 5, ......, f(10) = 9</p>
<p>Now, $$f(g(n)) = \left\{ {\matrix{
{n + 1,} & {if\,n\,is\,odd} ... | integer | jee-main-2022-online-27th-june-evening-shift | 6,209 |
1l58ex9ng | maths | functions | composite-functions | <p>Let f : R $$\to$$ R be defined as f (x) = x $$-$$ 1 and g : R $$-$$ {1, $$-$$1} $$\to$$ R be defined as $$g(x) = {{{x^2}} \over {{x^2} - 1}}$$.</p>
<p>Then the function fog is :</p> | [{"identifier": "A", "content": "one-one but not onto"}, {"identifier": "B", "content": "onto but not one-one"}, {"identifier": "C", "content": "both one-one and onto"}, {"identifier": "D", "content": "neither one-one nor onto"}] | ["D"] | null | <p>$$f:R \to R$$ defined as</p>
<p>$$f(x) = x - 1$$ and $$g:R \to \{ 1, - 1\} \to R,\,g(x) = {{{x^2}} \over {{x^2} - 1}}$$</p>
<p>Now $$fog(x) = {{{x^2}} \over {{x^2} - 1}} - 1 = {1 \over {{x^2} - 1}}$$</p>
<p>$$\therefore$$ Domain of $$fog(x) = R - \{ - 1,1\} $$</p>
<p>And range of $$fog(x) = ( - \infty , - 1] \cup ... | mcq | jee-main-2022-online-26th-june-evening-shift | 6,211 |
1l5aijk91 | maths | functions | composite-functions | <p>Let $$f:R \to R$$ and $$g:R \to R$$ be two functions defined by $$f(x) = {\log _e}({x^2} + 1) - {e^{ - x}} + 1$$ and $$g(x) = {{1 - 2{e^{2x}}} \over {{e^x}}}$$. Then, for which of the following range of $$\alpha$$, the inequality $$f\left( {g\left( {{{{{(\alpha - 1)}^2}} \over 3}} \right)} \right) > f\left( {g\l... | [{"identifier": "A", "content": "(2, 3)"}, {"identifier": "B", "content": "($$-$$2, $$-$$1)"}, {"identifier": "C", "content": "(1, 2)"}, {"identifier": "D", "content": "($$-$$1, 1)"}] | ["A"] | null | <p>$$f(x) = {\log _e}({x^2} + 1) - {e^{ - x}} + 1$$</p>
<p>$$f'(x) = {{2x} \over {{x^2} + 1}} + {e^{ - x}}$$</p>
<p>$$ = {2 \over {x + {1 \over x}}} + {e^{ - x}} > 0\,\,\forall x \in R$$</p>
<p>$$g(x) = {e^{ - x}} - 2{e^x}$$</p>
<p>$$g'(x) - - {e^{ - x}} - 2{e^x} < 0\,\,\,\,\forall x \in R$$</p>
<p>$$\Rightarrow$$ f(x... | mcq | jee-main-2022-online-25th-june-morning-shift | 6,212 |
1l5ajukpk | maths | functions | composite-functions | <p>Let $$f:R \to R$$ be a function defined by <br/><br/>$$f(x) = {\left( {2\left( {1 - {{{x^{25}}} \over 2}} \right)(2 + {x^{25}})} \right)^{{1 \over {50}}}}$$. If the function $$g(x) = f(f(f(x))) + f(f(x))$$, then the greatest integer less than or equal to g(1) is ____________.</p> | [] | null | 2 | <p>Given,</p>
<p>$$f(x) = {\left( {2\left( {1 - {{{x^{25}}} \over 2}} \right)\left( {2 + {x^{25}}} \right)} \right)^{{1 \over {50}}}}$$</p>
<p>and $$g(x) = f\left( {f\left( {f\left( x \right)} \right)} \right) + f\left( {f\left( x \right)} \right)$$</p>
<p>$$\therefore$$ $$g(1) = f\left( {f\left( {f\left( 1 \right)} \r... | integer | jee-main-2022-online-25th-june-morning-shift | 6,213 |
1l6f37b3x | maths | functions | composite-functions | <p>Let $$f(x)$$ be a quadratic polynomial with leading coefficient 1 such that $$f(0)=p, p \neq 0$$, and $$f(1)=\frac{1}{3}$$. If the equations $$f(x)=0$$ and $$f \circ f \circ f \circ f(x)=0$$ have a common real root, then $$f(-3)$$ is equal to ________________.</p> | [] | null | 25 | <p>Let $$f(x) = (x - \alpha )(x - \beta )$$</p>
<p>It is given that $$f(0) = p \Rightarrow \alpha \beta = p$$</p>
<p>and $$f(1) = {1 \over 3} \Rightarrow (1 - \alpha )(1 - \beta ) = {1 \over 3}$$</p>
<p>Now, let us assume that, $$\alpha$$ is the common root of $$f(x) = 0$$ and $$fofofof(x) = 0$$</p>
<p>$$fofofof(x) = ... | integer | jee-main-2022-online-25th-july-evening-shift | 6,214 |
1l6jb0fby | maths | functions | composite-functions | <p>Let $$f, g: \mathbb{N}-\{1\} \rightarrow \mathbb{N}$$ be functions defined by $$f(a)=\alpha$$, where $$\alpha$$ is the maximum of the powers of those primes $$p$$ such that $$p^{\alpha}$$ divides $$a$$, and $$g(a)=a+1$$, for all $$a \in \mathbb{N}-\{1\}$$. Then, the function $$f+g$$ is</p> | [{"identifier": "A", "content": "one-one but not onto"}, {"identifier": "B", "content": "onto but not one-one"}, {"identifier": "C", "content": "both one-one and onto"}, {"identifier": "D", "content": "neither one-one nor onto"}] | ["D"] | null | <p>$$f,g:N - \{ 1\} \to N$$ defined as</p>
<p>$$f(a) = \alpha $$, where $$\alpha$$ is the maximum power of those primes p such that p<sup>$$\alpha$$</sup> divides a.</p>
<p>$$g(a) = a + 1$$,</p>
<p>Now,</p>
<p>$$\matrix{
{f(2) = 1,} & {g(2) = 3} & \Rightarrow & {(f + g)\,(2) = 4} \cr
{f(3) = 1,} & {g(3) = 4}... | mcq | jee-main-2022-online-27th-july-morning-shift | 6,215 |
1l6m66fbv | maths | functions | composite-functions | <p>Let $$\alpha, \beta$$ and $$\gamma$$ be three positive real numbers. Let $$f(x)=\alpha x^{5}+\beta x^{3}+\gamma x, x \in \mathbf{R}$$ and $$g: \mathbf{R} \rightarrow \mathbf{R}$$ be such that $$g(f(x))=x$$ for all $$x \in \mathbf{R}$$. If $$\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots, \mathrm{a}_{\mathrm{... | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "27"}] | ["A"] | null | <p>$$f\left( {g\left( {{1 \over n}\sum\limits_{i = 1}^n {f({a_i})} } \right)} \right)$$</p>
<p>$${{{a_1} + {a_2} + {a_3}\, + \,......\, + \,{a_n}} \over n} = 0$$</p>
<p>$$\therefore$$ First and last term, second and second last and so on are equal in magnitude but opposite in sign.</p>
<p>$$f(x) = \alpha {x^5} + \beta ... | mcq | jee-main-2022-online-28th-july-morning-shift | 6,216 |
1ldv2y8dq | maths | functions | composite-functions | <p>For some a, b, c $$\in\mathbb{N}$$, let $$f(x) = ax - 3$$ and $$\mathrm{g(x)=x^b+c,x\in\mathbb{R}}$$. If $${(fog)^{ - 1}}(x) = {\left( {{{x - 7} \over 2}} \right)^{1/3}}$$, then $$(fog)(ac) + (gof)(b)$$ is equal to ____________.</p> | [] | null | 2039 | $f(x)=a x-3$
<br/><br/>
$g(x)=x^{b}+c$
<br/><br/>
$(fog)^{-1}=\left(\frac{x-7}{2}\right)^{\frac{1}{3}}$
<br/><br/>
$(fog)^{-1}(x)=\left(\frac{x+3-c a}{a}\right)^{\frac{1}{b}}=\left(\frac{x-7}{2}\right)^{\frac{1}{3}}$
<br/><br/>
$\Rightarrow a=2, b=3, c=5$
<br/><br/>
$fog(a c)+gof(b)$
<br/><br/>
$\because f(x)=2 x-3$
<b... | integer | jee-main-2023-online-25th-january-morning-shift | 6,217 |
lsaoplpj | maths | functions | composite-functions | Let $f: \mathbf{R} \rightarrow \mathbf{R}$ and $g: \mathbf{R} \rightarrow \mathbf{R}$ be defined as
<br/><br/>$f(x)=\left\{\begin{array}{ll}\log _{\mathrm{e}} x, & x>0 \\ \mathrm{e}^{-x}, & x \leq 0\end{array}\right.$ and
<br/><br/>$g(x)=\left\{\begin{array}{ll}x, & x \geqslant 0 \\ \mathrm{e}^x, &... | [{"identifier": "A", "content": "one-one but not onto"}, {"identifier": "B", "content": "neither one-one nor onto"}, {"identifier": "C", "content": "onto but not one-one"}, {"identifier": "D", "content": "both one-one and onto"}] | ["B"] | null | Given, $f(x)=\left\{\begin{array}{ll}\log _{\mathrm{e}} x, & x>0 \\ \mathrm{e}^{-x}, & x \leq 0\end{array}\right.$ and
<br><br>$g(x)=\left\{\begin{array}{ll}x, & x \geqslant 0 \\ \mathrm{e}^x, & x<0\end{array}\right.$
<br><br>then $g \circ f(x)$ $=g(f(x))$
<br><br>$\begin{aligned} & \mathrm{g... | mcq | jee-main-2024-online-1st-february-morning-shift | 6,219 |
jaoe38c1lscna9if | maths | functions | composite-functions | <p>Let $$f: \mathbf{R}-\left\{\frac{-1}{2}\right\} \rightarrow \mathbf{R}$$ and $$g: \mathbf{R}-\left\{\frac{-5}{2}\right\} \rightarrow \mathbf{R}$$ be defined as $$f(x)=\frac{2 x+3}{2 x+1}$$ and $$g(x)=\frac{|x|+1}{2 x+5}$$. Then, the domain of the function fog is :</p> | [{"identifier": "A", "content": "$$\\mathbf{R}-\\left\\{-\\frac{7}{4}\\right\\}$$\n"}, {"identifier": "B", "content": "$$\\mathbf{R}$$\n"}, {"identifier": "C", "content": "$$\\mathbf{R}-\\left\\{-\\frac{5}{2},-\\frac{7}{4}\\right\\}$$\n"}, {"identifier": "D", "content": "$$\\mathbf{R}-\\left\\{-\\frac{5}{2}\\right\\}$$... | ["D"] | null | <p>$$\begin{aligned}
& f(x)=\frac{2 x+3}{2 x+1} ; x \neq-\frac{1}{2} \\
& g(x)=\frac{|x|+1}{2 x+5}, x \neq-\frac{5}{2}
\end{aligned}$$</p>
<p>Domain of $$f(g(x))$$</p>
<p>$$f(g(x))=\frac{2 g(x)+3}{2 g(x)+1}$$</p>
<p>$$x \neq-\frac{5}{2}$$ and $$\frac{|x|+1}{2 x+5} \neq-\frac{1}{2}$$</p>
<p>$$x \in R-\left\{-\frac{5}{2}... | mcq | jee-main-2024-online-27th-january-evening-shift | 6,220 |
jaoe38c1lse52f1r | maths | functions | composite-functions | <p>If $$f(x)=\frac{4 x+3}{6 x-4}, x \neq \frac{2}{3}$$ and $$(f \circ f)(x)=g(x)$$, where $$g: \mathbb{R}-\left\{\frac{2}{3}\right\} \rightarrow \mathbb{R}-\left\{\frac{2}{3}\right\}$$, then $$(g ogog)(4)$$ is equal to</p> | [{"identifier": "A", "content": "$$-4$$"}, {"identifier": "B", "content": "$$\\frac{19}{20}$$"}, {"identifier": "C", "content": "$$-\\frac{19}{20}$$"}, {"identifier": "D", "content": "4"}] | ["D"] | null | <p>To find $$(g \circ g \circ g)(4),$$ we first need to understand the composition of $$f$$ with itself, i.e., $$(f \circ f)(x) = f(f(x)) = g(x).$$ We can then repeatedly apply $$g$$ to get the given expression.</p>
<p>First, let's calculate $$(f \circ f)(x) = g(x):$$</p>
<p>$$g(x) = (f \circ f)(x) = f(f(x))$$</p>
<... | mcq | jee-main-2024-online-31st-january-morning-shift | 6,221 |
jaoe38c1lseypzri | maths | functions | composite-functions | <p>If $$f(x)=\left\{\begin{array}{cc}2+2 x, & -1 \leq x < 0 \\ 1-\frac{x}{3}, & 0 \leq x \leq 3\end{array} ; g(x)=\left\{\begin{array}{cc}-x, & -3 \leq x \leq 0 \\ x, & 0 < x \leq 1\end{array}\right.\right.$$, then range of $$(f o g)(x)$$ is</p> | [{"identifier": "A", "content": "$$[0,1)$$\n"}, {"identifier": "B", "content": "$$[0,3)$$\n"}, {"identifier": "C", "content": "$$(0,1]$$\n"}, {"identifier": "D", "content": "$$[0,1]$$"}] | ["D"] | null | <p>$$f(g(x)) = \left\{ {\matrix{
{2 + 2g(x),} & { - 1 \le g(x) < 0} & {.....(1)} \cr
{1 - {{g(x)} \over 3},} & {0 \le g(x) \le 3} & {.....(2)} \cr
} } \right.$$</p>
<p>$$\text { By (1) } x \in \phi$$</p>
<p>And by (2) $$x \in[-3,0]$$ and $$x \in[0,1]$$</p>
<p><img src="https://app-content... | mcq | jee-main-2024-online-29th-january-morning-shift | 6,222 |
lv2erz8l | maths | functions | composite-functions | <p>Consider the function $$f: \mathbb{R} \rightarrow \mathbb{R}$$ defined by $$f(x)=\frac{2 x}{\sqrt{1+9 x^2}}$$. If the composition of $$f, \underbrace{(f \circ f \circ f \circ \cdots \circ f)}_{10 \text { times }}(x)=\frac{2^{10} x}{\sqrt{1+9 \alpha x^2}}$$, then the value of $$\sqrt{3 \alpha+1}$$ is equal to _______... | [] | null | 1024 | <p>$$f(x)=\frac{2 x}{\sqrt{1+9 x^2}}$$</p>
<p>$$(f \circ f)(x)=\frac{2 f(x)}{\sqrt{1+9(f(x))^2}}=\frac{\frac{4 x}{\sqrt{1+9 x^2}}}{\sqrt{1+9 \times \frac{4 x^2}{1+9 x^2}}}=\frac{4 x}{\sqrt{1+45 x^2}}$$</p>
<p>$$(f \circ f \circ f)(x)=\frac{4 \times \frac{2 x}{\sqrt{1+9 x^2}}}{\sqrt{1+45 \times \frac{4 x^2}{1+9 x^2}}}=\... | integer | jee-main-2024-online-4th-april-evening-shift | 6,223 |
lv9s1zso | maths | functions | composite-functions | <p>Let $$f, g: \mathbf{R} \rightarrow \mathbf{R}$$ be defined as :</p>
<p>$$f(x)=|x-1| \text { and } g(x)= \begin{cases}\mathrm{e}^x, & x \geq 0 \\ x+1, & x \leq 0 .\end{cases}$$</p>
<p>Then the function $$f(g(x))$$ is</p> | [{"identifier": "A", "content": "neither one-one nor onto.\n"}, {"identifier": "B", "content": "one-one but not onto.\n"}, {"identifier": "C", "content": "both one-one and onto.\n"}, {"identifier": "D", "content": "onto but not one-one."}] | ["A"] | null | <p>$$\begin{aligned}
& f(x)= \begin{cases}x-1, & x \geq 1 \\
1-x & x<0\end{cases} \\
& g(x)=\left\{\begin{array}{cc}
e^x & ; \quad x \geq 0 \\
x+1 & ; \quad x \leq 0
\end{array}\right.
