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The value of $$\sum\limits_{k = 1}^{10} {\left( {\sin {{2k\pi } \over {11}} + i\,\,\cos {{2k\pi } \over {11}}} \right)} $$ is :
Options:
[{"identifier": "A", "content": "i"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "- 1"}, {"identifier": "D", "content": "- i"}] | ["D"]
Explanation:
$$\sum\limits_{k = 1}^{10} {\left( {\sin {{2k\pi } \over {11}} + i\cos {{2k\pi } \over {11}}} \right)} $$
<br><br>$$ = i\sum\limits_{k = 1}^{10} {\left( {\cos {{2k\pi } \over {11}} - i\,\sin {{2k\pi } \over {11}}} \right)} $$
<br><br>$$ = i\sum\limits_{k = 1}^{10} {{e^{ - {{2k\pi } \over {11}}}}} i ... |
Let $$\alpha $$ and $$\beta $$ be two roots of the equation x<sup>2</sup> + 2x + 2 = 0 , then $$\alpha ^{15}$$ + $$\beta ^{15}$$ is equal to :
Options:
[{"identifier": "A", "content": "-256"}, {"identifier": "B", "content": "512"}, {"identifier": "C", "content": "-512"}, {"identifier": "D", "content": "256"}] | ["A"]
Explanation:
Given equation,
<br><br>x<sup>2</sup> + 2x + 2 = 0
<br><br>$$ \therefore $$ x = $${{ - 2 \pm \sqrt {4 - 4.1.2} } \over {2.1}}$$
<br><br>x = $$-$$ 1 $$ \pm $$ i
<br><br>$$ \therefore $$ $$\alpha $$ = $$-$$ 1 + i
<br><br>and $$\beta $$ = $$-$$ 1 $$-$$ i
<br><br><u>Note </u> :... |
If $$\alpha $$ and $$\beta $$ be the roots of the equation
x<sup>2</sup> – 2x + 2 = 0, then the least value of n for which $${\left( {{\alpha \over \beta }} \right)^n} = 1$$ is :
Options:
[{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "4"}, {"identifier": "D... | ["C"]
Explanation:
x<sup>2</sup> – 2x + 2 = 0
<br><br>$$ \therefore $$ x = $${{2 \pm \sqrt { - 4} } \over 2} = 1 \pm i$$
<br><br>Now, $${\alpha \over \beta } = {{1 + i} \over {1 - i}} = {{{{\left( {1 + i} \right)}^2}} \over {1 - {i^2}}} = i$$
<br><br>or $${\alpha \over \beta } = {{1 - i} \over {1 + i}} = {{{{\left( ... |
If $${\left( {{{1 + i} \over {1 - i}}} \right)^{{m \over 2}}} = {\left( {{{1 + i} \over {1 - i}}} \right)^{{n \over 3}}} = 1$$, (m, n
$$ \in $$ N) then the
greatest common divisor of the least values of
m and n is _______ .
Options:
[] | 4
Explanation:
$${\left( {{{1 + i} \over {1 - i}}} \right)^{m/2}} = {\left( {{{1 + i} \over {1 - i}}} \right)^{n/3}} = 1$$<br><br>$$ \Rightarrow {\left( {{{{{\left( {1 + i} \right)}^2}} \over 2}} \right)^{m/2}} = {\left( {{{{{\left( {1 + i} \right)}^2}} \over { - 2}}} \right)^{n/3}} = 1$$<br><br>$$ \Rightarrow {(i)^{m... |
Let $$z = {{1 - i\sqrt 3 } \over 2}$$, $$i = \sqrt { - 1} $$. Then the value of $$21 + {\left( {z + {1 \over z}} \right)^3} + {\left( {{z^2} + {1 \over {{z^2}}}} \right)^3} + {\left( {{z^3} + {1 \over {{z^3}}}} \right)^3} + .... + {\left( {{z^{21}} + {1 \over {{z^{21}}}}} \right)^3}$$ is ______________.
Options:
[] | 13
Explanation:
$$z = {{1 - \sqrt {3i} } \over 2} = {e^{ - i{\pi \over 3}}}$$<br><br>$${z^r} + {1 \over {{z^r}}} = 2\cos \left( { - {\pi \over 3}} \right)r = 2\cos {{r\pi } \over 3}$$<br><br>$$ \Rightarrow 21 + \sum\limits_{r = 1}^{21} {{{\left( {{z^r} + {1 \over {{z^r}}}} \right)}^3} = 8\left( {{{\cos }^3}{{r\pi } ... |
<p>If $$z$$ is a complex number, then the number of common roots of the equations $$z^{1985}+z^{100}+1=0$$ and $$z^3+2 z^2+2 z+1=0$$, is equal to</p>
Options:
[{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "3"}] | ["B"]
Explanation:
<p>$$\begin{array}{ll}
\text { } & z^{1985}+z^{100}+1=0 \& z^3+2 z^2+2 z+1=0 \\
& (z+1)\left(z^2-z+1\right)+2 z(z+1)=0 \\
& (z+1)\left(z^2+z+1\right)=0 \\
\Rightarrow \quad & z=-1, \quad z=w, w^2 \\
& \text { Now putting } z=-1 \text { not satisfy } \\
& \text { Now put } z=w \\
\Rightarrow \quad & ... |
