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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$$ f(x... | mcq | jee-main-2022-online-25th-june-morning-shift |
1l5ajukpk | maths | functions | composite-functions | <p>Let $$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 |
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>$$fofofof(x) = ... | integer | jee-main-2022-online-25th-july-evening-shift |
1l6jb0fby | maths | functions | composite-functions | <p>Let $$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 |
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 |
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 |
1lgpyd2rn | maths | functions | composite-functions | <p>For $$x \in \mathbb{R}$$, two real valued functions $$f(x)$$ and $$g(x)$$ are such that, $$g(x)=\sqrt{x}+1$$ and $$f \circ g(x)=x+3-\sqrt{x}$$. Then $$f(0)$$ is equal to</p> | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "$$-$$3"}, {"identifier": "D", "content": "1"}] | ["A"] | null | $$
\begin{aligned}
& g(x)=\sqrt{x}+1 \\\\
& \operatorname{fog}(x)=x+3-\sqrt{x} \\\\
& =(\sqrt{x}+1)^2-3(\sqrt{x}+1)+5 \\\\
& =g^2(x)-3 g(x)+5 \\\\
& \Rightarrow f(x)=x^2-3 x+5 \\\\
& \therefore f(0)=5
\end{aligned}
$$
<br/><br/>But, if we consider the domain of the composite function $f \circ g(x)$ then in that case $f... | mcq | jee-main-2023-online-13th-april-morning-shift |
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 |
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>$$x \in R-\left\{-\frac{5}{2}... | mcq | jee-main-2024-online-27th-january-evening-shift |
jaoe38c1lse52f1r | maths | functions | composite-functions | <p>If $$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))$$</p>
<... | mcq | jee-main-2024-online-31st-january-morning-shift |
jaoe38c1lseypzri | maths | functions | composite-functions | <p>If $$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]$$</p>
<p><img src="https://app-content... | mcq | jee-main-2024-online-29th-january-morning-shift |
lv2erz8l | maths | functions | composite-functions | <p>Consider the function $$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 |
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 |
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 $$\,\,\,\,x \in \left[... | mcq | aieee-2002 |
jKHkNWI2VphTKzW4 | maths | functions | domain | Domain of definition of the function f(x) = $${3 \over {4 - {x^2}}}$$ + $${\log _{10}}\left( {{x^3} - x} \right)$$, is | [{"identifier": "A", "content": "(-1, 0)$$ \\cup $$(1, 2)$$ \\cup $$(2, $$\\infty $$)"}, {"identifier": "B", "content": "(1, 2)"}, {"identifier": "C", "content": "(-1, 0) $$ \\cup $$ (1, 2)"}, {"identifier": "D", "content": "(1, 2)$$ \\cup $$(2, $$\\infty $$)"}] | ["A"] | null | $$f\left( x \right) = {3 \over {4 - {x^2}}} + {\log _{10}}\left( {{x^3} - x} \right)$$
<br><br>$$4 - {x^2} \ne 0;\,\,\,{x^3} - x > 0;$$
<br><br>$$x \ne \pm \sqrt 4 $$ and $$ - 1 < x < 0$$
<br><br>or $$\,\,\,1 < x < \infty $$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/... | mcq | aieee-2003 |
htONhxhNRHk21CjA | maths | functions | domain | The domain of the function
<br/>$$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 |
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 |
IuZ2hfWBOzl3se7H | maths | functions | domain | The domain of the function f(x) = $${1 \over {\sqrt {\left| x \right| - x} }}$$ is | [{"identifier": "A", "content": "$$\\left( {0,\\infty } \\right)$$"}, {"identifier": "B", "content": "$$\\left( { - \\infty ,0} \\right)$$"}, {"identifier": "C", "content": "$$\\left( { - \\infty ,\\infty } \\right) - \\left\\{ 0 \\right\\}$$"}, {"identifier": "D", "content": "$$\\left( { - \\infty ,\\infty } \\right)$... | ["B"] | null | $$f\left( x \right) = {1 \over {\sqrt {\left| x \right| - x} }},\,\,$$ define if $$\,\,\,\left| x \right| - x > 0$$
<br><br>$$ \Rightarrow \left| x \right| > x,\,\,\, \Rightarrow x < 0$$
<br><br>Hence domain of $$f\left( x \right)$$ is $$\left( { - \infty ,0} \right)$$ | mcq | aieee-2011 |