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwelk3uw/46e00973-b9... | mcq | jee-main-2024-online-5th-april-evening-shift | 6,224 |
xkNnw5MPjtXPCYQ0 | maths | functions | domain | The domain of $${\sin ^{ - 1}}\left[ {{{\log }_3}\left( {{x \over 3}} \right)} \right]$$ is | [{"identifier": "A", "content": "[1, 9]"}, {"identifier": "B", "content": "[-1, 9]"}, {"identifier": "C", "content": "[9, 1]"}, {"identifier": "D", "content": "[-9, -1]"}] | ["A"] | null | $$f\left( x \right) = {\sin ^{ - 1}}\left( {{{\log }_3}\left( {{x \over 3}} \right)} \right)$$ exists
<br><br>if $$\,\,\,\, - 1 \le {\log _3}\left( {{x \over 3}} \right) \le 1$$
<br><br>$$ \Leftrightarrow {3^{ - 1}} \le {x \over 3} \le {3^1}$$
<br><br>$$ \Leftrightarrow 1 \le x \le 9$$
<br><br>or $$\,\,\,\,x \in \left[... | mcq | aieee-2002 | 6,225 |
htONhxhNRHk21CjA | maths | functions | domain | The domain of the function
<br/>$$f\left( x \right) = {{{{\sin }^{ - 1}}\left( {x - 3} \right)} \over {\sqrt {9 - {x^2}} }}$$ | [{"identifier": "A", "content": "[1, 2]"}, {"identifier": "B", "content": "[2, 3)"}, {"identifier": "C", "content": "[1, 2)"}, {"identifier": "D", "content": "[2, 3]"}] | ["B"] | null | $$f\left( x \right) = {{{{\sin }^{ - 1}}\left( {x - 3} \right)} \over {\sqrt {9 - {x^2}} }}$$ is defined
<br><br>if $$(i)$$ $$\,\,\, - 1 \le x - 3 \le 1 \Rightarrow 2 \le x \le 4$$
<br><br>and $$(ii)$$ $$9 - {x^2} > 0 \Rightarrow - 3 < x < 3$$
<br><br>Taking common solution of $$\left( i \right)$$ and $$\left... | mcq | aieee-2004 | 6,227 |
d90DFXOHvtSgLUje | maths | functions | domain | The largest interval lying in $$\left( { - {\pi \over 2},{\pi \over 2}} \right)$$ for which the function
<br/><br/>$$f\left( x \right) = {4^{ - {x^2}}} + {\cos ^{ - 1}}\left( {{x \over 2} - 1} \right)$$$$ + \log \left( {\cos x} \right)$$,
<br/><br/>is defined, is | [{"identifier": "A", "content": "$$\\left[ { - {\\pi \\over 4},{\\pi \\over 2}} \\right)$$"}, {"identifier": "B", "content": "$$\\left[ {0,{\\pi \\over 2}} \\right)$$"}, {"identifier": "C", "content": "$$\\left[ {0,\\pi } \\right]$$"}, {"identifier": "D", "content": "$$\\left( { - {\\pi \\over 2},{\\pi \\over 2}} ... | ["B"] | null | $$f\left( x \right) = {4^{ - {x^2}}} + {\cos ^{ - 1}}\left( {{x \over 2} - 1} \right) + \log \left( {\cos \,x} \right)$$
<br><br>$$f\left( x \right)$$ is defined if $$ - 1 \le \left( {{x \over 2} - 1} \right) \le 1$$ and $$\cos \,x > 0$$
<br><br>or $$\,\,\,\,0 \le {x \over 2} \le 2\,\,$$ and $$\,\, - {\pi \over 2} ... | mcq | aieee-2007 | 6,228 |
w8Zuk7uFMhknNRsWwF18hoxe66ijvww16jt | maths | functions | domain | The domain of the definition of the function
<br/><br/>$$f(x) = {1 \over {4 - {x^2}}} + {\log _{10}}({x^3} - x)$$ is | [{"identifier": "A", "content": "(-1, 0) $$ \\cup $$ (1, 2) $$ \\cup $$ (2, $$\\infty $$)"}, {"identifier": "B", "content": "(-2, -1) $$ \\cup $$ (-1,0) $$ \\cup $$ (2, $$\\infty $$)"}, {"identifier": "C", "content": "(1, 2) $$ \\cup $$ (2, $$\\infty $$)"}, {"identifier": "D", "content": "(-1, 0) $$ \\cup $$ (1,2) $$ \... | ["A"] | null | Given $$f(x) = {1 \over {4 - {x^2}}} + {\log _{10}}({x^3} - x)$$
<br><br>Let f<sub>1</sub>(x) = $${1 \over {4 - {x^2}}}$$ and f<sub>2</sub>(x) = $${\log _{10}}({x^3} - x)$$
<br><br>Here in f<sub>1</sub>(x) denominator $$ \ne $$ 0
<br><br>4 - x<sup>2</sup> $$ \ne $$ 0
<br><br>$$ \Rightarrow $$ x $$ \ne $$ $$ \pm $$ 2 ..... | mcq | jee-main-2019-online-9th-april-evening-slot | 6,230 |
gQxMfOaeoMVTTKD7A61kmlivagb | maths | functions | domain | The real valued function <br/>$$f(x) = {{\cos e{c^{ - 1}}x} \over {\sqrt {x - [x]} }}$$, where [x] denotes the greatest integer less than or equal to x, is defined for all x belonging to : | [{"identifier": "A", "content": "all real except integers"}, {"identifier": "B", "content": "all non-integers except the interval [ $$-$$1, 1 ]"}, {"identifier": "C", "content": "all integers except 0, $$-$$1, 1"}, {"identifier": "D", "content": "all real except the interval [ $$-$$1, 1 ]"}] | ["B"] | null | Domain of $$\cos e{c^{ - 1}}x$$ :<br><br>$$x \in ( - \infty , - 1] \cup [1,\infty )$$<br><br>and, $$x - [x] > 0$$<br><br>$$ \Rightarrow \{ x\} > 0$$<br><br>$$ \Rightarrow x \ne I$$<br><br>$$ \therefore $$ Required domain = $$( - \infty , - 1] \cup [1,\infty ) - I$$ | mcq | jee-main-2021-online-18th-march-morning-shift | 6,231 |
1krpv18s8 | maths | functions | domain | Let [ x ] denote the greatest integer $$\le$$ x, where x $$\in$$ R. If the domain of the real valued function $$f(x) = \sqrt {{{\left| {[x]} \right| - 2} \over {\left| {[x]} \right| - 3}}} $$ is ($$-$$ $$\infty$$, a) $$]\cup$$ [b, c) $$\cup$$ [4, $$\infty$$), a < b < c, then the value of a + b + c is : | [{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$$-$$2"}, {"identifier": "D", "content": "$$-$$3"}] | ["C"] | null | For domain,<br><br>$${{{\left| {[x]} \right| - 2} \over {\left| {[x]} \right| - 3}}}$$ $$\ge$$ 0<br><br>Case I :<br><br>When $${\left| {[x]} \right| - 2}$$ $$\ge$$ 0<br><br>and $${\left| {[x]} \right| - 3}$$ > 0<br><br>$$\therefore$$ x $$\in$$ ($$-$$ $$\infty$$, $$-$$3) $$\cup$$ [4, $$\infty$$) ...... (1)<br><br>Cas... | mcq | jee-main-2021-online-20th-july-morning-shift | 6,232 |
1ldsvexaq | maths | functions | domain | <p>The domain of $$f(x) = {{{{\log }_{(x + 1)}}(x - 2)} \over {{e^{2{{\log }_e}x}} - (2x + 3)}},x \in \mathbb{R}$$ is</p> | [{"identifier": "A", "content": "$$( - 1,\\infty ) - \\{ 3\\} $$"}, {"identifier": "B", "content": "$$\\mathbb{R} - \\{ - 1,3)$$"}, {"identifier": "C", "content": "$$(2,\\infty ) - \\{ 3\\} $$"}, {"identifier": "D", "content": "$$\\mathbb{R} - \\{ 3\\} $$"}] | ["C"] | null | $x-2>0 \Rightarrow x>2$
<br/><br/>
$\mathrm{x}+1>0 \Rightarrow \mathrm{x}>-1$
<br/><br/>
$x+1 \neq 1 \Rightarrow x \neq 0$ and $x>0$
<br/><br/>
Denominator
<br/><br/>
$\mathrm{x}^{2}-2 \mathrm{x}-3 \neq 0$
<br/><br/>
$(x-3)(x+1) \neq 0$
<br/><br/>
$\mathrm{x} \neq-1,3$
<br/><br/>
So Ans $(2, \infty)-\{3\}$ | mcq | jee-main-2023-online-29th-january-morning-shift | 6,233 |
1lgrgf67d | maths | functions | domain | <p>Let $$\mathrm{D}$$ be the domain of the function $$f(x)=\sin ^{-1}\left(\log _{3 x}\left(\frac{6+2 \log _{3} x}{-5 x}\right)\right)$$. If the range of the function $$\mathrm{g}: \mathrm{D} \rightarrow \mathbb{R}$$ defined by $$\mathrm{g}(x)=x-[x],([x]$$ is the greatest integer function), is $$(\alpha, \beta)$$, then... | [{"identifier": "A", "content": "45"}, {"identifier": "B", "content": "136"}, {"identifier": "C", "content": "46"}, {"identifier": "D", "content": " nearly 135"}] | ["D"] | null | <p>First, the function $f(x) = \sin^{-1}(\log_{3x}(\frac{6 + 2 \log_3{x}}{-5x}))$ has several restrictions :</p>
<ol>
<li><p>Since the arcsine function $\sin^{-1}(x)$ is only defined for $-1\leq x\leq 1$, this means that $\log_{3x}(\frac{6 + 2 \log _3 x}{-5 x})$ must be between -1 and 1.</p>
</li>
<li><p>For the logari... | mcq | jee-main-2023-online-12th-april-morning-shift | 6,234 |
1lgsw05gr | maths | functions | domain | <p>The domain of the function $$f(x)=\frac{1}{\sqrt{[x]^{2}-3[x]-10}}$$ is : ( where $$[\mathrm{x}]$$ denotes the greatest integer less than or equal to $$x$$ )</p> | [{"identifier": "A", "content": "$$(-\\infty,-2) \\cup[6, \\infty)$$"}, {"identifier": "B", "content": "$$(-\\infty,-3] \\cup[6, \\infty)$$"}, {"identifier": "C", "content": "$$(-\\infty,-2) \\cup(5, \\infty)$$"}, {"identifier": "D", "content": "$$(-\\infty,-3] \\cup(5, \\infty)$$"}] | ["A"] | null | $$
f(x)=\frac{1}{\sqrt{[x]^2-3[x]-10}}
$$
<br/><br/>For Domain $[x]^2-3[x]-10>0$
<br/><br/>$$
\begin{aligned}
& \Rightarrow ([x]-5)([x]+2)>0 \\\\
& \Rightarrow [x] \in(-\infty,-2) \cup(5, \infty) \\\\
& \therefore x \in(-\infty,-2) \cup[6, \infty)
\end{aligned}
$$ | mcq | jee-main-2023-online-11th-april-evening-shift | 6,235 |
1lgyq24t0 | maths | functions | domain | <p>If domain of the function $$\log _{e}\left(\frac{6 x^{2}+5 x+1}{2 x-1}\right)+\cos ^{-1}\left(\frac{2 x^{2}-3 x+4}{3 x-5}\right)$$ is $$(\alpha, \beta) \cup(\gamma, \delta]$$, then $$18\left(\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2}\right)$$ is equal to ______________.</p> | [] | null | 20 | Domain of $\log _e\left(\frac{6 x^2+5 x+1}{2 x-1}\right)$
<br/><br/>So, $\frac{6 x^2+5 x+1}{2 x-1}>0$
<br/><br/>$$
\begin{aligned}
& \Rightarrow \frac{(3 x+1)(2 x+1)}{2 x-1}>0 \\\\
& \Rightarrow x \in\left(\frac{-1}{2}, \frac{-1}{3}\right) \cup\left(\frac{1}{2}, \infty\right)
\end{aligned}
$$
<br/><br/>Domain of
$$
\co... | integer | jee-main-2023-online-8th-april-evening-shift | 6,236 |
lsam05oo | maths | functions | domain | If the domain of the function