<p>If the set $$R=\{(a, b): a+5 b=42, a, b \in \mathbb{N}\}$$ has $$m$$ elements and $$\sum_\limits{n=1}^m\left(1-i^{n !}\right)=x+i y$$, where $$i=\sqrt{-1}$$, then the value of $$m+x+y$$ is</p>
Options:
[{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "8"}, {... | ["A"]
Explanation:
<p>$$R=\{(a, b): a+5 b=42\}$$</p>
<p>Then $$R=\{(2,8),(7,7),(12,6),(17,5),(22,4),(27, 3),(32,2),(37,1)\}$$</p>
<p>$$\begin{aligned}
& \text { and } \sum_{n=1}^{\substack{m=8}}\left(1-i^{n!}\right)=x+i y \\
& \therefore \sum_{n=1}^8\left(1-i^{n!}\right)=8-\left(i+i^2+i^6+1+1+1+1+1\right) \\
& =5-i \\... |
The imaginary part of
<br/>$${\left( {3 + 2\sqrt { - 54} } \right)^{{1 \over 2}}} - {\left( {3 - 2\sqrt { - 54} } \right)^{{1 \over 2}}}$$ can be :
Options:
[{"identifier": "A", "content": "-2$$\\sqrt 6 $$"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "$$\\sqrt 6 $$"}, {"identifier": "D", "con... | ["A"]
Explanation:
$$3 + 2\sqrt { - 54} $$<br><br>
$$ = 9 - 6 + 2\sqrt { - 54} $$<br><br>
$$ = 9 + {\left( {\sqrt 6 i} \right)^2} + 2.3.\sqrt 6 i$$<br><br>
$$ = {3^2} + {\left( {\sqrt 6 i} \right)^2} + 2.3.\sqrt 6 i$$<br><br>
$$ = {\left( {3 + \sqrt 6 i} \right)^2}$$<br><br>
Similarly, $$\left( {3 - 2\sqrt { - 54} } \... |
$$\mathop {\lim }\limits_{n \to \infty } {{{1^p} + {2^p} + {3^p} + ..... + {n^p}} \over {{n^{p + 1}}}}$$ is
Options:
[{"identifier": "A", "content": "$${1 \\over {p + 1}}$$"}, {"identifier": "B", "content": "$${1 \\over {1 - p}}$$"}, {"identifier": "C", "content": "$${1 \\over p} - {1 \\over {p - 1}}$$"}, {"identifier... | ["A"]
Explanation:
We have $$\mathop {\lim }\limits_{x \to \infty } {{{1^p} + {2^p} + .... + {n^p}} \over {{n^{p + 1}}}};$$
<br><br>$$\mathop {\lim }\limits_{x \to \infty } \sum\limits_{r = 1}^n {{{{r^p}} \over {{n^p}.n}}} = \int\limits_0^1 {{x^p}} dx$$
<br><br>$$ = {\left[ {{{{x^{p + 1}}} \over {p + 1}}} \right]_0} ... |
$$\mathop {\lim }\limits_{n \to \infty } {{1 + {2^4} + {3^4} + .... + {n^4}} \over {{n^5}}}$$ - $$\mathop {\lim }\limits_{n \to \infty } {{1 + {2^3} + {3^3} + .... + {n^3}} \over {{n^5}}}$$
Options:
[{"identifier": "A", "content": "$${1 \\over 5}$$"}, {"identifier": "B", "content": "$${1 \\over 30}$$"}, {"identifier":... | ["A"]
Explanation:
The given expression can be written as
<br><br>$$\mathop {\lim }\limits_{n \to \infty } {1 \over n}{\sum\limits_{r = 1}^n {\left( {{r \over n}} \right)} ^4} - \mathop {\lim }\limits_{n \to \infty } {1 \over n}.\mathop {\lim }\limits_{n \to \infty } {1 \over n}{\left( {{r \over n}} \right)^3}$$
<br><... |
$$\mathop {Lim}\limits_{n \to \infty } \sum\limits_{r = 1}^n {{1 \over n}{e^{{r \over n}}}} $$ is
Options:
[{"identifier": "A", "content": "$$e+1$$ "}, {"identifier": "B", "content": "$$e-1$$ "}, {"identifier": "C", "content": "$$1-e$$"}, {"identifier": "D", "content": "$$e$$"}] | ["B"]
Explanation:
$$\mathop {Lim}\limits_{n \to \infty } \sum\limits_{r = 1}^n {{1 \over n}} {e^{{r \over n}}}\,\,$$
<br><br>$$\left[ {} \right.$$ Using definite integrals as limit of sum $$\left. {} \right]$$
<br><br>$$ = \int\limits_\theta ^1 {{e^x}} dx = e - 1$$ |
$$\mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over {{n^2}}}{{\sec }^2}{1 \over {{n^2}}} + {2 \over {{n^2}}}{{\sec }^2}{4 \over {{n^2}}}.... + {1 \over n}{{\sec }^2}1} \right]$$
<br/>equals
Options:
[{"identifier": "A", "content": "$${1 \\over 2}\\sec 1$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$cosec... | ["D"]
Explanation:
$$\mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over {{n^2}}}{{\sec }^2}{1 \over {{n^2}}} + {2 \over {{n^2}}}{{\sec }^2}{4 \over {{n^2}}} + {3 \over {{n^2}}}se{c^2}{9 \over {{n^2}}} + ... + {1 \over n}{{\sec }^2}1} \right]$$