w8Zuk7uFMhknNRsWwF18hoxe66ijvww16jt | maths | functions | domain | The domain of the definition of the function
<br/><br/>$$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 |
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 |
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 |
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 |
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 |
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 |
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
$$
\co... | integer | jee-main-2023-online-8th-april-evening-shift |
lsam05oo | maths | functions | domain | If the domain of the function
<br/><br/>$f(x)=\frac{\sqrt{x^2-25}}{\left(4-x^2\right)}+\log _{10}\left(x^2+2 x-15\right)$ is $(-\infty, \alpha) \cup[\beta, \infty)$, then $\alpha^2+\beta^3$ is equal to : | [{"identifier": "A", "content": "140"}, {"identifier": "B", "content": "175"}, {"identifier": "C", "content": "125"}, {"identifier": "D", "content": "150"}] | ["D"] | null | <p>To find the domain of the function
$$f(x) = \frac{\sqrt{x^2-25}}{(4-x^2)}+\log_{10}(x^2+2x-15),$$
we need to consider the domain conditions for both the square root function and the logarithmic function.</p>
<p>The square root function $\sqrt{x^2-25}$ requires that the argument of the square root be non-negative, ... | mcq | jee-main-2024-online-1st-february-evening-shift |
1lsg4mhuk | maths | functions | domain | <p>If the domain of the function $$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}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsoxumdb/8342ad24-c8e3... | mcq | jee-main-2024-online-30th-january-evening-shift |
1lsgb11sw | maths | functions | domain | <p>If the domain of the function $$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$$ .... (2... | mcq | jee-main-2024-online-30th-january-morning-shift |
luy6z50x | maths | functions | domain | <p>If the domain of the function $$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}
$$</p>
<p><img src="https://a... | mcq | jee-main-2024-online-9th-april-morning-shift |
eDxALxbmMByHC8s7 | maths | functions | even-and-odd-functions | The function $$f\left( x \right)$$ $$ = \log \left( {x + \sqrt {{x^2} + 1} } \right)$$, is | [{"identifier": "A", "content": "neither an even nor an odd function"}, {"identifier": "B", "content": "an even function"}, {"identifier": "C", "content": "an odd function"}, {"identifier": "D", "content": "a periodic function"}] | ["C"] | null | $$f\left( x \right) = \log \left( {x + \sqrt {{x^2} + 1} } \right)$$
<br><br>$$f\left( { - x} \right) = \log \left\{ { - x + \sqrt {{x^2} + 1} } \right\}$$
<br><br>$$ = \log \left\{ {{{ - {x^2} + {x^2} + 1} \over {x + \sqrt {{x^2} + 1} }}} \right\}$$
<br><br>$$ = - \log \left( {x + \sqrt {{x^2} + 1} } \right) = - f\l... | mcq | aieee-2003 |
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$$ as shown in the figure.
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263707/exam_images/uiwdp9jmte0l8nkstssm.webp" loading="lazy" alt="AIEEE 2004 Mathematics - Functions Question 131 English Explanation">
<br>... | mcq | aieee-2004 |
7Gxl5AoOoGNHBVk7crwGB | maths | functions | even-and-odd-functions | Let ƒ(x) = a<sup>x</sup>
(a > 0) be written as
<br/>ƒ(x) = ƒ<sub>1</sub>
(x) + ƒ<sub>2</sub>
(x), where ƒ<sub>1</sub>
(x) is an even
function of ƒ<sub>2</sub>
(x) is an odd function. <br/>Then
ƒ<sub>1</sub>
(x + y) + ƒ<sub>1</sub>
(x – y) equals | [{"identifier": "A", "content": "2\u0192<sub>1</sub>\n(x)\u0192<sub>1</sub>\n(y)\n"}, {"identifier": "B", "content": "2\u0192<sub>1</sub>\n(x + y)\u0192<sub>1</sub>\n(x \u2013 y)"}, {"identifier": "C", "content": "2\u0192<sub>1</sub>\n(x)\u0192<sub>2</sub>\n(y)"}, {"identifier": "D", "content": "2\u0192<sub>1</sub>\n(x... | ["A"] | null | f(x) = a<sup>x</sup>
<br><br>As f<sub>1</sub>(x) is even function then
<br><br>f<sub>1</sub>(x) = $${{{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 |