<br/><br/>$f(x)=\frac{\sqrt{x^2-25}}{\left(4-x^2\right)}+\log _{10}\left(x^2+2 x-15\right)$ is $(-\infty, \alpha) \cup[\beta, \infty)$, then $\alpha^2+\beta^3$ is equal to : | [{"identifier": "A", "content": "140"}, {"identifier": "B", "content": "175"}, {"identifier": "C", "content": "125"}, {"identifier": "D", "content": "150"}] | ["D"] | null | <p>To find the domain of the function
$$f(x) = \frac{\sqrt{x^2-25}}{(4-x^2)}+\log_{10}(x^2+2x-15),$$
we need to consider the domain conditions for both the square root function and the logarithmic function.</p>
<p>The square root function $\sqrt{x^2-25}$ requires that the argument of the square root be non-negative, ... | mcq | jee-main-2024-online-1st-february-evening-shift | 6,237 |
1lsg4mhuk | maths | functions | domain | <p>If the domain of the function $$f(x)=\log _e\left(\frac{2 x+3}{4 x^2+x-3}\right)+\cos ^{-1}\left(\frac{2 x-1}{x+2}\right)$$ is $$(\alpha, \beta]$$, then the value of $$5 \beta-4 \alpha$$ is equal to</p> | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "11"}, {"identifier": "D", "content": "10"}] | ["B"] | null | <p>$$\begin{aligned}
& \frac{2 x+3}{4 x^2+x-3}>0 \text { and }-1 \leq \frac{2 x-1}{x+2} \leq 1 \\
& \frac{2 x+3}{(4 x-3)(x+1)}>0 \quad \frac{3 x+1}{x+2} \geq 0 \& \frac{x-3}{x+2} \leq 0
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsoxumdb/8342ad24-c8e3... | mcq | jee-main-2024-online-30th-january-evening-shift | 6,238 |
1lsgb11sw | maths | functions | domain | <p>If the domain of the function $$f(x)=\cos ^{-1}\left(\frac{2-|x|}{4}\right)+\left\{\log _e(3-x)\right\}^{-1}$$ is $$[-\alpha, \beta)-\{\gamma\}$$, then $$\alpha+\beta+\gamma$$ is equal to :</p> | [{"identifier": "A", "content": "11"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "8"}] | ["A"] | null | <p>$$\begin{aligned}
& -1 \leq\left|\frac{2-|x|}{4}\right| \leq 1 \\
& \Rightarrow\left|\frac{2-|x|}{4}\right| \leq 1 \\
& -4 \leq 2-|x| \leq 4 \\
& -6 \leq-|x| \leq 2 \\
& -2 \leq|x| \leq 6 \\
& |x| \leq 6
\end{aligned}$$</p>
<p>$$\Rightarrow x \in[-6,6]$$ .... (1)</p>
<p>Now, $$3-x\ne 1$$</p>
<p>And $$x\ne2$$ .... (2... | mcq | jee-main-2024-online-30th-january-morning-shift | 6,239 |
luy6z50x | maths | functions | domain | <p>If the domain of the function $$f(x)=\sin ^{-1}\left(\frac{x-1}{2 x+3}\right)$$ is $$\mathbf{R}-(\alpha, \beta)$$, then $$12 \alpha \beta$$ is equal to :</p> | [{"identifier": "A", "content": "40"}, {"identifier": "B", "content": "36"}, {"identifier": "C", "content": "24"}, {"identifier": "D", "content": "32"}] | ["D"] | null | <p>$$
\begin{array}{ll}
f(x)=\sin ^{-1}\left(\frac{x-1}{2 x+3}\right) & \\
-1 \leq \frac{x-1}{2 x+3} \leq 1 & \frac{x-1}{2 x+3}+1 \geq 0 \\
\frac{x-1}{2 x+3}-1 \leq 0 & \frac{x-1+2 x+3}{2 x+3} \geq 0 \\
\frac{x-1-2 x-3}{2 x+3} \leq 0 & \frac{3 x+2}{2 x+3} \geq 0
\end{array}
$$</p>
<p><img src="https://a... | mcq | jee-main-2024-online-9th-april-morning-shift | 6,240 |
7lH4aEIStXIhkbBM | maths | functions | even-and-odd-functions | The graph of the function y = f(x) is symmetrical about the line x = 2, then | [{"identifier": "A", "content": "$$f\\left( x \\right) = - f\\left( { - x} \\right)$$ "}, {"identifier": "B", "content": "$$f\\left( {2 + x} \\right) = f\\left( {2 - x} \\right)$$"}, {"identifier": "C", "content": "$$f\\left( x \\right) = f\\left( { - x} \\right)$$ "}, {"identifier": "D", "content": "$$f\\left( {x + ... | ["B"] | null | Let us consider a graph symm. with respect to line $$x=2$$ as shown in the figure.
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263707/exam_images/uiwdp9jmte0l8nkstssm.webp" loading="lazy" alt="AIEEE 2004 Mathematics - Functions Question 131 English Explanation">
<br>... | mcq | aieee-2004 | 6,242 |
7Gxl5AoOoGNHBVk7crwGB | maths | functions | even-and-odd-functions | Let ƒ(x) = a<sup>x</sup>
(a > 0) be written as
<br/>ƒ(x) = ƒ<sub>1</sub>
(x) + ƒ<sub>2</sub>
(x), where ƒ<sub>1</sub>
(x) is an even
function of ƒ<sub>2</sub>
(x) is an odd function. <br/>Then
ƒ<sub>1</sub>
(x + y) + ƒ<sub>1</sub>
(x – y) equals | [{"identifier": "A", "content": "2\u0192<sub>1</sub>\n(x)\u0192<sub>1</sub>\n(y)\n"}, {"identifier": "B", "content": "2\u0192<sub>1</sub>\n(x + y)\u0192<sub>1</sub>\n(x \u2013 y)"}, {"identifier": "C", "content": "2\u0192<sub>1</sub>\n(x)\u0192<sub>2</sub>\n(y)"}, {"identifier": "D", "content": "2\u0192<sub>1</sub>\n(x... | ["A"] | null | f(x) = a<sup>x</sup>
<br><br>As f<sub>1</sub>(x) is even function then
<br><br>f<sub>1</sub>(x) = $${{{f\left( x \right) + f\left( { - x} \right)} \over 2}}$$
<br><br>= $${{{a^x} + {a^{ - x}}} \over 2}$$
<br><br>As f<sub>2</sub>(x) is odd function then
<br><br>f<sub>2</sub>(x) = $${{{f\left( x \right) - f\left( { - x} ... | mcq | jee-main-2019-online-8th-april-evening-slot | 6,243 |
lv3ve421 | maths | functions | even-and-odd-functions | <p>Let $$f(x)=\left\{\begin{array}{ccc}-\mathrm{a} & \text { if } & -\mathrm{a} \leq x \leq 0 \\ x+\mathrm{a} & \text { if } & 0< x \leq \mathrm{a}\end{array}\right.$$ where $$\mathrm{a}> 0$$ and $$\mathrm{g}(x)=(f(|x|)-|f(x)|) / 2$$. Then the function $$g:[-a, a] \rightarrow[-a, a]$$ is</p> | [{"identifier": "A", "content": "neither one-one nor onto.\n"}, {"identifier": "B", "content": "both one-one and onto.\n"}, {"identifier": "C", "content": "one-one.\n"}, {"identifier": "D", "content": "onto"}] | ["A"] | null | <p>$$\begin{aligned}
& f(x)=\left\{\begin{array}{l}
-a \quad \text { if }-a \leq x \leq 0 \\
x+a \quad \text { if } 0< x \leq a
\end{array}\right. \\
& f(|x|)=\left\{\begin{array}{cc}
-a & -a \leq|x| \leq 0 \\
|x|+a & \text { if } 0 < |x| \leq a
\end{array}\right.
\end{aligned}$$</p>
<p>$$|x|<0... | mcq | jee-main-2024-online-8th-april-evening-shift | 6,244 |
8TmUW15AHe2hgz5a | maths | functions | functional-equations | If $$f:R \to R$$ satisfies $$f$$(x + y) = $$f$$(x) + $$f$$(y), for all x, y $$ \in $$ R and $$f$$(1) = 7, then $$\sum\limits_{r = 1}^n {f\left( r \right)} $$ is | [{"identifier": "A", "content": "$${{7n\\left( {n + 1} \\right)} \\over 2}$$"}, {"identifier": "B", "content": "$${{7n} \\over 2}$$"}, {"identifier": "C", "content": "$${{7\\left( {n + 1} \\right)} \\over 2}$$"}, {"identifier": "D", "content": "$$7n + \\left( {n + 1} \\right)$$"}] | ["A"] | null | $$f\left( {x + y} \right) = f\left( x \right) + f\left( y \right).$$
<br><br>Function should be $$f(x)=mx$$
<br><br>$$f\left( 1 \right) = 7;$$
<br><br>$$\therefore$$ $$m=7,$$ $$f\left( x \right) = 7x$$
<br><br>$$\sum\limits_{r = 1}^n {f\left( r \right)} = 7\sum\limits_1^n {r = {{7n\left( {n + 1} \right)} \over 2}} ... | mcq | aieee-2003 | 6,245 |
0SD6FFBfLQleiCCW | maths | functions | functional-equations | A real valued function f(x) satisfies the functional equation
<br/><br/>f(x - y) = f(x)f(y) - f(a - x)f(a + y)
<br/><br/>where a is given constant and f(0) = 1, f(2a - x) is equal to | [{"identifier": "A", "content": "- f(x)"}, {"identifier": "B", "content": "f(x)"}, {"identifier": "C", "content": "f(a) + f(a - x)"}, {"identifier": "D", "content": "f(- x)"}] | ["A"] | null | $$f\left( {2a - x} \right) = f\left( {a - \left( {x - a} \right)} \right)$$
<br><br>$$ = f\left( a \right)f\left( {x - a} \right) - f\left( 0 \right)f\left( x \right)$$
<br><br>$$ = f\left( a \right)f\left( {x - a} \right) - f\left( x \right)$$
<br><br>$$ = - f\left( x \right)$$
<br><br>$$\left[ {} \right.$$ as $$x = ... | mcq | aieee-2005 | 6,246 |
l87cb3gu | maths | functions | functional-equations | If $f(x)+2 f\left(\frac{1}{x}\right)=3 x, x \neq 0$, and $\mathrm{S}=\{x \in \mathbf{R}: f(x)=f(-x)\}$; then $\mathrm{S}:$ | [{"identifier": "A", "content": "is an empty set."}, {"identifier": "B", "content": "contains exactly one element."}, {"identifier": "C", "content": "contains exactly two elements."}, {"identifier": "D", "content": "contains more than two elements."}] | ["C"] | null | We have, $f(x)+2 f\left(\frac{1}{x}\right)=3 x, \quad x \neq 0$
$\ldots$ (i)<br/><br/>
On replacing $x$ by $\frac{1}{x}$ in the above equation, we get<br/><br/>
$$
\begin{aligned}
& f\left(\frac{1}{x}\right)+2 f(x) =\frac{3}{x} \\\\
\Rightarrow & \,\, 2 f(x)+f\left(\frac{1}{x}\right) =\frac{3}{x} \,\,\,\,\,... | mcq | jee-main-2016-offline | 6,247 |
2v2CovVtbLHzzXns | maths | functions | functional-equations | Let $$a$$, b, c $$ \in R$$. If $$f$$(x) = ax<sup>2</sup> + bx + c is such that
<br/>$$a$$ + b + c = 3 and $$f$$(x + y) = $$f$$(x) + $$f$$(y) + xy, $$\forall x,y \in R,$$
<br/><br/>then $$\sum\limits_{n = 1}^{10} {f(n)} $$ is equal to | [{"identifier": "A", "content": "165"}, {"identifier": "B", "content": "190"}, {"identifier": "C", "content": "255"}, {"identifier": "D", "content": "330"}] | ["D"] | null | f(x) = ax<sup>2</sup> + bx + c
<br><br>f(1) = a + b + c = 3 $$ \Rightarrow $$ f (1) = 3
<br><br>Now f(x + y) = f(x) + f(y) + xy ...(1)
<br><br>Put x = y = 1 in eqn (1)
<br><br>f(2) = f(1) + f(1) + 1
<br><br>= 2f(1) + 1
<br><br>$$ \Rightarrow $$ f(2) = 7
<br><br>Similarly f(3) = 12
<br><br>f(4) = 18