<br><br>$$ = \mathop {\lim }\limits_{n \to \infty } {r \over {{n^2}}}{... |
$$\mathop {\lim }\limits_{n \to \infty } {\left( {{{\left( {n + 1} \right)\left( {n + 2} \right)...3n} \over {{n^{2n}}}}} \right)^{{1 \over n}}}$$ is equal to:
Options:
[{"identifier": "A", "content": "$${9 \\over {{e^2}}}$$ "}, {"identifier": "B", "content": "$$3\\,\\log \\,3 - 2$$ "}, {"identifier": "C", "content": ... | ["D"]
Explanation:
$$y = \mathop {\lim }\limits_{n \to \infty } {\left( {{{\left( {n + 1} \right)\left( {n + 2} \right)...3n} \over {{n^{2n}}}}} \right)^{{1 \over n {}}}}$$
<br><br>$$\ln \,y = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\ln \left( {1 + {1 \over n}} \right)\left( {1 + {2 \over n}} \right)$$
<br><... |
If $$\mathop {\lim }\limits_{n \to \infty } \,\,{{{1^a} + {2^a} + ...... + {n^a}} \over {{{(n + 1)}^{a - 1}}\left[ {\left( {na + 1} \right) + \left( {na + 2} \right) + ..... + \left( {na + n} \right)} \right]}} = {1 \over {60}}$$
<br/><br/>for some positive real number a, then a is equal to :
Options:
[{"identifie... | ["A"]
Explanation:
$$\mathop {\lim }\limits_{n \to \infty } {{{1 \over {\left( {a + 1} \right)}}\,.\,{n^{a + 1}} + {a_1}{n^a} + {a_2}{n^{a - 1}} + .\,.\,.\,.} \over {{{\left( {n + 1} \right)}^{a - 1}}.\,{n^2}\left( {a + {{1 + {1 \over n}} \over 2}} \right)}} = {1 \over {60}}$$
<br><br>$$ \Rightarrow $$ $$\... |
$$\mathop {\lim }\limits_{x \to \infty } \left( {{n \over {{n^2} + {1^2}}} + {n \over {{n^2} + {2^2}}} + {n \over {{n^2} + {3^2}}} + ..... + {1 \over {5n}}} \right)$$ is equal to :
Options:
[{"identifier": "A", "content": "tan<sup>\u20131 </sup>(2)"}, {"identifier": "B", "content": "tan<sup>\u20131 </sup>(3)"}, {"ide... | ["A"]
Explanation:
$$\mathop {\lim }\limits_{x \to \infty } \sum\limits_{r = 1}^{2n} {{n \over {{n^2} + {r^2}}}} $$
<br><br>$$\mathop {\lim }\limits_{x \to \infty } \sum\limits_{r = 1}^{2n} {{1 \over {n\left( {1 + {{{r^2}} \over {{n^2}}}} \right)}} = \int\limits_0^2 {{{dx} \over {1 + {x^2}}}} } = {\tan ^{ - 1}}2$$ |
$$\mathop {\lim }\limits_{n \to \infty } \left( {{{{{(n + 1)}^{1/3}}} \over {{n^{4/3}}}} + {{{{(n + 2)}^{1/3}}} \over {{n^{4/3}}}} + ....... + {{{{(2n)}^{1/3}}} \over {{n^{4/3}}}}} \right)$$
<br/>is equal to :
Options:
[{"identifier": "A", "content": "$${4 \\over 3}{\\left( 2 \\right)^{3/4}}$$"}, {"identifier": "B", "... | ["B"]
Explanation:
$$\mathop {\lim }\limits_{n \to \infty } {{{{(n + 1)}^{{1 \over 3}}} + {{(n + 2)}^{{1 \over 3}}} + ..... + {{\left( {n + n} \right)}^{{1 \over 3}}}} \over {n{{(n)}^{{1 \over 3}}}}}$$<br><br>
$$ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{{{{(n + r)}^{{1 \over 3}}}} \ov... |
$$\mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over n} + {n \over {{{(n + 1)}^2}}} + {n \over {{{(n + 2)}^2}}} + ........ + {n \over {{{(2n + 1)}^2}}}} \right]$$ is equal to :
Options:
[{"identifier": "A", "content": "$${{1 \\over 2}}$$"}, {"identifier": "B", "content": "$${{1 \\over 3}}$$"}, {"identifier": "C",... | ["A"]
Explanation:
$$\mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over n} + {n \over {{{(n + 1)}^2}}} + {n \over {{{(n + 2)}^2}}} + ... + {n \over {{{(2n - 1)}^2}}}} \right]$$
<br><br>$$\mathop {\lim }\limits_{n \to \infty } \left[ {{n \over {{{\left( {n + 0} \right)}^2}}} + {n \over {{{\left( {n + 1} \right)}^2... |
Let f : (0, 2) $$ \to $$ R be defined as f(x) = log<sub>2</sub>$$\left( {1 + \tan \left( {{{\pi x} \over 4}} \right)} \right)$$. Then, $$\mathop {\lim }\limits_{n \to \infty } {2 \over n}\left( {f\left( {{1 \over n}} \right) + f\left( {{2 \over n}} \right) + ... + f(1)} \right)$$ is equal to ___________.