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>$$|x|<0... | mcq | jee-main-2024-online-8th-april-evening-shift |
8TmUW15AHe2hgz5a | maths | functions | functional-equations | If $$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 |
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 $$x = ... | mcq | aieee-2005 |
l87cb3gu | maths | functions | functional-equations | If $f(x)+2 f\left(\frac{1}{x}\right)=3 x, x \neq 0$, and $\mathrm{S}=\{x \in \mathbf{R}: f(x)=f(-x)\}$; then $\mathrm{S}:$ | [{"identifier": "A", "content": "is an empty set."}, {"identifier": "B", "content": "contains exactly one element."}, {"identifier": "C", "content": "contains exactly two elements."}, {"identifier": "D", "content": "contains more than two elements."}] | ["C"] | null | We have, $f(x)+2 f\left(\frac{1}{x}\right)=3 x, \quad x \neq 0$
$\ldots$ (i)<br/><br/>
On replacing $x$ by $\frac{1}{x}$ in the above equation, we get<br/><br/>
$$
\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 |
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>$$\sum\limit... | mcq | jee-main-2017-offline |
7T8qHanHgmWJe870g3Bhs | maths | functions | functional-equations | If $$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 |
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 |
yAWIEwUDWnjs2sVq2ejMn | maths | functions | functional-equations | Let f<sub>k</sub>(x) = $${1 \over k}\left( {{{\sin }^k}x + {{\cos }^k}x} \right)$$ for k = 1, 2, 3, ... Then for all x $$ \in $$ R, the value of f<sub>4</sub>(x) $$-$$ f<sub>6</sub>(x) is equal to | [{"identifier": "A", "content": "$${1 \\over 4}$$ "}, {"identifier": "B", "content": "$${5 \\over {12}}$$"}, {"identifier": "C", "content": "$${{ - 1} \\over {12}}$$"}, {"identifier": "D", "content": "$${1 \\over {12}}$$"}] | ["D"] | null | f<sub>4</sub>(x) $$-$$ f<sub>6</sub>(x)
<br><br>= $${1 \over 4}$$ (sin<sup>4</sup> x + cos<sup>4</sup> x) $$-$$ $${1 \over 6}$$ (sin<sup>6</sup> x + cos<sup>6</sup> x)
<br><br>= $${1 \over 4}$$ (1$$-$$ $${1 \over 2}$$ sin<sup>2</sup> 2x) $$-$$ $${1 \over 6}$$ (1 $$-$$ $${3 \over 4}$$ sin<sup>2</sup> 2x) = $${1 \over ... | mcq | jee-main-2019-online-11th-january-morning-slot |
kmmWFv4B3ZGx0zoZFq7k9k2k5khqkpe | maths | functions | functional-equations | Let a – 2b + c = 1.<br/><br/>
If $$f(x)=\left| {\matrix{
{x + a} & {x + 2} & {x + 1} \cr
{x + b} & {x + 3} & {x + 2} \cr
{x + c} & {x + 4} & {x + 3} \cr
} } \right|$$, then: | [{"identifier": "A", "content": "\u0192(50) = 1"}, {"identifier": "B", "content": "\u0192(\u201350) = \u20131"}, {"identifier": "C", "content": "\u0192(50) = \u2013501"}, {"identifier": "D", "content": "\u0192(\u201350) = 501"}] | ["A"] | null | R<sub>1</sub> $$ \to $$ R<sub>1</sub> + R<sub>3</sub> – 2R<sub>2</sub>
<br><br>f(x) = $$\left| {\matrix{
{a + c - 2b} & 0 & 0 \cr
{x + b} & {x + 3} & {x + 2} \cr
{x + c} & {x + 4} & {x + 3} \cr
} } \right|$$
<br><br>= (a + c – 2b) ((x + 3)<sup>2</sup> – (x + 2)(x + 4))
<br><br>=... | mcq | jee-main-2020-online-9th-january-evening-slot |
oOLSAdvLWP0U3MIENGjgy2xukez67a9z | maths | functions | functional-equations | Let f : R $$ \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)} $$
<br><br>= f(1) + f(2) +.........+ f(n - 1)
<br><br>= 2 + 4 + 6 ... | mcq | jee-main-2020-online-2nd-september-evening-slot |
o26Zk0Ad7pZ3fy2Cu9jgy2xukfuvv0hp | maths | functions | functional-equations | If f(x + y) = f(x)f(y) and $$\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 $$ = 2 ....(1)
<br><br>On f(x + y) = f(x) f(y)
<br>* Put x = 1, y = 1
<br>f(2) = (f(1))<sup>2</sup>
<br>* Put x = 2, y = 1
<br>f(3) = f(2). f(1) = f((1))<sup>3</sup>
<br>*... | mcq | jee-main-2020-online-6th-september-morning-slot |
xRBLs2qkeW9u7OBLNTjgy2xukg4n20of | maths | functions | functional-equations | Suppose that a function f : R $$ \to $$ R satisfies<br/>
f(x + y) = f(x)f(y) for all x, y $$ \in $$ R and f(1) = 3.<br/> If $$\sum\limits_{i = 1}^n {f(i)} = 363$$ then n is equal to ________ . | [] | null | 5 | f(x + y) = f(x) f(y)