<br><br>$$\sum\limit... | mcq | jee-main-2017-offline | 6,248 |
7T8qHanHgmWJe870g3Bhs | maths | functions | functional-equations | If $$f(x) = {\log _e}\left( {{{1 - x} \over {1 + x}}} \right)$$, $$\left| x \right| < 1$$ then $$f\left( {{{2x} \over {1 + {x^2}}}} \right)$$ is equal to | [{"identifier": "A", "content": "2f(x<sup>2</sup>)"}, {"identifier": "B", "content": "2f(x)"}, {"identifier": "C", "content": "(f(x))<sup>2</sup>"}, {"identifier": "D", "content": "-2f(x)"}] | ["B"] | null | Given, $$f(x) = {\log _e}\left( {{{1 - x} \over {1 + x}}} \right)$$
<br><br>$$f\left( {{{2x} \over {1 + {x^2}}}} \right)$$ = $$\ln \left( {{{1 - {{2x} \over {1 + {x^2}}}} \over {1 + {{2x} \over {1 + {x^2}}}}}} \right)$$
<br><br>= $$\ln \left( {{{{x^2} - 2x + 1} \over {{x^2} + 2x + 1}}} \right)$$
<br><br>= $$\ln {\left(... | mcq | jee-main-2019-online-8th-april-morning-slot | 6,249 |
vZDInSndsBCwo3mCTX18hoxe66ijvwp4erx | maths | functions | functional-equations | Let $$\sum\limits_{k = 1}^{10} {f(a + k) = 16\left( {{2^{10}} - 1} \right)} $$ where the function
ƒ satisfies
<br/>ƒ(x + y) = ƒ(x)ƒ(y) for all natural
numbers x, y and ƒ(1) = 2. then the natural
number 'a' is | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "16"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "3"}] | ["D"] | null | Given ƒ(1) = 2
<br><br> and ƒ(x + y) = ƒ(x)ƒ(y)
<br><br>When x = 1 and y = 1 then,
<br><br>ƒ(1 + 1) = ƒ(1)ƒ(1)
<br><br>$$ \Rightarrow $$ f(2) = (f(1))<sup>2</sup> = 2<sup>2</sup>
<br><br>Also when x = 2 and y = 1 then,
<br><br>ƒ(2 + 1) = ƒ(2)ƒ(1)
<br><br>$$ \Rightarrow $$ f(3) = 2<sup>3</sup>
<br><br>$$ \therefore $$ S... | mcq | jee-main-2019-online-9th-april-morning-slot | 6,250 |
oOLSAdvLWP0U3MIENGjgy2xukez67a9z | maths | functions | functional-equations | Let f : R $$ \to $$ R be a function which satisfies
<br/>f(x + y) = f(x) + f(y) $$\forall $$ x, y $$ \in $$ R. If f(1) = 2 and
<br/>g(n) = $$\sum\limits_{k = 1}^{\left( {n - 1} \right)} {f\left( k \right)} $$, n $$ \in $$ N then the value of n, for
which g(n) = 20, is : | [{"identifier": "A", "content": "20"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "4"}] | ["C"] | null | Given f(1) = 2 ;
<br><br>f(x + y) = f(x) + f(y)
<br><br>When x = y = 1 $$ \Rightarrow $$
f(2) = 2 + 2 = 4
<br><br>When x = 2, y = 1 $$ \Rightarrow $$ f(3) = 4 + 2 = 6
<br><br> g(n) = $$\sum\limits_{k = 1}^{\left( {n - 1} \right)} {f\left( k \right)} $$
<br><br>= f(1) + f(2) +.........+ f(n - 1)
<br><br>= 2 + 4 + 6 ... | mcq | jee-main-2020-online-2nd-september-evening-slot | 6,253 |
o26Zk0Ad7pZ3fy2Cu9jgy2xukfuvv0hp | maths | functions | functional-equations | If f(x + y) = f(x)f(y) and $$\sum\limits_{x = 1}^\infty {f\left( x \right)} = 2$$ , x, y $$ \in $$ N, where N is the set of all natural number, then the
value of
$${{f\left( 4 \right)} \over {f\left( 2 \right)}}$$ is : | [{"identifier": "A", "content": "$${2 \\over 3}$$"}, {"identifier": "B", "content": "$${1 \\over 9}$$"}, {"identifier": "C", "content": "$${1 \\over 3}$$"}, {"identifier": "D", "content": "$${4 \\over 9}$$"}] | ["D"] | null | f(x + y) = f(x)f(y)
<br><br>$$\sum\limits_{x = 1}^\infty {f\left( x \right)} = 2$$
<br><br>$$ \Rightarrow $$ f(1) + f(2) + f(3) + ........$$\infty $$ = 2 ....(1)
<br><br>On f(x + y) = f(x) f(y)
<br>* Put x = 1, y = 1
<br>f(2) = (f(1))<sup>2</sup>
<br>* Put x = 2, y = 1
<br>f(3) = f(2). f(1) = f((1))<sup>3</sup>
<br>*... | mcq | jee-main-2020-online-6th-september-morning-slot | 6,254 |
TBP1H0CdJEYh9UShXc1klt7f0ou | maths | functions | functional-equations | A function f(x) is given by $$f(x) = {{{5^x}} \over {{5^x} + 5}}$$, then the sum of the series $$f\left( {{1 \over {20}}} \right) + f\left( {{2 \over {20}}} \right) + f\left( {{3 \over {20}}} \right) + ....... + f\left( {{{39} \over {20}}} \right)$$ is equal to : | [{"identifier": "A", "content": "$${{{39} \\over 2}}$$"}, {"identifier": "B", "content": "$${{{19} \\over 2}}$$"}, {"identifier": "C", "content": "$${{{49} \\over 2}}$$"}, {"identifier": "D", "content": "$${{{29} \\over 2}}$$"}] | ["A"] | null | $$f(x) = {{{5^x}} \over {{5^x} + 5}}$$ ..... (i)<br><br>$$f(2 - x) = {{{5^{2 - x}}} \over {{5^{2 - x}} + 5}}$$<br><br>$$f(2 - x) = {5 \over {{5^x} + 5}}$$ .... (ii)<br><br>Adding equation (i) and (ii) <br><br>$$f(x) + f(2 - x) = 1$$<br><br>$$f\left( {{1 \over {20}}} \right) + f\left( {{{39} \over {20}}} \right) = 1$$<b... | mcq | jee-main-2021-online-25th-february-evening-slot | 6,257 |
GZkr2iQcFKkQTUhoWJ1kmm3zbjp | maths | functions | functional-equations | If f(x) and g(x) are two polynomials such that the polynomial P(x) = f(x<sup>3</sup>) + x g(x<sup>3</sup>) is divisible by x<sup>2</sup> + x + 1, then P(1) is equal to ___________. | [] | null | 0 | Given, p(x) = f(x<sup>3</sup>) + xg(x<sup>3</sup>)<br><br>We know, x<sup>2</sup> + x + 1 = (x $$-$$ $$\omega$$) (x $$-$$ $$\omega$$<sup>2</sup>)<br><br>Given, p(x) is divisible by x<sup>2</sup> + x + 1. So, roots of p(x) is $$\omega$$ and $$\omega$$<sup>2</sup>.<br><br>As root satisfy the equation,<br><br>So, put x = $... | integer | jee-main-2021-online-18th-march-evening-shift | 6,258 |
1ks0d47kv | maths | functions | functional-equations | Let S = {1, 2, 3, 4, 5, 6, 7}. Then the number of possible functions f : S $$\to$$ S <br/>such that f(m . n) = f(m) . f(n) for every m, n $$\in$$ S and m . n $$\in$$ S is equal to _____________. | [] | null | 490 | F(mn) = f(m) . f(n)<br><br>Put m = 1 f(n) = f(1) . f(n) $$\Rightarrow$$ f(1) = 1<br><br>Put m = n = 2<br><br>$$f(4) = f(2).f(2)\left\{ \matrix{
f(2) = 1 \Rightarrow f(4) = 1 \hfill \cr
or \hfill \cr
f(2) = 2 \Rightarrow f(4) = 4 \hfill \cr} \right.$$<br><br>Put m = 2, n = 3<br><br>$$f(6) = f(2).f(3)\left\{ \ma... | integer | jee-main-2021-online-27th-july-morning-shift | 6,260 |
1l546a0kk | maths | functions | functional-equations | <p>Let c, k $$\in$$ R. If $$f(x) = (c + 1){x^2} + (1 - {c^2})x + 2k$$ and $$f(x + y) = f(x) + f(y) - xy$$, for all x, y $$\in$$ R, then the value of $$|2(f(1) + f(2) + f(3) + \,\,......\,\, + \,\,f(20))|$$ is equal to ____________.</p> | [] | null | 3395 | <p>f(x) is polynomial</p>
<p>Put y = 1/x in given functional equation we get</p>
<p>$$f\left( {x + {1 \over x}} \right) = f(x) + f\left( {{1 \over x}} \right) - 1$$</p>
<p>$$ \Rightarrow (c + 1){\left( {x + {1 \over x}} \right)^2} + (1 - {c^2})\left( {x + {1 \over x}} \right) + 2K$$</p>
<p>$$ = (c + 1){x^2} + (1 - {c^2... | integer | jee-main-2022-online-29th-june-morning-shift | 6,262 |
1l5ahps1l | maths | functions | functional-equations | <p>Let f : N $$\to$$ R be a function such that $$f(x + y) = 2f(x)f(y)$$ for natural numbers x and y. If f(1) = 2, then the value of $$\alpha$$ for which</p>
<p>$$\sum\limits_{k = 1}^{10} {f(\alpha + k) = {{512} \over 3}({2^{20}} - 1)} $$</p>
<p>holds, is :</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "6"}] | ["C"] | null | <p>Given,</p>
<p>$$f(x + y) = 2f(x)f(y)$$</p>
<p>and $$f(1) = 2$$</p>
<p>For x = 1 and y = 1,</p>
<p>$$f(1 + 1) = 2f(1)f(1)$$</p>
<p>$$ \Rightarrow f(2) = 2{\left( {f(1)} \right)^2} = 2{(2)^2} = {2^3}$$</p>
<p>For x = 1, y = 2,</p>
<p>$$f(1 + 2) = 2f(1)y(2)$$</p>
<p>$$ \Rightarrow f(3) = 2\,.\,2\,.\,{2^3} = {2^5}$$</p>... | mcq | jee-main-2022-online-25th-june-morning-shift | 6,264 |
1l6jeicx4 | maths | functions | functional-equations | <p> Let $$f(x)=2 x^{2}-x-1$$ and $$\mathrm{S}=\{n \in \mathbb{Z}:|f(n)| \leq 800\}$$. Then, the value of $$\sum\limits_{n \in S} f(n)$$ is equal to ___________.</p> | [] | null | 10620 | <p>$$\because$$ $$\left| {f(n)} \right| \le 800$$</p>
<p>$$ \Rightarrow - 800 \le 2{n^2} - n - 1 \le 800$$</p>
<p>$$ \Rightarrow 2{n^2} - n - 801 \le 0$$</p>
<p>$$\therefore$$ $$n \in \left[ {{{ - \sqrt {6409} + 1} \over 4},{{\sqrt {6409} + 1} \over 4}} \right]$$ and $$n \in z$$</p>
<p>$$\therefore$$ $$n = - 19, - ... | integer | jee-main-2022-online-27th-july-morning-shift | 6,265 |
1l6m6qe90 | maths | functions | functional-equations | <p>For $$\mathrm{p}, \mathrm{q} \in \mathbf{R}$$, consider the real valued function $$f(x)=(x-\mathrm{p})^{2}-\mathrm{q}, x \in \mathbf{R}$$ and $$\mathrm{q}>0$$. Let $$\mathrm{a}_{1}$$, $$\mathrm{a}_{2^{\prime}}$$ $$\mathrm{a}_{3}$$ and $$\mathrm{a}_{4}$$ be in an arithmetic progression with mean $$\mathrm{p}$$ and... | [] | null | 50 | <p>$$\because$$ $${a_1},{a_2},{a_3},{a_4}$$</p>
<p>$$\therefore$$ $${a_2} = p - 3d,\,{a_2} = p - d,\,{a_3} = p + d$$ and $${a_4} = p + 3d$$</p>
<p>Where $$d > 0$$</p>
<p>$$\because$$ $$\left| {f({a_i})} \right| = 500$$</p>
<p>$$ \Rightarrow |9{d^2} - q| = 500$$</p>
<p>and $$|{d^2} - q| = 500$$ ..... (i)</p>
<p>either $... | integer | jee-main-2022-online-28th-july-morning-shift | 6,266 |