Options:
[] | 1
Explanation:
$$E = 2\mathop {\lim }\limits_{x \to \infty } \sum\limits_{r = 1}^n {{1 \over n}} f\left( {{r \over n}} \right)$$<br><br>$$E = {2 \over {\ln 2}}\int_0^1 {\ln \left( {1 + \tan {{\pi x} \over 4}} \right)dx} $$ ..... (i)<br><br>replacing x $$ \to $$ 1 $$-$$ x<br><br>$$E = {2 \over {\ln 2}}\int_0^1 {\ln \le... |
The value of $$\mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{j = 1}^n {{{(2j - 1) + 8n} \over {(2j - 1) + 4n}}} $$ is equal to :
Options:
[{"identifier": "A", "content": "$$5 + {\\log _e}\\left( {{3 \\over 2}} \\right)$$"}, {"identifier": "B", "content": "$$2 - {\\log _e}\\left( {{2 \\over 3}} \\right... | ["D"]
Explanation:
$$\mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{j = 1}^n {{{\left( {{{2j} \over n} - {1 \over n} + 8} \right)} \over {\left( {{{2j} \over n} - {1 \over n} + 4} \right)}}} $$<br><br>$$\int\limits_0^1 {{{2x + 8} \over {2x + 4}}dx = \int\limits_0^1 {dx + \int\limits_0^1 {{4 \over {2x +... |
The value of <br/><br/>$$\mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 0}^{2n - 1} {{{{n^2}} \over {{n^2} + 4{r^2}}}} $$ is :
Options:
[{"identifier": "A", "content": "$${1 \\over 2}{\\tan ^{ - 1}}(2)$$"}, {"identifier": "B", "content": "$${1 \\over 2}{\\tan ^{ - 1}}(4)$$"}, {"identifier": "C", "c... | ["B"]
Explanation:
$$L = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 0}^{2n - 1} {{1 \over {1 + 4{{\left( {{r \over n}} \right)}^2}}}} $$<br><br>$$ \Rightarrow L = \int\limits_0^2 {{1 \over {1 + 4{x^2}}}dx} $$<br><br>$$ \Rightarrow L = \left. {{1 \over 2}{{\tan }^{ - 1}}(2x)} \right|_0^2 \Righta... |
If $${U_n} = \left( {1 + {1 \over {{n^2}}}} \right)\left( {1 + {{{2^2}} \over {{n^2}}}} \right)^2.....\left( {1 + {{{n^2}} \over {{n^2}}}} \right)^n$$, then $$\mathop {\lim }\limits_{n \to \infty } {({U_n})^{{{ - 4} \over {{n^2}}}}}$$ is equal to :
Options:
[{"identifier": "A", "content": "$${{{e^2}} \\over {16}}$$"},... | ["A"]
Explanation:
$${U_n} = \prod\limits_{r = 1}^n {{{\left( {1 + {{{r^2}} \over {{n^2}}}} \right)}^r}} $$<br><br>$$L = \mathop {\lim }\limits_{n \to \infty } {({U_n})^{ - 4/{n^2}}}$$<br><br>$$\log L = \mathop {\lim }\limits_{n \to \infty } {{ - 4} \over {{n^2}}}\sum\limits_{r = 1}^n {\log {{\left( {1 + {{{r^2}} \ove... |
<p>$$\mathop {\lim }\limits_{n \to \infty } \left( {{{{n^2}} \over {({n^2} + 1)(n + 1)}} + {{{n^2}} \over {({n^2} + 4)(n + 2)}} + {{{n^2}} \over {({n^2} + 9)(n + 3)}} + \,\,....\,\, + \,\,{{{n^2}} \over {({n^2} + {n^2})(n + n)}}} \right)$$ is equal to :</p>
Options:
[{"identifier": "A", "content": "$${\\pi \\over 8} ... | ["A"]
Explanation:
<p>$$\mathop {\lim }\limits_{n \to \infty } \left( {{{{n^2}} \over {({n^2} + 1)(n + 1)}} + {{{n^2}} \over {({n^2} + 4)(n + 2)}} + \,\,...\,\, + \,\,{{{n^2}} \over {({n^2} + {n^2})(n + n)}}} \right)$$</p>
<p>$$ = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{{{n^2}} \over {({n^2} + {... |
<p>$$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{r \over {2{r^2} - 7rn + 6{n^2}}}} $$ is equal to :</p>
Options:
[{"identifier": "A", "content": "$${\\log _e}\\left( {{{\\sqrt 3 } \\over 2}} \\right)$$"}, {"identifier": "B", "content": "$${\\log _e}\\left( {{{3\\sqrt 3 } \\over 4}} \\right)$$"}, {"i... | ["B"]
Explanation:
<p>$$\mathop {\lim }\limits_{n \to \alpha } \sum\limits_{r = 1}^n {{r \over {2{r^2} - 7rn + 6{n^2}}}} $$</p>
<p>$$ = \mathop {\lim }\limits_{n \to \alpha } {1 \over n}\sum\limits_{r = 1}^n {{{\left( {{r \over n}} \right)} \over {2{{\left( {{r \over n}} \right)}^2} - 7\left( {{r \over n}} \right) + 6... |
<p>$$
\begin{aligned}
&\text { If } \lim _{n \rightarrow \infty} \frac{(n+1)^{k-1}}{n^{k+1}}[(n k+1)+(n k+2)+\ldots+(n k+n)] \\
&=33 \cdot \lim _{n \rightarrow \infty} \frac{1}{n^{k+1}} \cdot\left[1^{k}+2^{k}+3^{k}+\ldots+n^{k}\right]
\end{aligned}$$,
then the integral value of $$\mathrm{k}$$ is equal to _____... | 5
Explanation:
$\lim\limits_{n \rightarrow \infty}\left(\frac{n+1}{n}\right)^{k-1} \frac{1}{n} \sum_{r=1}^{n}\left(k+\frac{r}{n}\right)=33 \lim\limits_{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n}\left(\frac{r}{n}\right)^{k}$