<br><br>put x = y = 1
<br>$$ \therefore $$ f(2) = (ƒ(1))<sup>2</sup>
= 3<sup>2</sup>
<br><br>put x = 2, y = 1
<br>$$ \therefore $$ f(3) = (ƒ(1))<sup>3</sup>
= 3<sup>3</sup>
<br><br>Similarly f(x) = 3<sup>x</sup>
<br>$$ \Rightarrow $$ f(i) = 3<sup>i</sup>
<br><br>Given, $$\sum\limits_{i = 1}^n {... | integer | jee-main-2020-online-6th-september-evening-slot |
k8qq4kBTBPpAjRvtRg1klrmmoe0 | maths | functions | functional-equations | If a + $$\alpha$$ = 1, b + $$\beta$$ = 2 and $$af(x) + \alpha f\left( {{1 \over x}} \right) = bx + {\beta \over x},x \ne 0$$, then the value of the expression $${{f(x) + f\left( {{1 \over x}} \right)} \over {x + {1 \over x}}}$$ is __________. | [] | null | 2 | $$af(x) + \alpha f\left( {{1 \over x}} \right) = bx + {\beta \over x}$$ ........(i)<br><br>Replace $$x $$ with $$ {1 \over x}$$<br><br>$$af\left( {{1 \over x}} \right) + af(x) = {b \over x} + \beta x$$ ..... (ii)<br><br>(i) + (ii)<br><br>$$(a + \alpha )\left[ {f(x) + f\left( {{1 \over x}} \right)} \right] = \left( {x ... | integer | jee-main-2021-online-24th-february-evening-slot |
TBP1H0CdJEYh9UShXc1klt7f0ou | maths | functions | functional-equations | A function f(x) is given by $$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$$<b... | mcq | jee-main-2021-online-25th-february-evening-slot |
GZkr2iQcFKkQTUhoWJ1kmm3zbjp | maths | functions | functional-equations | If f(x) and g(x) are two polynomials such that the polynomial P(x) = f(x<sup>3</sup>) + x g(x<sup>3</sup>) is divisible by x<sup>2</sup> + x + 1, then P(1) is equal to ___________. | [] | null | 0 | Given, p(x) = f(x<sup>3</sup>) + xg(x<sup>3</sup>)<br><br>We know, x<sup>2</sup> + x + 1 = (x $$-$$ $$\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$$<sup>2</sup>.<br><br>As root satisfy the equation,<br><br>So, put x = $... | integer | jee-main-2021-online-18th-march-evening-shift |
1krxhvhxq | maths | functions | functional-equations | Let f : R $$\to$$ R be defined as $$f(x + y) + f(x - y) = 2f(x)f(y),f\left( {{1 \over 2}} \right) = - 1$$. Then, the value of $$\sum\limits_{k = 1}^{20} {{1 \over {\sin (k)\sin (k + f(k))}}} $$ is equal to : | [{"identifier": "A", "content": "cosec<sup>2</sup>(21) cos(20) cos(2)"}, {"identifier": "B", "content": "sec<sup>2</sup>(1) sec(21) cos(20)"}, {"identifier": "C", "content": "cosec<sup>2</sup>(1) cosec(21) sin(20)"}, {"identifier": "D", "content": "sec<sup>2</sup>(21) sin(20) sin(2)"}] | ["C"] | null | f(x) = cos$$\lambda$$x<br><br>$$\because$$ $$f\left( {{1 \over 2}} \right)$$ = $$-$$1<br><br>So, $$-$$1 = cos$${\lambda \over 2}$$<br><br>$$\Rightarrow$$ $$\lambda$$ = 2$$\pi$$<br><br>Thus f(x) = cos2$$\pi$$x<br><br>Now k is natural number<br><br>Thus f(k) = 1<br><br>$$\sum\limits_{k = 1}^{20} {{1 \over {\sin k\sin (k... | mcq | jee-main-2021-online-27th-july-evening-shift |
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>$$f(6) = f(2).f(3)\left\{ \ma... | integer | jee-main-2021-online-27th-july-morning-shift |
1ktk52hwc | maths | functions | functional-equations | Let f : N $$\to$$ N be a function such that f(m + n) = f(m) + f(n) for every m, n$$\in$$N. If f(6) = 18, then f(2) . f(3) is equal to : | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "54"}, {"identifier": "C", "content": "18"}, {"identifier": "D", "content": "36"}] | ["B"] | null | f(m + n) = f(m) + f(n)<br><br>Put m = 1, n = 1<br><br>f(2) = 2f(1)<br><br>Put m = 2, n = 1<br><br>f(3) = f(2) + f(1) = 3f(1)<br><br>Put m = 3, n = 3<br><br>f(6) = 2f(3) $$\Rightarrow$$ f(3) = 9<br><br>$$\Rightarrow$$ f(1) = 3, f(2) = 6<br><br>f(2) . f(3) = 6 $$\times$$ 9 = 54 | mcq | jee-main-2021-online-31st-august-evening-shift |
1l546a0kk | maths | functions | functional-equations | <p>Let c, k $$\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 |