1ldo6xo14 | maths | functions | functional-equations | <p>Let $$f:\mathbb{R}-{0,1}\to \mathbb{R}$$ be a function such that $$f(x)+f\left(\frac{1}{1-x}\right)=1+x$$. Then $$f(2)$$ is equal to</p> | [{"identifier": "A", "content": "$$\\frac{9}{4}$$"}, {"identifier": "B", "content": "$$\\frac{7}{4}$$"}, {"identifier": "C", "content": "$$\\frac{7}{3}$$"}, {"identifier": "D", "content": "$$\\frac{9}{2}$$"}] | ["A"] | null | $\begin{aligned} & \mathrm{f}(\mathrm{x})+\mathrm{f}\left(\frac{1}{1-\mathrm{x}}\right)=1+\mathrm{x} \\\\ & \mathrm{x}=2 \Rightarrow \mathrm{f}(2)+\mathrm{f}(-1)=3 ........(1) \\\\ & \mathrm{x}=-1 \Rightarrow \mathrm{f}(-1)+\mathrm{f}\left(\frac{1}{2}\right)=0 .........(2) \\\\ & \mathrm{x}=\frac{1}{2} \Rightarrow \mat... | mcq | jee-main-2023-online-1st-february-evening-shift | 6,268 |
ldo9i6yu | maths | functions | functional-equations | The absolute minimum value, of the function
<br/><br/>$f(x)=\left|x^{2}-x+1\right|+\left[x^{2}-x+1\right]$,
<br/><br/>where $[t]$ denotes the greatest integer function, in the interval $[-1,2]$, is : | [{"identifier": "A", "content": "$\\frac{3}{4}$"}, {"identifier": "B", "content": "$\\frac{3}{2}$"}, {"identifier": "C", "content": "$\\frac{1}{4}$"}, {"identifier": "D", "content": "$\\frac{5}{4}$"}] | ["A"] | null | $\mathrm{f}(\mathrm{x})=\left|\mathrm{x}^{2}-\mathrm{x}+1\right|+\left[\mathrm{x}^{2}-\mathrm{x}+1\right] ; \mathrm{x} \in[-1,2]$
<br/><br/>Let $g(x)=x^{2}-x+1$
<br/><br/>$$
=\left(x-\frac{1}{2}\right)^{2}+\frac{3}{4}
$$
<br/><br/>$$
\because\left|\mathrm{x}^{2}-\mathrm{x}+1\right| \text { and }\left[\mathrm{x}^{2... | mcq | jee-main-2023-online-31st-january-evening-shift | 6,269 |
1ldon97rl | maths | functions | functional-equations | <p>Let $$f(x) = \left| {\matrix{
{1 + {{\sin }^2}x} & {{{\cos }^2}x} & {\sin 2x} \cr
{{{\sin }^2}x} & {1 + {{\cos }^2}x} & {\sin 2x} \cr
{{{\sin }^2}x} & {{{\cos }^2}x} & {1 + \sin 2x} \cr
} } \right|,\,x \in \left[ {{\pi \over 6},{\pi \over 3}} \right]$$. If $$\alpha$$ and $$... | [{"identifier": "A", "content": "$${\\alpha ^2} - {\\beta ^2} = 4\\sqrt 3 $$"}, {"identifier": "B", "content": "$${\\beta ^2} - 2\\sqrt \\alpha = {{19} \\over 4}$$"}, {"identifier": "C", "content": "$${\\beta ^2} + 2\\sqrt \\alpha = {{19} \\over 4}$$"}, {"identifier": "D", "content": "$${\\alpha ^2} + {\\beta ^2} =... | ["B"] | null | $$f(x) = \left| {\matrix{
{1 + {{\sin }^2}x} & {{{\cos }^2}x} & {\sin 2x} \cr
{{{\sin }^2}x} & {1 + {{\cos }^2}x} & {\sin 2x} \cr
{{{\sin }^2}x} & {{{\cos }^2}x} & {1 + \sin 2x} \cr
} } \right|$$
<br/><br/>$C_{1} \rightarrow C_{1}+C_{2}+C_{3}$
<br/><br/>$$
\begin{aligned}
& = (2+\sin 2 x)\left|\begin{... | mcq | jee-main-2023-online-1st-february-morning-shift | 6,270 |
ldqyb0zm | maths | functions | functional-equations | Let $A=\{1,2,3,5,8,9\}$. Then the number of possible functions $f: A \rightarrow A$ such that $f(m \cdot n)=f(m) \cdot f(n)$ for every $m, n \in A$ with $m \cdot n \in A$ is equal to ___________. | [] | null | 432 | <p>$$f(1.n)=f(1).f(n)\Rightarrow f(1)=1$$.</p>
<p>$$f(3.3)=(f(3))^2$$</p>
<p>Hence, the possibilities for $$(t(3),(9))$$ are $$(1,1)$$ and $$(3,9)$$.</p>
<p>Other three i.e. $$f(2),f(5),f(8)$$</p>
<p>Can be chosen in 6$$^3$$ ways.</p>
<p>Hence, total number of functions</p>
<p>$$6^3\times2=432$$</p> | integer | jee-main-2023-online-30th-january-evening-shift | 6,271 |
1ldsf1d3h | maths | functions | functional-equations | <p>Consider a function $$f:\mathbb{N}\to\mathbb{R}$$, satisfying $$f(1)+2f(2)+3f(3)+....+xf(x)=x(x+1)f(x);x\ge2$$ with $$f(1)=1$$. Then $$\frac{1}{f(2022)}+\frac{1}{f(2028)}$$ is equal to</p> | [{"identifier": "A", "content": "8000"}, {"identifier": "B", "content": "8400"}, {"identifier": "C", "content": "8100"}, {"identifier": "D", "content": "8200"}] | ["C"] | null | <p>$$f(1) + 2f(2) + 3f(3)\, + \,...\, + \,nf(n) = n(n + 1) + (n)$$ ..... (i)</p>
<p>$$n \to n + 1$$</p>
<p>$$f(1) + 2f(2)\, + \,...\, + \,(n + 1)f(n + 1) = (n + 1)(n + 2)f(n + 1)$$ ...... (ii)</p>
<p>(i) and (ii) gives</p>
<p>$$3f(3) - 2f(2) = 0$$</p>
<p>$$4f(4) - 3f(3) = 0$$</p>
<p>$$ \vdots $$</p>
<p>$$(n + 1)f(n + 1... | mcq | jee-main-2023-online-29th-january-evening-shift | 6,272 |
1ldswo1xf | maths | functions | functional-equations | <p>Suppose $$f$$ is a function satisfying $$f(x + y) = f(x) + f(y)$$ for all $$x,y \in N$$ and $$f(1) = {1 \over 5}$$. If $$\sum\limits_{n = 1}^m {{{f(n)} \over {n(n + 1)(n + 2)}} = {1 \over {12}}} $$, then $$m$$ is equal to __________.</p> | [] | null | 10 | $\because f(1)=\frac{1}{5} ~\therefore f(2)=f(1)+f(1)=\frac{2}{5}$
<br/><br/>
$f(2)=\frac{2}{5} \quad\quad f(3)=f(2)+f(1)=\frac{3}{5}$
<br/><br/>
$f(3)=\frac{3}{5}$
<br/><br/>
$\therefore \sum\limits_{n=1}^{m} \frac{f(n)}{n(n+1)(n+2)}$
<br/><br/>
$=\frac{1}{5} \sum\limits_{n=1}^{m}\left(\frac{1}{n+1}-\frac{1}{n+2}\righ... | integer | jee-main-2023-online-29th-january-morning-shift | 6,273 |
1ldu5gt5g | maths | functions | functional-equations | <p>Let $$f:\mathbb{R}\to\mathbb{R}$$ be a function defined by $$f(x) = {\log _{\sqrt m }}\{ \sqrt 2 (\sin x - \cos x) + m - 2\} $$, for some $$m$$, such that the range of $$f$$ is [0, 2]. Then the value of $$m$$ is _________</p> | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "2"}] | ["C"] | null | We know that $\sin x-\cos x \in[-\sqrt{2}, \sqrt{2}]$
<br/><br/>
$$
\begin{aligned}
& \log _{\sqrt{M}}(\sqrt{2}(\sin x-\cos ) +M-2) \\\\
&\quad\quad\in {\left[\log _{\sqrt{M}}(M-4), \log _{\sqrt{M}} M\right] }
\end{aligned}
$$
<br/><br/>
$\Rightarrow \log _{\sqrt{M}}(M-4)=0 \Rightarrow M=5$
| mcq | jee-main-2023-online-25th-january-evening-shift | 6,274 |
1ldu5xulj | maths | functions | functional-equations | <p>Let $$f(x) = 2{x^n} + \lambda ,\lambda \in R,n \in N$$, and $$f(4) = 133,f(5) = 255$$. Then the sum of all the positive integer divisors of $$(f(3) - f(2))$$ is</p> | [{"identifier": "A", "content": "60"}, {"identifier": "B", "content": "58"}, {"identifier": "C", "content": "61"}, {"identifier": "D", "content": "59"}] | ["A"] | null | $f(x)=2 x^{n}+\lambda, \lambda \in \mathbb{R}, n \in \mathbb{N}$
<br/><br/>
$f(4)=2 \cdot 4^{n}+\lambda=133, f(5)=2 \cdot 5^{n}+\lambda=255$
<br/><br/>
$f(5)-f(4)=2 \cdot\left(5^{n} \cdot 4^{n}\right)=122 \Rightarrow n=3$
<br/><br/>
$\Rightarrow f(3)-f(2)=2 \cdot\left(3^{n} \cdot 2^{n}\right)=2 \cdot\left(3^{3}-2^{3}\r... | mcq | jee-main-2023-online-25th-january-evening-shift | 6,275 |
1ldwvz91z | maths | functions | functional-equations | <p>Let $$f(x)$$ be a function such that $$f(x+y)=f(x).f(y)$$ for all $$x,y\in \mathbb{N}$$. If $$f(1)=3$$ and $$\sum\limits_{k = 1}^n {f(k) = 3279} $$, then the value of n is</p> | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "8"}] | ["B"] | null | $$
\begin{aligned}
& \mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x}) \cdot \mathrm{f}(\mathrm{y}) \forall \mathrm{x}, \mathrm{y} \in \mathrm{N}, \mathrm{f}(1)=3 \\\\
& \mathrm{f}(2)=\mathrm{f}^2(1)=3^2 \\\\
& \mathrm{f}(3)=\mathrm{f}(1) \mathrm{f}(2)=3^3 \\\\
& \mathrm{f}(4)=3^4 \\\\
& \mathrm{f}(\mathrm{k})=3... | mcq | jee-main-2023-online-24th-january-evening-shift | 6,276 |
1lgxszp3h | maths | functions | functional-equations | <p>If $$f(x) = {{(\tan 1^\circ )x + {{\log }_e}(123)} \over {x{{\log }_e}(1234) - (\tan 1^\circ )}},x > 0$$, then the least value of $$f(f(x)) + f\left( {f\left( {{4 \over x}} \right)} \right)$$ is :</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "8"}] | ["B"] | null | Given that $f(x)=\frac{\left(\tan 1^{\circ}\right) x+\log _e(123)}{x \log _e(1234)-\left(\tan 1^{\circ}\right)}$
<br/><br/>Let us consider a similar function of $(x)$,
<br/><br/>$\therefore f(x)=\frac{A x+B}{C x-A}$
<br/><br/>$\text { Now, } $
<br/><br/>$$
\begin{aligned}
&f(f(x)) =\frac{A\left(\frac{A x+B}{C x-A}\rig... | mcq | jee-main-2023-online-10th-april-morning-shift | 6,278 |
luy9clgk | maths | functions | functional-equations | <p>If a function $$f$$ satisfies $$f(\mathrm{~m}+\mathrm{n})=f(\mathrm{~m})+f(\mathrm{n})$$ for all $$\mathrm{m}, \mathrm{n} \in \mathbf{N}$$ and $$f(1)=1$$, then the largest natural number $$\lambda$$ such that $$\sum_\limits{\mathrm{k}=1}^{2022} f(\lambda+\mathrm{k}) \leq(2022)^2$$ is equal to _________.</p> | [] | null | 1010 | <p>$$\begin{aligned}
& f(m+n)=f(m)+f(n) \\
& f(x)=k x \\
& \because f(1)=1 \\
& \Rightarrow k=1 \\
& \Rightarrow f(x)=x
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \sum_{k=1}^{2022} f(\lambda+k)=\sum_{k=1}^{2022}(\lambda+k)=\underbrace{\lambda+\lambda+\ldots+\lambda}_{2022}+(1+2+\ldots+2022) \\
& =2022 \lambda+\frac{202... | integer | jee-main-2024-online-9th-april-morning-shift | 6,279 |
lv7v47tt | maths | functions | functional-equations | <p>If $$S=\{a \in \mathbf{R}:|2 a-1|=3[a]+2\{a \}\}$$, where $$[t]$$ denotes the greatest integer less than or equal to $$t$$ and $$\{t\}$$ represents the fractional part of $$t$$, then $$72 \sum_\limits{a \in S} a$$ is equal to _________.</p> | [] | null | 18 | <p>$$\begin{aligned}