<br/><br/>
$$
\begin{aligned}
&\Rightarrow \int_{0}^{1}(k+x) d x=33 \int_{0}^{1} x^{k} d ... |
<p>$$\mathop {\lim }\limits_{n \to \infty } {1 \over {{2^n}}}\left( {{1 \over {\sqrt {1 - {1 \over {{2^n}}}} }} + {1 \over {\sqrt {1 - {2 \over {{2^n}}}} }} + {1 \over {\sqrt {1 - {3 \over {{2^n}}}} }} + \,\,...\,\, + \,\,{1 \over {\sqrt {1 - {{{2^n} - 1} \over {{2^n}}}} }}} \right)$$ is equal to</p>
Options:
[{"ident... | ["C"]
Explanation:
<p>$$I = \mathop {\lim }\limits_{n \to \infty } {1 \over {{2^n}}}\left( {{1 \over {\sqrt {1 - {1 \over {{2^n}}}} }} + {1 \over {\sqrt {1 - {2 \over {{2^n}}}} }} + {1 \over {\sqrt {1 - {3 \over {{2^n}}}} }} + \,\,.....\,\, + \,{1 \over {\sqrt {1 - {{{2^n} - 1} \over {{2^n}}}} }}} \right)$$</p>
<p>Let... |
<p>If $$a = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {{{2n} \over {{n^2} + {k^2}}}} $$ and $$f(x) = \sqrt {{{1 - \cos x} \over {1 + \cos x}}} $$, $$x \in (0,1)$$, then :</p>
Options:
[{"identifier": "A", "content": "$$2\\sqrt 2 f\\left( {{a \\over 2}} \\right) = f'\\left( {{a \\over 2}} \\right)$$"... | ["C"]
Explanation:
<p>$$a = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {{{2n} \over {{n^2} + {k^2}}}} $$</p>
<p>$$ = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{k = 1}^n {{2 \over {1 + {{\left( {{k \over n}} \right)}^2}}}} $$</p>
<p>$$a = \int\limits_0^1 {{2 \over {1 + {x^2}}}dx = ... |
<p>$$\mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over {1 + n}} + {1 \over {2 + n}} + {1 \over {3 + n}}\, + \,...\, + \,{1 \over {2n}}} \right]$$ is equal to</p>
Options:
[{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "$${\\log _e}2$$"}, {"identifier": "C", "content": "$${\\log _e}\\left( {... | ["B"]
Explanation:
$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left[\frac{1}{1+n}+\frac{1}{2+n}+\frac{1}{3+n}+\ldots \ldots+\frac{1}{2 n}\right] \\\\
& =\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{r+n} \\\\
& =\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{n}\left(\frac{1}{\frac{r}{n}+1}\right) \\\\
&... |
$\lim\limits_{n \rightarrow \infty} \frac{3}{n}\left\{4+\left(2+\frac{1}{n}\right)^2+\left(2+\frac{2}{n}\right)^2+\ldots+\left(3-\frac{1}{n}\right)^2\right\}$ is equal to :
Options:
[{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "$\\frac{19}{3}$"}, {"identifier": "C", "content": "19"}, {"identifi... | ["C"]
Explanation:
<p>$$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 0}^{n - 1} {{3 \over n}{{\left( {2 + {r \over n}} \right)}^2}} $$</p>
<p>$$ = \int_0^1 {3{{(2 + x)}^2}\,dx} $$</p>
<p>$$ = \left. {3\,.\,{{{{(2 + x)}^3}} \over 3}} \right|_0^1$$</p>
<p>$$ = {3^3} - {2^3} = 19$$</p> |
<p>Among</p>
<p>(S1): $$\lim_\limits{n \rightarrow \infty} \frac{1}{n^{2}}(2+4+6+\ldots \ldots+2 n)=1$$</p>
<p>(S2) : $$\lim_\limits{n \rightarrow \infty} \frac{1}{n^{16}}\left(1^{15}+2^{15}+3^{15}+\ldots \ldots+n^{15}\right)=\frac{1}{16}$$</p>
Options:
[{"identifier": "A", "content": "Only (S1) is true"}, {"identifie... | ["B"]
Explanation:
$$
\begin{aligned}
& S_1: \lim _{n \rightarrow \infty} \frac{1}{n^2}[2+4+6+\ldots+2 n] \\\\
& \lim _{n \rightarrow \infty} 2 \frac{n(n+1)}{2 n^2}=1 \\\\
& S_2: \lim _{n \rightarrow \infty} \frac{1}{n^{16}}\left(\sum r^{15}\right)=\lim _{n \rightarrow \infty} \frac{1}{n} \sum\left(\frac{r}{n}\right)... |
<p>Let $$f(x)=\frac{x}{\left(1+x^{n}\right)^{\frac{1}{n}}}, x \in \mathbb{R}-\{-1\}, n \in \mathbb{N}, n > 2$$.</p>
<p>If $$f^{n}(x)=\left(f \circ f \circ f \ldots .\right.$$. upto $$n$$ times) $$(x)$$, then
<br/><br/>$$\lim _\limits{n \rightarrow \infty} \int_\limits{0}^{1} x^{n-2}\left(f^{n}(x)\right) d x$$ is eq... | 0
Explanation:
$$
\begin{aligned}
& \text { We have, } f(x)=\frac{x}{\left(1+x^n\right)^{1 / n}} \\\\
& \therefore f(f(x))=\frac{f(x)}{\left(1+\left[f(x)^n\right]^{1 / n}\right.}\\\\
&=\frac{\frac{x}{\left(1+x^n\right)^{1 / n}}}{\left(1+\frac{x^n}{1+x^n}\right)^{1 / n}}\\\\
&=\frac{x}{\left(1+2 x^n\right)^{1 / n}} \\\... |
<p>The value of $$\lim _\limits{n \rightarrow \infty} \sum_\limits{k=1}^n \frac{n^3}{\left(n^2+k^2\right)\left(n^2+3 k^2\right)}$$ is :</p>
Options:
[{"identifier": "A", "content": "$$\\frac{\\pi}{8(2 \\sqrt{3}+3)}$$\n"}, {"identifier": "B", "content": "$$\\frac{(2 \\sqrt{3}+3) \\pi}{24}$$\n"}, {"identifier": "C", "co... | ["C"]
Explanation:
<p>$$\begin{aligned}
& \lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{n^3}{n^4\left(1+\frac{k^2}{n^2}\right)\left(1+\frac{3 k^2}{n^2}\right)} \\
& =\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \frac{n^3}{\left(1+\frac{k^2}{n^2}\right)\left(1+\frac{3 k^2}{n^2}\right)}
\end{aligned}$$</p>
<p... |
<p>Let $$\lim _\limits{n \rightarrow \infty}\left(\frac{n}{\sqrt{n^4+1}}-\frac{2 n}{\left(n^2+1\right) \sqrt{n^4+1}}+\frac{n}{\sqrt{n^4+16}}-\frac{8 n}{\left(n^2+4\right) \sqrt{n^4+16}}\right.$$ $$\left.+\ldots+\frac{n}{\sqrt{n^4+n^4}}-\frac{2 n \cdot n^2}{\left(n^2+n^2\right) \sqrt{n^4+n^4}}\right)$$ be $$\frac{\pi}{k... | 32
Explanation:
<p>$$\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\frac{n}{\sqrt{n^4+1}}+\frac{n}{\sqrt{n^4+16}}+\ldots \frac{n}{\sqrt{n^4+n^4}}\right) \\
& -\lim _{n \rightarrow \infty}\left(\frac{2 n}{\left(n^2+1\right)\left(\sqrt{n^4+1}\right)}\right)+\frac{8 n}{\left(n^2+4\right) \sqrt{n^4+1}}+\cdots \frac{... |
The value of $$\mathop {\lim }\limits_{x \to 0} {{\int\limits_0^{{x^2}} {{{\sec }^2}tdt} } \over xsinx}$$ is
Options:
[{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "1"}] | ["D"]
Explanation:
$$\mathop {\lim }\limits_{x \to 0} {{{d \over {dx}}\int\limits_0^{{x^2}} {{{\sec }^2}tdt} } \over {{d \over x}\left( {x\sin x} \right)}}$$
<br><br>$$ = \mathop {\lim }\limits_{x \to 0} {{{{\sec }^2}{x^2}.2x} \over {\sin \,x + x\,\cos \,x}}$$ (by $$L'$$ Hospital rule)
<br><br>$$\mathop {\lim }\limits... |
Let $$f:R \to R$$ be a differentiable function having $$f\left( 2 \right) = 6$$,
<br/>$$f'\left( 2 \right) = \left( {{1 \over {48}}} \right)$$. Then $$\mathop {\lim }\limits_{x \to 2} \int\limits_6^{f\left( x \right)} {{{4{t^3}} \over {x - 2}}dt} $$ equals :
Options:
[{"identifier": "A", "content": "$$24$$ "}, {"i... | ["D"]
Explanation:
$$\mathop {\lim }\limits_{x \to 2} \int\limits_6^{f\left( x \right)} {{{4{t^3}} \over {x - 2}}} dt$$
<br><br>$$ = \mathop {\lim }\limits_{x \to 0} {{\int\limits_6^{f\left( x \right)} {4{t^3}dt} } \over {x - 2}}$$
<br/><br/>This limit resembles a derivative because the fraction has the form $0/0$ as ... |
For x $$ \in $$ <b>R</b>, x $$ \ne $$ 0, if y(x) is a differentiable function such that
<br/><br/>x $$\int\limits_1^x y $$ (t) dt = (x + 1) $$\int\limits_1^x ty $$ (t) dt, then y (x) equals :
<br/><br/>(where C is a constant.)
Options:
[{"identifier": "A", "content": "$${C \\over x}{e^{ - {1 \\over x}}}$$"}, {"ide... | ["C"]
Explanation:
$$x\int\limits_1^x {y\left( t \right)dt} = x\int\limits_1^x {ty} \left( t \right)dt + \int\limits_1^x {ty\left( t \right)} \,\,dt$$
<br><br>Differentiate w.r. to x.
<br><br>$$\int\limits_1^x {y\left( t \right)dt + x\left[ {y\left( x \right) - y\left( 1 \right)} \right]} $$
<br><br>$$ = \int\limits_... |
Let f be a differentiable function from
<br/><br/><b>R</b> to <b>R</b> such that $$\left| {f\left( x \right) - f\left( y \right)} \right| \le 2{\left| {x - y} \right|^{{3 \over 2}}},$$ <br/><br/>for all $$x,y \in $$ <b>R</b>.
<br/><br/>If $$f\left( 0 \right) = 1$$
<br/><br/>then $$\int\limits_0^1 {{f^2}} ... | ["A"]
Explanation:
$$\left| {f(x) - f(y)} \right| \le 2{\left[ {x - y} \right]^{3/2}}$$
<br><br>$$\left| {{{f(x) - f(y)} \over {x - y}}} \right| \le 2{\left| {x - y} \right|^{1/2}}$$
<br><br>$$\mathop {\lim }\limits_{y \to x} \left| {{{f(x) - f(y)} \over {x - y}}} \right| \le \mathop {\lim }\limits_{y \to x} 2{\left| ... |
If $$\int\limits_0^x \, $$f(t) dt = x<sup>2</sup> + $$\int\limits_x^1 \, $$ t<sup>2</sup>f(t) dt then f '$$\left( {{1 \over 2}} \right)$$ is -
Options:
[{"identifier": "A", "content": "$${{18} \\over {25}}$$"}, {"identifier": "B", "content": "$${{6} \\over {25}}$$"}, {"identifier": "C", "content": "$${{24} \\over {... | ["C"]
Explanation:
$$\int\limits_0^x \, $$f(t) dt = x<sup>2</sup> + $$\int\limits_x^1 \, $$ t<sup>2</sup>f(t) dt f '$$\left( {{1 \over 2}} \right)$$ = ?