1l57ovytd | maths | functions | functional-equations | <p>Let f : R $$\to$$ R be a function defined by $$f(x) = {{2{e^{2x}}} \over {{e^{2x}} + e}}$$. Then $$f\left( {{1 \over {100}}} \right) + f\left( {{2 \over {100}}} \right) + f\left( {{3 \over {100}}} \right) + \,\,\,.....\,\,\, + \,\,\,f\left( {{{99} \over {100}}} \right)$$ is equal to ______________.</p> | [] | null | 99 | <p>Given,</p>
<p>$$f(x) = {{2{e^{2x}}} \over {{e^{2x}} + e}}$$</p>
<p>$$\therefore$$ $$f(1 - x) = {{2{e^{2(1 - x)}}} \over {{e^{2(1 - x)}} + e}}$$</p>
<p>$$ = {{2\,.\,{{{e^2}} \over {{e^{2x}}}}} \over {{{{e^2}} \over {{e^{2x}}}} + e}}$$</p>
<p>$$ = {{2{e^2}} \over {{e^2} + {e^{2x}}\,.\,e}}$$</p>
<p>$$ = {{2{e^2}} \over... | integer | jee-main-2022-online-27th-june-morning-shift |
1l5ahps1l | maths | functions | functional-equations | <p>Let f : N $$\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}$$</p>... | mcq | jee-main-2022-online-25th-june-morning-shift |
1l6jeicx4 | maths | functions | functional-equations | <p> Let $$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 |
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 |
1l6nld95e | maths | functions | functional-equations | <p>$$
\text { Let } f(x)=a x^{2}+b x+c \text { be such that } f(1)=3, f(-2)=\lambda \text { and } $$ $$f(3)=4$$. If $$f(0)+f(1)+f(-2)+f(3)=14$$, then $$\lambda$$ is equal to :</p> | [{"identifier": "A", "content": "$$-$$4"}, {"identifier": "B", "content": "$$\\frac{13}{2}$$"}, {"identifier": "C", "content": "$$\\frac{23}{2}$$"}, {"identifier": "D", "content": "4"}] | ["D"] | null | <p>$$f(1) = a + b + c = 3$$ ..... (i)</p>
<p>$$f(3) = 9a + 3b + c = 4$$ .... (ii)</p>
<p>$$f(0) + f(1) + f( - 2) + f(3) = 14$$</p>
<p>OR $$c + 3 + (4a - 2b + c) + 4 = 14$$</p>
<p>OR $$4a - 2b + 2c = 7$$ ..... (iii)</p>
<p>From (i) and (ii) $$8a + 2b = 1$$ ..... (iv)</p>
<p>From (iii) $$ - (2) \times $$ (i)</p>
<p>$$ \R... | mcq | jee-main-2022-online-28th-july-evening-shift |
1ldo6xo14 | maths | functions | functional-equations | <p>Let $$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}$$"}] | ["A"] | null | $\begin{aligned} & \mathrm{f}(\mathrm{x})+\mathrm{f}\left(\frac{1}{1-\mathrm{x}}\right)=1+\mathrm{x} \\\\ & \mathrm{x}=2 \Rightarrow \mathrm{f}(2)+\mathrm{f}(-1)=3 ........(1) \\\\ & \mathrm{x}=-1 \Rightarrow \mathrm{f}(-1)+\mathrm{f}\left(\frac{1}{2}\right)=0 .........(2) \\\\ & \mathrm{x}=\frac{1}{2} \Rightarrow \mat... | mcq | jee-main-2023-online-1st-february-evening-shift |
ldo9i6yu | maths | functions | functional-equations | The absolute minimum value, of the function
<br/><br/>$f(x)=\left|x^{2}-x+1\right|+\left[x^{2}-x+1\right]$,
<br/><br/>where $[t]$ denotes the greatest integer function, in the interval $[-1,2]$, is : | [{"identifier": "A", "content": "$\\frac{3}{4}$"}, {"identifier": "B", "content": "$\\frac{3}{2}$"}, {"identifier": "C", "content": "$\\frac{1}{4}$"}, {"identifier": "D", "content": "$\\frac{5}{4}$"}] | ["A"] | null | $\mathrm{f}(\mathrm{x})=\left|\mathrm{x}^{2}-\mathrm{x}+1\right|+\left[\mathrm{x}^{2}-\mathrm{x}+1\right] ; \mathrm{x} \in[-1,2]$
<br/><br/>Let $g(x)=x^{2}-x+1$
<br/><br/>$$
=\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 |
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/>$$
\begin{aligned}
& = (2+\sin 2 x)\left|\begin{... | mcq | jee-main-2023-online-1st-february-morning-shift |
ldqyb0zm | maths | functions | functional-equations | Let $A=\{1,2,3,5,8,9\}$. Then the number of possible functions $f: A \rightarrow A$ such that $f(m \cdot n)=f(m) \cdot f(n)$ for every $m, n \in A$ with $m \cdot n \in A$ is equal to ___________. | [] | null | 432 | <p>$$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$$</p> | integer | jee-main-2023-online-30th-january-evening-shift |
1ldsf1d3h | maths | functions | functional-equations | <p>Consider a function $$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 |
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 |
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 |
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 |
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 | $$
\begin{aligned}