& S:\{a \in R:|2 a-1|=3[a]+2\{a\}\} \\
& |2 a-1|=3[a]+2(a-[a]) \\
& |2 a-1|=[a]+2 a
\end{aligned}$$</p>
<p>Case I: If $$0 < a < \frac{1}{2}$$</p>
<p>$$\begin{aligned}
& 1-2 a=0+2 a \\
& \Rightarrow a=\frac{1}{4}
\end{aligned}$$</p>
<p>Case II: If $$\frac{1}{2} < a < 1$$</p>
<p>$$2 a-1=0+2 a$$</p>
<... | integer | jee-main-2024-online-5th-april-morning-shift | 6,280 |
S9ViRTELMsFpCdXCyu7k9k2k5gzd75n | maths | functions | inverse-functions | The inverse function of
<br/><br/>f(x) = $${{{8^{2x}} - {8^{ - 2x}}} \over {{8^{2x}} + {8^{ - 2x}}}}$$, x $$ \in $$ (-1, 1), is : | [{"identifier": "A", "content": "$${1 \\over 4}{\\log _e}\\left( {{{1 - x} \\over {1 + x}}} \\right)$$"}, {"identifier": "B", "content": "$${1 \\over 4}\\left( {{{\\log }_8}e} \\right){\\log _e}\\left( {{{1 - x} \\over {1 + x}}} \\right)$$"}, {"identifier": "C", "content": "$${1 \\over 4}\\left( {{{\\log }_8}e} \\right... | ["C"] | null | f(x) = $${{{8^{2x}} - {8^{ - 2x}}} \over {{8^{2x}} + {8^{ - 2x}}}}$$ = y
<br><br>$$ \therefore $$ $${{y + 1} \over {y - 1}} = {{{{2.8}^{2x}}} \over { - {{2.8}^{ - 2x}}}}$$
<br><br>$$ \Rightarrow $$ $${{1 + y} \over {1 - y}}$$ = 8<sup>4x</sup>
<br><br>$$ \Rightarrow $$ $${\log _e}\left( {{{1 + y} \over {1 - y}}} \right)... | mcq | jee-main-2020-online-8th-january-morning-slot | 6,282 |
3KiVygqDw488GbnoRq1kmjaxjig | maths | functions | inverse-functions | The inverse of $$y = {5^{\log x}}$$ is : | [{"identifier": "A", "content": "$$x = {5^{\\log y}}$$"}, {"identifier": "B", "content": "$$x = {y^{{1 \\over {\\log 5}}}}$$"}, {"identifier": "C", "content": "$$x = {5^{{1 \\over {\\log y}}}}$$"}, {"identifier": "D", "content": "$$x = {y^{\\log 5}}$$"}] | ["B"] | null | $$y = {5^{\log x}}$$<br><br>$$ \Rightarrow \log y = \log x.log5$$<br><br>$$ \Rightarrow \log x = {{\log y} \over {\log 5}} = {\log _5}y$$<br><br>$$ \Rightarrow x = {e^{{{\log }_5}y}}$$<br><br>$$ \Rightarrow x = {y^{{{\log }_5}e}}$$<br><br>$$ \Rightarrow x = {y^{{1 \over {\log 5}}}}$$ | mcq | jee-main-2021-online-17th-march-morning-shift | 6,283 |
LQM2NmZMuUOeUUPSBE1kmm3nlbt | maths | functions | inverse-functions | Let f : R $$-$$ {3} $$ \to $$ R $$-$$ {1} be defined by f(x) = $${{x - 2} \over {x - 3}}$$.<br/><br/>Let g : R $$ \to $$ R be given as g(x) = 2x $$-$$ 3. Then, the sum of all the values of x for which f<sup>$$-$$1</sup>(x) + g<sup>$$-$$1</sup>(x) = $${{13} \over 2}$$ is equal to : | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "7"}] | ["B"] | null | Finding inverse of f(x)<br><br>$$y = {{x - 2} \over {x - 3}} \Rightarrow xy - 3y = x - 2 \Rightarrow x(y - 1) = 3y - 2$$<br><br>$$ \therefore $$ $${f^{ - 1}}(x) = {{3x - 2} \over {x - 1}}$$<br><br>Similarly for $${g^{ - 1}}(x)$$<br><br>$$y = 2x - 3 \Rightarrow x = {{y + 3} \over 2} \Rightarrow {g^{ - 1}}(x) = {{x + 3} ... | mcq | jee-main-2021-online-18th-march-evening-shift | 6,284 |
1lgoxs5ux | maths | functions | inverse-functions | <p>The range of $$f(x)=4 \sin ^{-1}\left(\frac{x^{2}}{x^{2}+1}\right)$$ is</p> | [{"identifier": "A", "content": "$$[0,2 \\pi]$$"}, {"identifier": "B", "content": "$$[0,2 \\pi)$$"}, {"identifier": "C", "content": "$$[0, \\pi)$$"}, {"identifier": "D", "content": "$$[0, \\pi]$$"}] | ["B"] | null | $$
\begin{aligned}
& \frac{x^2}{1+x^2}=1-\frac{1}{1+x^2}<1 \\\\
\therefore & 0 \leq \frac{x^2}{1+x^2}<1 \\\\
\Rightarrow & 0 \leq \sin ^{-1}\left(\frac{x^2}{1+x^2}\right)<\frac{\pi}{2} \\\\
\Rightarrow & 0 \leq 4 \sin ^{-1}\left(\frac{x^2}{1+x^2}\right)<2 \pi
\end{aligned}
$$ | mcq | jee-main-2023-online-13th-april-evening-shift | 6,285 |
bIuxRNjZnQV19sc9 | maths | functions | periodic-functions | The period of $${\sin ^2}\theta $$ is | [{"identifier": "A", "content": "$${\\pi ^2}$$ "}, {"identifier": "B", "content": "$$\\pi $$ "}, {"identifier": "C", "content": "$$2\\pi $$ "}, {"identifier": "D", "content": "$$\\pi /2$$ "}] | ["B"] | null | The period of $${\sin ^2}\theta $$ is = $$\pi $$
<br><br><b>Note :</b>
<br>(1) When $$n$$ is odd then the period of $${\sin ^n}\theta $$, $${\cos ^n}\theta $$, $${\csc ^n}\theta $$, $${\sec ^n}\theta $$ = $$2\pi $$
<br><br>(2) When $$n$$ is even then the period of $${\sin ^n}\theta $$, $${\cos ^n}\theta $$, $${\csc ^n}... | mcq | aieee-2002 | 6,286 |
twJaTXmPzxEKgG6b | maths | functions | range | The range of the function f(x) = $${}^{7 - x}{P_{x - 3}}$$ is | [{"identifier": "A", "content": "{1, 2, 3, 4, 5}"}, {"identifier": "B", "content": "{1, 2, 3, 4, 5, 6}"}, {"identifier": "C", "content": "{1, 2, 3, 4}"}, {"identifier": "D", "content": "{1, 2, 3}"}] | ["D"] | null | The range of the function $f(x) = {}^{7-x}P_{x-3}$ can be found by considering the possible values of $f(x)$ as $x$ varies over its domain.
<br/><br/>
The domain of $f(x)$ is the set of all real numbers such that
<br/><br/>(i) $x \geq 3$ (since the permutation function is only defined for non-negative integers)
<br/>... | mcq | aieee-2004 | 6,288 |
m5lbIUG3trWiugfMUNHOx | maths | functions | range | Let f : R $$ \to $$ R be defined by f(x) = $${x \over {1 + {x^2}}},x \in R$$. Then the range of f is : | [{"identifier": "A", "content": "$$\\left[ { - {1 \\over 2},{1 \\over 2}} \\right]$$"}, {"identifier": "B", "content": "$$R - \\left[ { - {1 \\over 2},{1 \\over 2}} \\right]$$"}, {"identifier": "C", "content": "($$-$$ 1, 1) $$-$$ {0}"}, {"identifier": "D", "content": "R $$-$$ [$$-$$1, 1]"}] | ["A"] | null | f(0) = 0 & f(x) is odd
<br><br>Further, if x > 0 then
<br><br>f(x) = $$f(x) = {1 \over {x + {1 \over x}}} \in \left( {0,{1 \over 2}} \right]$$
<br><br>Hence, $$f(x) \in \left[ { - {1 \over 2},{1 \over 2}} \right]$$ | mcq | jee-main-2019-online-11th-january-morning-slot | 6,289 |
ohIDlBis3k5AggYqUF7k9k2k5hjmi7l | maths | functions | range | Let ƒ : (1, 3) $$ \to $$ R be a function defined by<br/>
$$f(x) = {{x\left[ x \right]} \over {1 + {x^2}}}$$ , where [x] denotes the greatest
integer $$ \le $$ x. Then the range of ƒ is | [{"identifier": "A", "content": "$$\\left( {{2 \\over 5},{1 \\over 2}} \\right) \\cup \\left( {{3 \\over 4},{4 \\over 5}} \\right]$$"}, {"identifier": "B", "content": "$$\\left( {{3 \\over 5},{4 \\over 5}} \\right)$$"}, {"identifier": "C", "content": "$$\\left( {{2 \\over 5},{4 \\over 5}} \\right]$$"}, {"identifier": "... | ["A"] | null | f(x) = $$\left\{ {\matrix{
{{x \over {{x^2} + 1}},} & {1 < x < 2} \cr
{{{2x} \over {{x^2} + 1}},} & {2 \le x < 3} \cr
} } \right.$$
<br><br>$$ \therefore $$ f(x) is decreasing function
<br><br>$$ \therefore $$ Range is $$\left( {{2 \over 5},{1 \over 2}} \right) \cup \left( {{3 \over 4},{4 ... | mcq | jee-main-2020-online-8th-january-evening-slot | 6,290 |
17P9YSWU4u46PPcqJx1kmhxmh7h | maths | functions | range | The range of a$$\in$$R for which the <br/><br/>function f(x) = (4a $$-$$ 3)(x + log<sub>e</sub> 5) + 2(a $$-$$ 7) cot$$\left( {{x \over 2}} \right)$$ sin<sup>2</sup>$$\left( {{x \over 2}} \right)$$, x $$\ne$$ 2n$$\pi$$, n$$\in$$N has critical points, is : | [{"identifier": "A", "content": "[1, $$\\infty $$)"}, {"identifier": "B", "content": "($$-$$3, 1)"}, {"identifier": "C", "content": "$$\\left[ { - {4 \\over 3},2} \\right]$$"}, {"identifier": "D", "content": "($$-$$$$\\infty $$, $$-$$1]"}] | ["C"] | null | $$f(x) = (4a - 3)(x + \ln 5) + 2(a - 7)\left( {{{\cos {x \over 2}} \over {\sin {x \over 2}}}.{{\sin }^2}{x \over 2}} \right)$$<br><br>$$f(x) = (4a - 3)(x + \ln 5) + (a - 7)\sin x$$<br><br>$$f'(x) = (4a - 3) + (a - 7)\cos x = 0$$<br><br>$$\cos x = {{ - (4a - 3)} \over {a - 7}}$$<br><br>$$ - 1 \le - {{4a - 3} \over {a -... | mcq | jee-main-2021-online-16th-march-morning-shift | 6,291 |
1kto99h4t | maths | functions | range | The range of the function, <br/><br/>$$f(x) = {\log _{\sqrt 5 }}\left( {3 + \cos \left( {{{3\pi } \over 4} + x} \right) + \cos \left( {{\pi \over 4} + x} \right) + \cos \left( {{\pi \over 4} - x} \right) - \cos \left( {{{3\pi } \over 4} - x} \right)} \right)$$ is : | [{"identifier": "A", "content": "$$\\left( {0,\\sqrt 5 } \\right)$$"}, {"identifier": "B", "content": "[$$-$$2, 2]"}, {"identifier": "C", "content": "$$\\left[ {{1 \\over {\\sqrt 5 }},\\sqrt 5 } \\right]$$"}, {"identifier": "D", "content": "[0, 2]"}] | ["D"] | null | $$f(x) = {\log _{\sqrt 5 }}\left( {3 + \cos \left( {{{3\pi } \over 4} + x} \right) + \cos \left( {{\pi \over 4} + x} \right) + \cos \left( {{\pi \over 4} - x} \right) - \cos \left( {{{3\pi } \over 4} - x} \right)} \right)$$<br><br>$$f(x) = {\log _{\sqrt 5 }}\left[ {3 + 2\cos \left( {{\pi \over 4}} \right)\cos (x) - ... | mcq | jee-main-2021-online-1st-september-evening-shift | 6,292 |
ldo7i3kx | maths | functions | range | Let $f: \mathbb{R}-\{2,6\} \rightarrow \mathbb{R}$ be real valued function<br/><br/> defined as $f(x)=\frac{x^2+2 x+1}{x^2-8 x+12}$.