<br><br>Differentiate w.r.t. 'x'
<br><br>f(x) = 2x + 0 $$-$$ x<sup>2</sup> f(x)
<br><br>f(x) = $${{2x} \o... |
If f : R $$ \to $$ R is a differentiable function and
f(2) = 6,<br/> then $$\mathop {\lim }\limits_{x \to 2} {{\int\limits_6^{f\left( x \right)} {2tdt} } \over {\left( {x - 2} \right)}}$$ is :-
Options:
[{"identifier": "A", "content": "2f'(2)"}, {"identifier": "B", "content": "24f'(2)"}, {"identifier": "C", "content":... | ["D"]
Explanation:
$$\mathop {\lim }\limits_{x \to 2} {{\int\limits_6^{f\left( x \right)} {2tdt} } \over {\left( {x - 2} \right)}}$$
<br><br>This is $${0 \over 0}$$ form so we use L – Hospital Rule.
<br><br>= $$\mathop {\lim }\limits_{x \to 2} {{2f\left( x \right).f'\left( x \right) - 0} \over 1}$$
<br><br>= $${2f\le... |
Let f : R $$ \to $$ R be a continuously differentiable function such that f(2) = 6 and f'(2) = $${1 \over {48}}$$. If $$\int\limits_6^{f\left( x \right)} {4{t^3}} dt$$ = (x - 2)g(x), then $$\mathop {\lim }\limits_{x \to 2} g\left( x \right)$$ is equal to :
Options:
[{"identifier": "A", "content": "18"}, {"identifier":... | ["A"]
Explanation:
Given $$\int\limits_6^{f(x)} {4{x^3}dx} = g(x).(x - 2)$$
<br><br>$$ \Rightarrow g(x) = $$ $${{\int\limits_0^{f\left( x \right)} {4{x^3}dx} } \over {x - 2}}$$<br><br>
$$ \therefore $$ $$\mathop {\lim }\limits_{x \to 2} g(x) = $$$$\mathop {\lim }\limits_{x \to 2} {{\int\limits_0^{f\left( x \right)} {... |
$$\mathop {\lim }\limits_{x \to 0} {{\int_0^x {t\sin \left( {10t} \right)dt} } \over x}$$ is equal to
Options:
[{"identifier": "A", "content": "$$ - {1 \\over 5}$$"}, {"identifier": "B", "content": "$$ - {1 \\over 10}$$"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "$$ {1 \\over 10}$$"}] | ["C"]
Explanation:
$$\mathop {\lim }\limits_{x \to 0} {{\int_0^x {t\sin \left( {10t} \right)dt} } \over x}$$
<br><br>This is in $${0 \over 0}$$ form.
<br><br>So apply newton leibniz rule
<br><br>$$\mathop {\lim }\limits_{x \to 0} {{x.\sin \left( {10x} \right) - 0} \over 1}$$ = 0 |
Let a function ƒ : [0, 5] $$ \to $$ R be continuous,
ƒ(1) = 3 and F be defined as :<br/><br/>
$$F(x) = \int\limits_1^x {{t^2}g(t)dt} $$ , where $$g(t) = \int\limits_1^t {f(u)du} $$ <br/><br/>Then for the function F, the point x = 1 is :
Options:
[{"identifier": "A", "content": "a point of inflection."}, {"identifier":... | ["C"]
Explanation:
$$F(x) = \int\limits_1^x {{t^2}g(t)dt} $$
<br><br>$$ \Rightarrow $$ F'(x) = x<sup>2</sup>g(x) = x<sup>2</sup>$$\int\limits_1^t {f(u)du} $$
<br><br>$$ \therefore $$ F'(1) = (1)(0) = 0
<br><br>Now, F''(x) = 2xg(x) + x<sup>2</sup>g'(x)
<br><br>F''(1) = 2g(1) + g'(1) = 0 + g'(1) = 3
<br><br>[ As g'(t) =... |
$$\mathop {\lim }\limits_{x \to 1} \left( {{{\int\limits_0^{{{\left( {x - 1} \right)}^2}} {t\cos \left( {{t^2}} \right)dt} } \over {\left( {x - 1} \right)\sin \left( {x - 1} \right)}}} \right)$$
Options:
[{"identifier": "A", "content": "is equal to 0"}, {"identifier": "B", "content": "is equal to $${1 \\over 2}$$"}, {... | ["A"]
Explanation:
$$\mathop {\lim }\limits_{x \to 1} \left( {{{\int\limits_0^{{{\left( {x - 1} \right)}^2}} {t\cos \left( {{t^2}} \right)dt} } \over {\left( {x - 1} \right)\sin \left( {x - 1} \right)}}} \right)$$$$\left( {{0 \over 0}} \right)$$
<br><br>Apply L Hospital Rule
<br><br>= $$\mathop {\lim }\limits_{x \to ... |
$$\mathop {\lim }\limits_{x \to 0} {{\int\limits_0^{{x^2}} {\left( {\sin \sqrt t } \right)dt} } \over {{x^3}}}$$ is equal to :
Options:
[{"identifier": "A", "content": "$${1 \\over {15}}$$"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "$${2 \\over 3}$$"}, {"identifier": "D", "content": "$${3 \... | ["C"]
Explanation:
$$\mathop {\lim }\limits_{x \to {0^ + }} {{\int\limits_0^{{x^2}} {\sin (\sqrt t )dt} } \over {{x^3}}}$$<br><br>This is in $${0 \over 0}$$ form, so use L' Hospital rule<br><br>$$ = \mathop {\lim }\limits_{x \to {0^ + }} {{{d \over {dx}}\left( {\int\limits_0^{{x^2}} {\sin (\sqrt t )dt} } \right)} \ove... |
Let $$f(x) = \int\limits_0^x {{e^t}f(t)dt + {e^x}} $$ be a differentiable function for all x$$\in$$R. Then f(x) equals :
Options:
[{"identifier": "A", "content": "$${e^{({e^{x - 1}})}}$$"}, {"identifier": "B", "content": "$$2{e^{{e^x}}} - 1$$"}, {"identifier": "C", "content": "$$2{e^{{e^x} - 1}} - 1$$"}, {"identifier"... | ["C"]
Explanation:
$$f(x) = \int\limits_0^x {{e^t}f(t)dt + {e^x}} $$ .... (1)<br><br>Differentiating both sides w.r.t. x<br><br>$$f'(x) = {e^x}.f(x) + {e^x}$$ (Using Newton L:eibnitz Theorem)<br><br>$$ \Rightarrow {{f'(x)} \over {f(x) + 1}} = {e^x}$$<br><br>Integrating w.r.t. x<br><br>$$\int {{{f'(x)} \over {f(x) + 1}... |
If the normal to the curve y(x) = $$\int\limits_0^x {(2{t^2} - 15t + 10)dt} $$ at a point (a, b) is parallel to the line x + 3y = $$-$$5, a > 1, then the value of | a + 6b | is equal to ___________.