& \mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x}) \cdot \mathrm{f}(\mathrm{y}) \forall \mathrm{x}, \mathrm{y} \in \mathrm{N}, \mathrm{f}(1)=3 \\\\
& \mathrm{f}(2)=\mathrm{f}^2(1)=3^2 \\\\
& \mathrm{f}(3)=\mathrm{f}(1) \mathrm{f}(2)=3^3 \\\\
& \mathrm{f}(4)=3^4 \\\\
& \mathrm{f}(\mathrm{k})=3... | mcq | jee-main-2023-online-24th-january-evening-shift |
1ldwwwfha | maths | functions | functional-equations | <p>If $$f(x) = {{{2^{2x}}} \over {{2^{2x}} + 2}},x \in \mathbb{R}$$, then $$f\left( {{1 \over {2023}}} \right) + f\left( {{2 \over {2023}}} \right)\, + \,...\, + \,f\left( {{{2022} \over {2023}}} \right)$$ is equal to</p> | [{"identifier": "A", "content": "2011"}, {"identifier": "B", "content": "2010"}, {"identifier": "C", "content": "1010"}, {"identifier": "D", "content": "1011"}] | ["D"] | null | $$
\begin{aligned}
& f(x)=\frac{4^x}{4^x+2} \\\\
& f(x)+f(1-x)=\frac{4^x}{4^x+2}+\frac{4^{1-x}}{4^{1-x}+2} \\\\
& =\frac{4^x}{4^x+2}+\frac{4}{4+2\left(4^x\right)} \\\\
& =\frac{4^x}{4^{\mathrm{x}}+2}+\frac{2}{2+4^{\mathrm{x}}} \\\\
& =1 \\\\
& \Rightarrow \mathrm{f}(\mathrm{x})+\mathrm{f}(1-\mathrm{x})=1 \\\\
& \text {... | mcq | jee-main-2023-online-24th-january-evening-shift |
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/>$$
\begin{aligned}
&f(f(x)) =\frac{A\left(\frac{A x+B}{C x-A}\rig... | mcq | jee-main-2023-online-10th-april-morning-shift |
luy9clgk | maths | functions | functional-equations | <p>If a function $$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 |
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 |
NmOxptYb4egNVmQdBRShM | maths | functions | inverse-functions | Let <i> f</i> : A $$ \to $$ B be a function defined as f(x) = $${{x - 1} \over {x - 2}},$$ Where A = <b>R</b> $$-$$ {2} and B = <b>R</b> $$-$$ {1}. Then <i>f</i> is : | [{"identifier": "A", "content": "invertible and $${f^{ - 1}}(y) = $$ $${{3y - 1} \\over {y - 1}}$$ "}, {"identifier": "B", "content": "invertible and $${f^{ - 1}}\\left( y \\right) = {{2y - 1} \\over {y - 1}}$$ "}, {"identifier": "C", "content": "invertible and $${f^{ - 1}}\\left( y \\right) = {{2y + 1} \\over {y - 1}... | ["B"] | null | Assume,
<br><br>y = f(x)
<br><br>$$ \Rightarrow $$ y = $${{x - 1} \over {x - 2}}$$
<br><br>$$ \Rightarrow $$ yx - 2y = x - 1
<br><br>$$ \Rightarrow $$ (y - 1)x = 2y - 1
<br><br>$$ \Rightarrow $$ x = $${{2y - 1} \over {y - 1}}$$ = f<sup> -1</sup>(y)
<br><br>As on the given domain the function is invertible and its inver... | mcq | jee-main-2018-online-15th-april-evening-slot |
S9ViRTELMsFpCdXCyu7k9k2k5gzd75n | maths | functions | inverse-functions | The inverse function of
<br/><br/>f(x) = $${{{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>$$ \Rightarrow $$ $${\log _e}\left( {{{1 + y} \over {1 - y}}} \right)... | mcq | jee-main-2020-online-8th-january-morning-slot |
3KiVygqDw488GbnoRq1kmjaxjig | maths | functions | inverse-functions | The inverse of $$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}}}}$$ | mcq | jee-main-2021-online-17th-march-morning-shift |
LQM2NmZMuUOeUUPSBE1kmm3nlbt | maths | functions | inverse-functions | Let f : R $$-$$ {3} $$ \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 |
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 |
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 $${\sin ^n}\theta $$, $${\cos ^n}\theta $$, $${\csc ^n}... | mcq | aieee-2002 |
rGh90lKpMIAeYj89 | maths | functions | periodic-functions | Which one is not periodic? | [{"identifier": "A", "content": "$$\\left| {\\sin 3x} \\right| + {\\sin ^2}x$$ "}, {"identifier": "B", "content": "$$\\cos \\sqrt x + {\\cos ^2}x$$ "}, {"identifier": "C", "content": "$$\\cos \\,4x + {\\tan ^2}x$$ "}, {"identifier": "D", "content": "$$cos\\,2x + \\sin x$$ "}] | ["B"] | null | $$\sqrt x $$ is non periodic function and $$\cos \left( {something} \right)$$ is a periodic function so here in $$\cos \sqrt x $$ $$ \to $$ inside periodic function there is non periodic function which always produce non periodic function.