<br/><br/>Then range of $f$ is | [{"identifier": "A", "content": "$ \\left(-\\infty,-\\frac{21}{4}\\right] \\cup[1, \\infty) $"}, {"identifier": "B", "content": "$\\left(-\\infty,-\\frac{21}{4}\\right) \\cup(0, \\infty) $"}, {"identifier": "C", "content": "$\\left(-\\infty,-\\frac{21}{4}\\right] \\cup[0, \\infty) $"}, {"identifier": "D", "content": "$... | ["C"] | null | $y=\frac{x^{2}+2 x+1}{x^{2}-8 x+12}$
<br/><br/>$\Rightarrow(y-1) x^{2}-(8 y+2) x+12 y-1=0$
<br/><br/>Let $y \neq 1$, then $D \geq 0$
<br/><br/>$$
4(4 y+1)^{2}-4(y-1)(12 y-1) \geq 0
$$
<br/><br/>$\Rightarrow 16 y^{2}+1+8 y-\left(12 y^{2}-13 y+1\right) \geq 0$
<br/><br/>$\Rightarrow 4 y^{2}+21 y \geq 0$
<br/><br/>$... | mcq | jee-main-2023-online-31st-january-evening-shift | 6,293 |
1ldprz63h | maths | functions | range | If the domain of the function $$f(x)=\frac{[x]}{1+x^{2}}$$, where $$[x]$$ is greatest integer $$\leq x$$, is $$[2,6)$$, then its range is | [{"identifier": "A", "content": "$$\\left(\\frac{5}{37}, \\frac{2}{5}\\right]-\\left\\{\\frac{9}{29}, \\frac{27}{109}, \\frac{18}{89}, \\frac{9}{53}\\right\\}$$"}, {"identifier": "B", "content": "$$\\left(\\frac{5}{37}, \\frac{2}{5}\\right]$$"}, {"identifier": "C", "content": "$$\\left(\\frac{5}{26}, \\frac{2}{5}\\righ... | ["B"] | null | $f(x)=\frac{k}{1+x^{2}}$ is a decreasing function where $k>0$
<br/><br/>$$
\begin{gathered}
\therefore \quad x \in[2,3) \Rightarrow f(x)=\frac{2}{1+x^{2}} \in\left(\frac{2}{10}, \frac{2}{5}\right]=R_{1} \\\\
x \in[3,4) \Rightarrow f(x)=\frac{3}{1+x^{2}} \in\left(\frac{3}{17}, \frac{3}{10}\right]=R_{2} \\\\
x \in[4,5) ... | mcq | jee-main-2023-online-31st-january-morning-shift | 6,294 |
ldqy1wog | maths | functions | range | The range of the function $f(x)=\sqrt{3-x}+\sqrt{2+x}$ is : | [{"identifier": "A", "content": "$[2 \\sqrt{2}, \\sqrt{11}]$"}, {"identifier": "B", "content": "$[\\sqrt{5}, \\sqrt{13}]$"}, {"identifier": "C", "content": "$[\\sqrt{2}, \\sqrt{7}]$"}, {"identifier": "D", "content": "$[\\sqrt{5}, \\sqrt{10}]$"}] | ["D"] | null | <p>$$f(x) = \sqrt {3 - x} + \sqrt {x + 2} $$</p>
<p>$$y' = {{ - 1} \over {2\sqrt 3 - x}} + {1 \over {2\sqrt {x + 2} }} = 0$$</p>
<p>$$ \Rightarrow \sqrt x + 2 = \sqrt 3 - x$$</p>
<p>$$ \Rightarrow x = {1 \over 2}$$</p>
<p>$$y\left( {{1 \over 2}} \right) = \sqrt {{5 \over 2}} + \sqrt {{5 \over 2}} = \sqrt {10} $$<... | mcq | jee-main-2023-online-30th-january-evening-shift | 6,295 |
1lh2ylef4 | maths | functions | range | <p>Let the sets A and B denote the domain and range respectively of the function $$f(x)=\frac{1}{\sqrt{\lceil x\rceil-x}}$$, where $$\lceil x\rceil$$ denotes the smallest integer greater than or equal to $$x$$. Then among the statements</p>
<p>(S1) : $$A \cap B=(1, \infty)-\mathbb{N}$$ and</p>
<p>(S2) : $$A \cup B=(1, ... | [{"identifier": "A", "content": "only $$(\\mathrm{S} 2)$$ is true"}, {"identifier": "B", "content": "only (S1) is true"}, {"identifier": "C", "content": "neither (S1) nor (S2) is true"}, {"identifier": "D", "content": "both (S1) and (S2) are true"}] | ["B"] | null | $$
f(x)=\frac{1}{\sqrt{\lceil x\rceil-x}}
$$
<br/><br/>If $\mathrm{x} \in \mathrm{I},\lceil\mathrm{x}\rceil=[\mathrm{x}]$ (greatest integer function)
<br/><br/>If $x \notin I,\lceil x\rceil=[x]+1$
<br/><br/>$$
\begin{aligned}
& \Rightarrow \mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}
\frac{1}{\sqrt{[\mathrm{x}]-\math... | mcq | jee-main-2023-online-6th-april-evening-shift | 6,296 |
luxwcbm5 | maths | functions | range | <p>Let the range of the function $$f(x)=\frac{1}{2+\sin 3 x+\cos 3 x}, x \in \mathbb{R}$$ be $$[a, b]$$. If $$\alpha$$ and $$\beta$$ ar respectively the A.M. and the G.M. of $$a$$ and $$b$$, then $$\frac{\alpha}{\beta}$$ is equal to</p> | [{"identifier": "A", "content": "$$\\pi$$\n"}, {"identifier": "B", "content": "$$\\sqrt{\\pi}$$\n"}, {"identifier": "C", "content": "$$\\sqrt{2}$$"}, {"identifier": "D", "content": "2"}] | ["C"] | null | <p>$$\begin{aligned}
& F(x)=\frac{1}{2+\sin 3 x+\cos 3 x}, x \in \mathbb{R} \\
& \sin 3 x+\cos 3 x \in[-\sqrt{2}, \sqrt{2}] \\
& 2+\sin 3 x+\cos 3 x \in[2-\sqrt{2}, 2+\sqrt{2}] \\
& \Rightarrow \frac{1}{2+\sin 3 x+\cos 3 x} \in\left[\frac{1}{2+\sqrt{2}}, \frac{1}{2-\sqrt{2}}\right] \\
& \Rightarrow a=\frac{1}{2+\sqrt{2... | mcq | jee-main-2024-online-9th-april-evening-shift | 6,297 |
lv5gt1wc | maths | functions | range | <p>If the range of $$f(\theta)=\frac{\sin ^4 \theta+3 \cos ^2 \theta}{\sin ^4 \theta+\cos ^2 \theta}, \theta \in \mathbb{R}$$ is $$[\alpha, \beta]$$, then the sum of the infinite G.P., whose first term is 64 and the common ratio is $$\frac{\alpha}{\beta}$$, is equal to __________.</p> | [] | null | 96 | <p>To determine the range of the function $$f(\theta)=\frac{\sin ^4 \theta+3 \cos ^2 \theta}{\sin ^4 \theta+\cos ^2 \theta}$$, let's start by simplifying the expression. Let $$\sin^2 \theta = x$$, so $$\cos^2 \theta = 1 - x$$. The function then transforms into:</p>
<p>$$ f(x) = \frac{x^2 + 3(1-x)}{x^2 + (1-x)} $$</p>
... | integer | jee-main-2024-online-8th-april-morning-shift | 6,298 |
lvb29471 | maths | functions | range | <p>If the function $$f(x)=\left(\frac{1}{x}\right)^{2 x} ; x>0$$ attains the maximum value at $$x=\frac{1}{\mathrm{e}}$$ then :</p> | [{"identifier": "A", "content": "$$\\mathrm{e}^\\pi<\\pi^{\\mathrm{e}}$$\n"}, {"identifier": "B", "content": "$$\\mathrm{e}^{2 \\pi}<(2 \\pi)^{\\mathrm{e}}$$\n"}, {"identifier": "C", "content": "$$(2 e)^\\pi>\\pi^{(2 e)}$$\n"}, {"identifier": "D", "content": "$$\\mathrm{e}^\\pi>\\pi^{\\mathrm{e}}$$"}] | ["D"] | null | <p>$$f\left(\frac{1}{\pi}\right)< f\left(\frac{1}{e}\right) \quad \text { as } \frac{1}{\pi}<\frac{1}{e}$$</p>
<p>$$\begin{aligned}
& \Rightarrow\left(\frac{1}{1}\right)^{\frac{2}{\pi}}<\left(\frac{1}{\frac{1}{e}}\right)^{\frac{2}{e}} \\
& \Rightarrow(\pi)^{\frac{2}{\pi}}<(e)^{\frac{2}{e}} \\
& \Rightarrow \pi^e < e^\p... | mcq | jee-main-2024-online-6th-april-evening-shift | 6,299 |
lvb294cr | maths | functions | range | <p>Let $$f(x)=\frac{1}{7-\sin 5 x}$$ be a function defined on $$\mathbf{R}$$. Then the range of the function $$f(x)$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\left[\\frac{1}{8}, \\frac{1}{5}\\right]$$\n"}, {"identifier": "B", "content": "$$\\left[\\frac{1}{7}, \\frac{1}{6}\\right]$$\n"}, {"identifier": "C", "content": "$$\\left[\\frac{1}{7}, \\frac{1}{5}\\right]$$\n"}, {"identifier": "D", "content": "$$\\left[\\frac{1}{8}, \\frac{1}{6}\\r... | ["D"] | null | <p>$$\begin{aligned}
& f(x)=\frac{1}{7-\sin 5 x} \\\\
& -1 \leq \sin 5 x \leq 1 \\\\
& -1 \leq-\sin 5 x \leq 1 \\\\
& -1+7 \leq 7-\sin 5 x \leq 1+7 \\\\
& 6 \leq 7-\sin 5 x \leq 8 \\\\
& \frac{1}{8} \leq \frac{1}{7-\sin 5 x} \leq \frac{1}{6} \\\\
& \frac{1}{8} \leq f(x) \leq \frac{1}{6} \\\\
& \text { Range }=\left[\fr... | mcq | jee-main-2024-online-6th-april-evening-shift | 6,300 |
tWejUYNwkeIGCAA9 | maths | height-and-distance | height-and-distance | A tower stands at the centre of a circular park. $$A$$ and $$B$$ are two points on the boundary of the park such that $$AB(=a)$$ subtends an angle of $${60^ \circ }$$ at the foot of the tower, and the angle of elevation of the top of the tower from $$A$$ or $$B$$ is $${30^ \circ }$$. The height of the tower is : | [{"identifier": "A", "content": "$$a/\\sqrt 3 $$ "}, {"identifier": "B", "content": "$$a\\sqrt 3 $$"}, {"identifier": "C", "content": "$$2a/\\sqrt 3 $$ "}, {"identifier": "D", "content": "$$2a\\sqrt 3 $$"}] | ["A"] | null | In the $$\Delta AOB,\,\,\angle AOB = {60^ \circ },$$ and
<br><br>$$\angle OBA = \angle OAB$$
<br><br>(since $$OA=OB=AB$$ radius of same circle).
<br><br>$$\therefore$$ $$\Delta AOB$$ is a equilateral triangle.
<br><br>Let the height of tower is $$h$$
<br><br><img class="question-image" src="https://res.cloudinary.... | mcq | aieee-2007 | 6,302 |
GR58oSFYu58GzaCR | maths | height-and-distance | height-and-distance | $$AB$$ is a vertical pole with $$B$$ at the ground level and $$A$$ at the top. $$A$$ man finds that the angle of elevation of the point $$A$$ from a certain point $$C$$ on the ground is $${60^ \circ }$$. He moves away from the pole along the line $$BC$$ to a point $$D$$ such that $$CD=7$$ m. From $$D$$ the angle of ele... | [{"identifier": "A", "content": "$${{7\\sqrt 3 } \\over 2} {1 \\over {\\sqrt {3 - 1} }}m$$ "}, {"identifier": "B", "content": "$${{7\\sqrt 3 } \\over 2}\\left( {\\sqrt {3 } + 1 } \\right)m$$ "}, {"identifier": "C", "content": "$${{7\\sqrt 3 } \\over 2}\\left( {\\sqrt {3 } - 1 } \\right)m$$"}, {"identifier": "D", "cont... | ["B"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266701/exam_images/izrw5r6ggvq1o2nql6cu.webp" loading="lazy" alt="AIEEE 2008 Mathematics - Height and Distance Question 38 English Explanation">
<br><br>In $$\Delta ABC$$
<br><br>$${h \over x} = \tan {60^ \circ } = \sqrt 3 $$
<br><b... | mcq | aieee-2008 | 6,303 |
XLrrI8GbRHPBpVqu | maths | height-and-distance | height-and-distance | $$ABCD$$ is a trapezium such that $$AB$$ and $$CD$$ are parallel and $$BC \bot CD.$$ If $$\angle ADB = \theta ,\,BC = p$$ and $$CD = q,$$ then AB is equal to: | [{"identifier": "A", "content": "$${{\\left( {{p^2} + {q^2}} \\right)\\sin \\theta } \\over {p\\cos \\theta + q\\sin \\theta }}$$ "}, {"identifier": "B", "content": "$${{{p^2} + {q^2}\\cos \\theta } \\over {p\\cos \\theta + q\\sin \\theta }}$$ "}, {"identifier": "C", "content": "$${{{p^2} + {q^2}} \\over {{p^2}\\cos ... | ["A"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266634/exam_images/yvyztb8wfjakqr2ttuvt.webp" loading="lazy" alt="JEE Main 2013 (Offline) Mathematics - Height and Distance Question 41 English Explanation">
<br><br>From Sine Rule
<br><br>$${{AB} \over {\sin \theta }} = {{\sqrt {{... | mcq | jee-main-2013-offline | 6,304 |
xU2eo9pTfAzBTHRg | maths | height-and-distance | height-and-distance | If the angles of elevation of the top of a tower from three collinear points $$A, B$$ and $$C,$$ on a line leading to the foot of the tower, are $${30^ \circ }$$, $${45^ \circ }$$ and $${60^ \circ }$$ respectively, then the ratio, $$AB:BC,$$ is : | [{"identifier": "A", "content": "$$1:\\sqrt 3 $$ "}, {"identifier": "B", "content": "$$2:3$$"}, {"identifier": "C", "content": "$$\\sqrt 3 :1$$ "}, {"identifier": "D", "content": "$$\\sqrt 3 :\\sqrt 2 $$ "}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l91njbgq/c332d856-caa2-46a4-8d1d-eda9ad96fe55/6437c2a0-47fc-11ed-8757-0f869593f41f/file-1l91njbgr.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l91njbgq/c332d856-caa2-46a4-8d1d-eda9ad96fe55/6437c2a0-47fc-11ed-8757-0f869593f41f/fi... | mcq | jee-main-2015-offline | 6,306 |
dkDTQQj1IIHo6pmfr63Q9 | maths | height-and-distance | height-and-distance | The angle of elevation of the top of a vertical tower from a point A, due east of it is 45<sup>o</sup>. The angle of elevation of the top of the same tower from a point B, due south of A is 30<sup>o</sup>. If the distance between A and B is $$54\sqrt 2 \,m,$$ then the height of the tower (in metres), is : | [{"identifier": "A", "content": "$$36\\sqrt 3 $$"}, {"identifier": "B", "content": "54"}, {"identifier": "C", "content": "$$54\\sqrt 3 $$ "}, {"identifier": "D", "content": "108"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267746/exam_images/sfbzoig20kahtq3la2hf.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2016 (Online) 10th April Morning Slot Mathematics - Height and Distance Question 30 English Explanation">
<br>... | mcq | jee-main-2016-online-10th-april-morning-slot | 6,307 |
l87d3ma7 | maths | height-and-distance | height-and-distance | A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is 30<sup>o</sup>. After walking for 10 minutes from A in the same direction, at a point B, he observes that the angle of elevation of the... | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "20"}, {"identifier": "D", "content": "5"}] | ["D"] | null | According to given information, we have the following figure<br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l87czgsa/9146332a-d177-4510-8b69-14ebf4d88eae/9d5eea90-3753-11ed-9417-1312e45a73c7/file-1l87czgsb.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l87czgsa/9146332... | mcq | jee-main-2016-offline | 6,308 |
DcXLZz7J4dXtDQAC | maths | height-and-distance | height-and-distance | Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point
on the ground such that AP = 2AB. If $$\angle $$BPC = $$\beta $$, then tan$$\beta $$ is equal to: | [{"identifier": "A", "content": "$${1 \\over 4}$$"}, {"identifier": "B", "content": "$${2 \\over 9}$$"}, {"identifier": "C", "content": "$${4 \\over 9}$$"}, {"identifier": "D", "content": "$${6 \\over 7}$$"}] | ["B"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263839/exam_images/lz3a9qs9vzwnjl9bmeiu.webp" loading="lazy" alt="JEE Main 2017 (Offline) Mathematics - Height and Distance Question 34 English Explanation">
<br><br>Let the height of tower $$AB = x$$ and $$LCPA = \propto $$
<br><... | mcq | jee-main-2017-offline | 6,309 |
ZnQ7c1CR8ZKBJ5XWE59SM | maths | height-and-distance | height-and-distance | An aeroplane flying at a constant speed, parallel to the horizontal ground, $$\sqrt 3 $$ kmabove it, is obsered at an elevation of $${60^o}$$ from a point on the ground. If, after five seconds, its elevation from the same point, is $${30^o}$$, then the speed (in km / hr) of the aeroplane, is : | [{"identifier": "A", "content": "1500"}, {"identifier": "B", "content": "1440"}, {"identifier": "C", "content": "750"}, {"identifier": "D", "content": "720"}] | ["B"] | null | For $$\Delta $$OA, A, OA<sub>1</sub> = $${{\sqrt 3 } \over {\tan {{60}^o}}}$$ = 1 km