<br><br>$${{{\cos }^2}x}$$ is a periodic function with period $$\pi $$
<br><br><... | mcq | aieee-2002 |
twJaTXmPzxEKgG6b | maths | functions | range | The range of the function f(x) = $${}^{7 - x}{P_{x - 3}}$$ is | [{"identifier": "A", "content": "{1, 2, 3, 4, 5}"}, {"identifier": "B", "content": "{1, 2, 3, 4, 5, 6}"}, {"identifier": "C", "content": "{1, 2, 3, 4}"}, {"identifier": "D", "content": "{1, 2, 3}"}] | ["D"] | null | The range of the function $f(x) = {}^{7-x}P_{x-3}$ can be found by considering the possible values of $f(x)$ as $x$ varies over its domain.
<br/><br/>
The domain of $f(x)$ is the set of all real numbers such that
<br/><br/>(i) $x \geq 3$ (since the permutation function is only defined for non-negative integers)
<br/>... | mcq | aieee-2004 |
m5lbIUG3trWiugfMUNHOx | maths | functions | range | Let f : R $$ \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]$$ | mcq | jee-main-2019-online-11th-january-morning-slot |
ohIDlBis3k5AggYqUF7k9k2k5hjmi7l | maths | functions | range | Let ƒ : (1, 3) $$ \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 |
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>$$ - 1 \le - {{4a - 3} \over {a -... | mcq | jee-main-2021-online-16th-march-morning-shift |
1kto99h4t | maths | functions | range | The range of the function, <br/><br/>$$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 |
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 |
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 |
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 |
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 |
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>$$\begin{aligned}
& F(x)=\frac{1}{2+\sin 3 x+\cos 3 x}, x \in \mathbb{R} \\
& \sin 3 x+\cos 3 x \in[-\sqrt{2}, \sqrt{2}] \\
& 2+\sin 3 x+\cos 3 x \in[2-\sqrt{2}, 2+\sqrt{2}] \\
& \Rightarrow \frac{1}{2+\sin 3 x+\cos 3 x} \in\left[\frac{1}{2+\sqrt{2}}, \frac{1}{2-\sqrt{2}}\right] \\
& \Rightarrow a=\frac{1}{2+\sqrt{2... | mcq | jee-main-2024-online-9th-april-evening-shift |
lv5gt1wc | maths | functions | range | <p>If the range of $$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)} $$</p>
... | integer | jee-main-2024-online-8th-april-morning-shift |
lvb29471 | maths | functions | range | <p>If the function $$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 |
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 |
LYcYPZMeePF9jKMj | maths | height-and-distance | height-and-distance | A person standing on the bank of a river observes that the angle of elevation of the top of a tree on the opposite bank of the river is $${60^ \circ }$$ and when he retires $$40$$ meters away from the tree the angle of elevation becomes $${30^ \circ }$$. The breadth of the river is : | [{"identifier": "A", "content": "$$60\\,\\,m$$ "}, {"identifier": "B", "content": "$$30\\,\\,m$$"}, {"identifier": "C", "content": "$$40\\,\\,m$$"}, {"identifier": "D", "content": "$$20\\,\\,m$$"}] | ["D"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265502/exam_images/egbnkczftvlxditjtcbz.webp" loading="lazy" alt="AIEEE 2004 Mathematics - Height and Distance Question 40 English Explanation">
<br><br>From the figure
<br><br>$$\tan {60^ \circ } = {y \over x}$$
<br><br>$$ \Rightar... | mcq | aieee-2004 |
tWejUYNwkeIGCAA9 | maths | height-and-distance | height-and-distance | A tower stands at the centre of a circular park. $$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$$
<br><br><img class="question-image" src="https://res.cloudinary.... | mcq | aieee-2007 |
GR58oSFYu58GzaCR | maths | height-and-distance | height-and-distance | $$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 $$
<br><b... | mcq | aieee-2008 |
XLrrI8GbRHPBpVqu | maths | height-and-distance | height-and-distance | $$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>$${{AB} \over {\sin \theta }} = {{\sqrt {{... | mcq | jee-main-2013-offline |
c3vyLJcH7Y9uv1nY | maths | height-and-distance | height-and-distance | A bird is sitting on the top of a vertical pole $$20$$ m high and its elevation from a point $$O$$ on the ground is $${45^ \circ }$$. It files off horizontally straight away from the point $$O$$. After one second, the elevation of the bird from $$O$$ is reduced to $${30^ \circ }$$. Then the speed (in m/s) of the bird i... | [{"identifier": "A", "content": "$$20\\sqrt 2 $$ "}, {"identifier": "B", "content": "$$20\\left( {\\sqrt 3 - 1} \\right)$$ "}, {"identifier": "C", "content": "$$40\\left( {\\sqrt 2 - 1} \\right)$$"}, {"identifier": "D", "content": "$$40\\left( {\\sqrt 3 - \\sqrt 2 } \\right)$$ "}] | ["B"] | null | Let the speed be $$y$$ $$m/sec$$.
<br><br>Let $$AC$$ be the vertical pole of height $$20$$ $$m.$$
<br><br>Let $$O$$ be the point on the ground such that $$\angle AOC = {45^ \circ }$$
<br><br>Let $$OC = x$$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267056/exam_image... | mcq | jee-main-2014-offline |
xU2eo9pTfAzBTHRg | maths | height-and-distance | height-and-distance | If the angles of elevation of the top of a tower from three collinear points $$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 $$ "}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l91njbgq/c332d856-caa2-46a4-8d1d-eda9ad96fe55/6437c2a0-47fc-11ed-8757-0f869593f41f/file-1l91njbgr.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l91njbgq/c332d856-caa2-46a4-8d1d-eda9ad96fe55/6437c2a0-47fc-11ed-8757-0f869593f41f/fi... | mcq | jee-main-2015-offline |
dkDTQQj1IIHo6pmfr63Q9 | maths | height-and-distance | height-and-distance | The angle of elevation of the top of a vertical tower from a point A, due east of it is 45<sup>o</sup>. The angle of elevation of the top of the same tower from a point B, due south of A is 30<sup>o</sup>. If the distance between A and B is $$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 $$ "}, {"identifier": "D", "content": "108"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267746/exam_images/sfbzoig20kahtq3la2hf.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2016 (Online) 10th April Morning Slot Mathematics - Height and Distance Question 30 English Explanation">
<br>... | mcq | jee-main-2016-online-10th-april-morning-slot |
l87d3ma7 | maths | height-and-distance | height-and-distance | A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is 30<sup>o</sup>. After walking for 10 minutes from A in the same direction, at a point B, he observes that the angle of elevation of the... | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "20"}, {"identifier": "D", "content": "5"}] | ["D"] | null | According to given information, we have the following figure<br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l87czgsa/9146332a-d177-4510-8b69-14ebf4d88eae/9d5eea90-3753-11ed-9417-1312e45a73c7/file-1l87czgsb.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l87czgsa/9146332... | mcq | jee-main-2016-offline |
DcXLZz7J4dXtDQAC | maths | height-and-distance | height-and-distance | Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point
on the ground such that AP = 2AB. If $$\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 $$
<br><... | mcq | jee-main-2017-offline |
u9KCmuptWLfgAMh2 | maths | height-and-distance | height-and-distance | PQR is a triangular park with PQ = PR = 200 m. A T.V. tower stands at the mid-point of QR. If the angles
of elevation of the top of the tower at P, Q and R are respectively 45$$^\circ $$, 30$$^\circ $$ and 30$$^\circ $$, then the height of the tower (in m) is : | [{"identifier": "A", "content": "$$50\\sqrt 2 $$"}, {"identifier": "B", "content": "100"}, {"identifier": "C", "content": "50"}, {"identifier": "D", "content": "$$100\\sqrt 3 $$"}] | ["B"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266852/exam_images/nlqfkpsnhv5bqrg6xjyo.webp" style="max-width:100%;height:auto;max-height:300px;margin: 0 auto" loading="lazy" alt="JEE Main 2018 (Offline) Mathematics - Height and Distance Question 35 English Explanation">
<br><b... | mcq | jee-main-2018-offline |
ZnQ7c1CR8ZKBJ5XWE59SM | maths | height-and-distance | height-and-distance | An aeroplane flying at a constant speed, parallel to the horizontal ground, $$\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 |
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