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266827/exam_images/o576rgrvrso9cx8btgxu.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2018 (Online) 15t... | mcq | jee-main-2018-online-15th-april-morning-slot | 6,311 |
QXs3r63kN8yLmLW88R12Y | maths | height-and-distance | height-and-distance | A tower T<sub>1</sub> of height 60 m is located exactly opposite to a tower T<sub>2</sub> of height 80 m on a straight road. Fromthe top of T<sub>1</sub>, if the angle of depression of the foot of T<sub>2</sub> is twice the angle of elevation of the top of T<sub>2</sub>, then the width (in m) of the road between the fe... | [{"identifier": "A", "content": "$$10\\sqrt 2 $$ "}, {"identifier": "B", "content": "$$10\\sqrt 3 $$"}, {"identifier": "C", "content": "$$20\\sqrt 3 $$"}, {"identifier": "D", "content": "$$20\\sqrt 2 $$"}] | ["C"] | null | Let the distance between T<sub>1</sub> and T<sub>2</sub> be x
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264796/exam_images/a99rugdbqpo21gvuvpmm.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2018 (Online) 15th April Evening Slot ... | mcq | jee-main-2018-online-15th-april-evening-slot | 6,312 |
szA1nx6Gyh7SRMBqUuAbX | maths | height-and-distance | height-and-distance | A man on the top of a vertical tower observes a car moving at a uniform speed towards the tower on a horizontal road. If it takes 18 min. for the angle of depression of the car to change from 30<sup>o</sup> to 45<sup>o</sup> ; then after this, the time taken (in min.) by the car to reach the foot of the tower, is : | [{"identifier": "A", "content": "$$9\\left( {1 + \\sqrt 3 } \\right)$$ "}, {"identifier": "B", "content": "$$18\\left( {1 + \\sqrt 3 } \\right)$$"}, {"identifier": "C", "content": "$$18\\left( {\\sqrt 3 - 1} \\right)$$"}, {"identifier": "D", "content": "$${9 \\over 2}\\left( {\\sqrt 3 - 1} \\right)$$"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267455/exam_images/gwuen82slkerzet1ztko.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2018 (Online) 16th April Morning Slot Mathematics - Height and Distance Question 31 English Explanation">
<br>... | mcq | jee-main-2018-online-16th-april-morning-slot | 6,313 |
tkQWnPZ6nK5w9g4gfgFcZ | maths | height-and-distance | height-and-distance | Consider a triangular plot ABC with sides AB = 7m, BC = 5m and CA = 6m. A vertical lamp-post at the mid point D of AC subtends an angle 30<sup>o</sup> at B. The height (in m) of the lamp-post is - | [{"identifier": "A", "content": "$$2\\sqrt {21} $$"}, {"identifier": "B", "content": "$${3 \\over 2}\\sqrt {21} $$"}, {"identifier": "C", "content": "$$7\\sqrt {3} $$"}, {"identifier": "D", "content": "$${2 \\over 3}\\sqrt {21} $$"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267190/exam_images/txlbrnxh7uu1cyfihe9v.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th January Morning Slot Mathematics - Height and Distance Question 29 English Explanation">
<b... | mcq | jee-main-2019-online-10th-january-morning-slot | 6,314 |
FrbvgcqsNWYOAzdb0T3rsa0w2w9jxb41ua0 | maths | height-and-distance | height-and-distance | The angle of elevation of the top of a vertical tower standing on a horizontal plane is observed to be 45<sup>o</sup> from
a point A on the plane. Let B be the point 30 m vertically above the point A. If the angle of elevation of the
top of the tower from B be 30<sup>o</sup>, then the distance (in m) of the foot of the... | [{"identifier": "A", "content": "$$15\\left( {1 + \\sqrt 3 } \\right)$$"}, {"identifier": "B", "content": "$$15\\left( {3 - \\sqrt 3 } \\right)$$"}, {"identifier": "C", "content": "$$15\\left( {3 + \\sqrt 3 } \\right)$$"}, {"identifier": "D", "content": "$$15\\left( {5 - \\sqrt 3 } \\right)$$"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267678/exam_images/ssjeqbfwbhvdrbkypihv.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th April Evening Slot Mathematics - Height and Distance Question 24 English Explanation">
$${x... | mcq | jee-main-2019-online-12th-april-evening-slot | 6,315 |
TRgZU92si7ThZAHGJW3rsa0w2w9jwy1ds9o | maths | height-and-distance | height-and-distance | ABC is a triangular park with AB = AC = 100 metres. A vertical tower is situated at the mid-point of BC. If
the angles of elevation of the top of the tower at A and B are cot<sup>–1</sup>
(3$$\sqrt 2 $$ ) and cosec<sup>–1</sup>
(2$$\sqrt 2 $$ ) respectively,
then the height of the tower (in metres) is : | [{"identifier": "A", "content": "$${{100} \\over {3\\sqrt 3 }}$$"}, {"identifier": "B", "content": "25"}, {"identifier": "C", "content": "20"}, {"identifier": "D", "content": "10$$\\sqrt 5 $$"}] | ["C"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267486/exam_images/ucqctvtzjka0ninslfkd.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265297/exam_images/mga2pd8jfduhngdumk72.webp"><img src="https://res.c... | mcq | jee-main-2019-online-10th-april-morning-slot | 6,316 |
pLTsdMGhjup9oQjoBN18hoxe66ijvwuynvi | maths | height-and-distance | height-and-distance | Two poles standing on a horizontal ground are of
heights 5m and 10 m respectively. The line joining
their tops makes an angle of 15º with ground. Then
the distance (in m) between the poles, is :- | [{"identifier": "A", "content": "$$5\\left( {2 + \\sqrt 3 } \\right)$$"}, {"identifier": "B", "content": "$${5 \\over 2}\\left( {2 + \\sqrt 3 } \\right)$$"}, {"identifier": "C", "content": "$$10\\left( {\\sqrt3 - 1 } \\right)$$"}, {"identifier": "D", "content": "$$5\\left( {\\sqrt3 + 1 } \\right)$$"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264351/exam_images/b08oyn8k1lljfispvted.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th April Evening Slot Mathematics - Height and Distance Question 26 English Explanation">
<br>
$$... | mcq | jee-main-2019-online-9th-april-evening-slot | 6,317 |
fNVBNnJtWYO5d0c1AjsBn | maths | height-and-distance | height-and-distance | Two vertical poles of heights, 20 m and 80 m stand
a part on a horizontal plane. The height (in meters)
of the point of intersection of the lines joining the
top of each pole to the foot of the other, from this
horizontal plane is : | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "16"}, {"identifier": "C", "content": "15"}, {"identifier": "D", "content": "18"}] | ["B"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264407/exam_images/wgqikoy7kceictq8u4qg.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266318/exam_images/ksjej4im98rcph4nc2mb.webp" style="max-width: 100%;height: auto;display: block;margi... | mcq | jee-main-2019-online-8th-april-evening-slot | 6,318 |
pxH6cut2HzLmnMIUy0jgy2xukg0cyxoq | maths | height-and-distance | height-and-distance | The angle of elevation of the summit of a
mountain from a point on the ground is 45°.
After climbing up one km towards the summit
at an inclination of 30° from the ground, the
angle of elevation of the summit is found to be
60°. Then the height (in km) of the summit from
the ground is : | [{"identifier": "A", "content": "$${1 \\over {\\sqrt 3 - 1}}$$"}, {"identifier": "B", "content": "$${{\\sqrt 3 + 1} \\over {\\sqrt 3 - 1}}$$"}, {"identifier": "C", "content": "$${1 \\over {\\sqrt 3 + 1}}$$"}, {"identifier": "D", "content": "$${{\\sqrt 3 - 1} \\over {\\sqrt 3 + 1}}$$"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265755/exam_images/vceiuoriishdtliybgle.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 6th September Evening Slot Mathematics - Height and Distance Question 20 English Explanation">
<br... | mcq | jee-main-2020-online-6th-september-evening-slot | 6,320 |
g3Go8rysjUigdUl2cnjgy2xukfw17ig0 | maths | height-and-distance | height-and-distance | Let AD and BC be two vertical poles <br/>at A and B respectively on a horizontal ground. <br/>If
AD = 8 m, BC = 11 m and AB = 10 m; then the distance<br/> (in meters) of a point M on AB from the point
A such<br/> that MD<sup>2</sup> + MC<sup>2</sup> is minimum is ______. | [] | null | 5 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266427/exam_images/aq1jmc4l3u0dm4zx0orb.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 6th September Morning Slot Mathematics - Height and Distance Question 22 English Explanation">
<br... | integer | jee-main-2020-online-6th-september-morning-slot | 6,321 |
lewrHJiFB4GaqO5A3fjgy2xukfxgnp8b | maths | height-and-distance | height-and-distance | The angle of elevation of the top of a hill from a point on the horizontal plane passing through the
foot of the hill is found to be 45<sup>o</sup>. After walking a distance of 80 meters towards the top, up a slope
inclined at an angle of 30<sup>o</sup> to the horizontal plane, the angle of elevation of the top of the ... | [] | null | 80 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264592/exam_images/wpspq1xvs3grsccoimwk.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 6th September Morning Slot Mathematics - Height and Distance Question 21 English Explanation">
<br... | integer | jee-main-2020-online-6th-september-morning-slot | 6,322 |
TYuBguLiKOQljYZ9Bbjgy2xukfakapsn | maths | height-and-distance | height-and-distance | The angle of elevation of a cloud C from a point P, 200 m above a still lake is 30°. If the angle of depression of the image of C in the lake from the point P is 60°,then PC (in m) is equal to :
| [{"identifier": "A", "content": "$$200\\sqrt 3 $$"}, {"identifier": "B", "content": "400"}, {"identifier": "C", "content": "100"}, {"identifier": "D", "content": "$$400\\sqrt 3 $$"}] | ["B"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267793/exam_images/nysg0p5njtrlfbcb1wam.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266112/exam_images/nyi5l1fouysd8mkjmgb3.webp"><source media="(max-wid... | mcq | jee-main-2020-online-4th-september-evening-slot | 6,323 |
MxCfgJ8IDqtwtaRJnP1klrga9a8 | maths | height-and-distance | height-and-distance | Two vertical poles are 150 m apart and the height of one is three times that of the other. If
from the middle point of the line joining their feet, an observer finds the angles of elevation of
their tops to be complementary, then the height of the shorter pole (in meters) is : | [{"identifier": "A", "content": "30"}, {"identifier": "B", "content": "25"}, {"identifier": "C", "content": "20$$\\sqrt 3 $$"}, {"identifier": "D", "content": "25$$\\sqrt 3 $$"}] | ["D"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266120/exam_images/yeugccdhox3noatlqnmc.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267418/exam_images/vespqs6qac89duryh7y9.webp"><source media="(max-wid... | mcq | jee-main-2021-online-24th-february-morning-slot | 6,324 |
cMNUUY1RfRfxsUTd461klrlwmtk | maths | height-and-distance | height-and-distance | The angle of elevation of a jet plane from a point A on the ground is 60$$^\circ$$. After a flight of 20 seconds at the speed of 432 km/hour, the angle of elevation changes to 30$$^\circ$$. If the jet plane is flying at a constant height, then its height is : | [{"identifier": "A", "content": "$$3600\\sqrt 3 $$ m"}, {"identifier": "B", "content": "$$1200\\sqrt 3 $$ m"}, {"identifier": "C", "content": "$$1800\\sqrt 3 $$ m"}, {"identifier": "D", "content": "$$2400\\sqrt 3 $$ m"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266806/exam_images/yei81ivy9h4mpf0cbjzv.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 24th February Evening Shift Mathematics - Height and Distance Question 18 English Explanation">
<b... | mcq | jee-main-2021-online-24th-february-evening-slot | 6,325 |
7LjqYLBFSIq2nWK5cW1kls4jij6 | maths | height-and-distance | height-and-distance | A man is observing, from the top of a tower, a boat speeding towards the lower from a certain point A, with uniform speed. At that point, angle of depression of the boat with the man's eye is 30$$^\circ$$ (Ignore man's height). After sailing for 20 seconds, towards the base of the tower (which is at the level of water)... | [{"identifier": "A", "content": "$$10(\\sqrt 3 + 1)$$"}, {"identifier": "B", "content": "$$10(\\sqrt 3 - 1)$$"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "$$10\\sqrt 3$$"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266282/exam_images/cyid3ux4etov3mikiruk.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 25th February Morning Shift Mathematics - Height and Distance Question 17 English Explanation">
<b... | mcq | jee-main-2021-online-25th-february-morning-slot | 6,326 |
y2vesUbzQ1pa4cPPZy1kmm2uv3f | maths | height-and-distance | height-and-distance | A pole stands vertically inside a triangular park ABC. Let the angle of elevation of the top of the pole from each corner of the park be $${\pi \over 3}$$. If the radius of the circumcircle of $$\Delta$$ABC is 2, then the height of the pole is equal to : | [{"identifier": "A", "content": "$${{1 \\over {\\sqrt 3 }}}$$"}, {"identifier": "B", "content": "2$${\\sqrt 3 }$$"}, {"identifier": "C", "content": "$${\\sqrt 3 }$$"}, {"identifier": "D", "content": "$${{{2\\sqrt 3 } \\over 3}}$$"}] | ["B"] | null | <picture><source media="(max-width: 1221px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263609/exam_images/jzim1jti8mbu5b7il5p0.webp"><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267155/exam_images/s0gaiujemitqjrle3a8g.webp"><source media="(max-wi... | mcq | jee-main-2021-online-18th-march-evening-shift | 6,327 |
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