Options:
[] | 406
Explanation:
Normal to the curve at point P(a, b) is parallel to the line x + 3y = $$-$$5.
<br><br>m<sub>normal</sub> = $$ - {1 \over 3}$$
<br><br>$$ \therefore $$ m<sub>tangent</sub> = 3 = $${{dy} \over {dx}}$$
<br><br>Given y(x) = $$\int\limits_0^x {(2{t^2} - 15t + 10)dt} $$
<br><br>$$ \Rightarrow $$ y'(x) = (2x... |
Let f : R $$ \to $$ R be defined as f(x) = e<sup>$$-$$x</sup>sinx. If F : [0, 1] $$ \to $$ R is a differentiable function with that F(x) = $$\int_0^x {f(t)dt} $$, then the value of $$\int_0^1 {(F'(x) + f(x)){e^x}dx} $$ lies in the interval
Options:
[{"identifier": "A", "content": "$$\\left[ {{{331} \\over {360}},{{334... | ["B"]
Explanation:
F(x) = $$\int_0^x {f(t)dt} $$
<br><br>$$ \Rightarrow $$ F'(x) = f(x) by Leibnitz theorem<br><br>I = $$\int\limits_0^1 {(F'(x) + f(x)){e^x}dx = \int\limits_0^1 {2f(x){e^x}dx} } $$<br><br>$$I = \int\limits_0^1 {2\sin x\,dx} $$<br><br>$$I = 2(1 - \cos 1)$$<br><br>$$ = 2\left\{ {1 - \left( {1 - {{{1^2}}... |
Let f : (a, b) $$\to$$ R be twice differentiable function such that $$f(x) = \int_a^x {g(t)dt} $$ for a differentiable function g(x). If f(x) = 0 has exactly five distinct roots in (a, b), then g(x)g'(x) = 0 has at least :
Options:
[{"identifier": "A", "content": "twelve roots in (a, b)"}, {"identifier": "B", "content... | ["C"]
Explanation:
$$f(x) = \int_a^x {g(t)dt} $$
<br><br>$$ \Rightarrow $$ f′(x) = g(x)
<br><br>$$ \Rightarrow $$ f′'(x) = g'(x)
<br><br>Given, g(x).g'(x) = 0
<br><br>$$ \Rightarrow $$ f′(x).f′'(x) = 0
<br><br>Also given f(x) has exactly 5 roots.
<br><br>So from Rolle's theorem we can say,
<br><br>f′(x) has 4 roots an... |
Let $$F:[3,5] \to R$$ be a twice differentiable function on (3, 5) such that <br/><br/>$$F(x) = {e^{ - x}}\int\limits_3^x {(3{t^2} + 2t + 4F'(t))dt} $$. If $$F'(4) = {{\alpha {e^\beta } - 224} \over {{{({e^\beta } - 4)}^2}}}$$, then $$\alpha$$ + $$\beta$$ is equal to _______________.
Options:
[] | 16
Explanation:
$$F(3) = 0$$<br><br>$${e^x}F(x) = \int\limits_3^x {(3{t^2} + 2t + 4F'(t))dt} $$<br><br>$${e^x}F(x) + {e^x}F'(x) = 3{x^2} + 2x + 4F'(x)$$<br><br>$$({e^x} - 4){{dy} \over {dx}} + {e^x}y = (3{x^2} + 2x)$$<br><br>$${{dy} \over {dx}} + {{{e^x}} \over {({e^x} - 4)}}y = {{(3{x^2} + 2x)} \over {({e^x} - 4)}}$$... |
Let f be a non-negative function in [0, 1] and twice differentiable in (0, 1). If $$\int_0^x {\sqrt {1 - {{(f'(t))}^2}} dt = \int_0^x {f(t)dt} } $$, $$0 \le x \le 1$$ and f(0) = 0, then $$\mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\int_0^x {f(t)dt} $$ :
Options:
[{"identifier": "A", "content": "equals 0"}, {"id... | ["D"]
Explanation:
$$\int_0^x {\sqrt {1 - {{(f'(t))}^2}} dt = \int_0^x {f(t)dt} } ,\,0 \le x \le 1$$<br><br>differentiating both the sides<br><br>$$\sqrt {1 - {{(f'(x))}^2}} = f(x)$$<br><br>$$ \Rightarrow 1 - {(f'(x))^2} = {f^2}(x)$$<br><br>$${{f'(x)} \over {\sqrt {1 - {f^2}(x)} }} = 1$$<br><br>$${\sin ^{ - 1}}f(x) =... |
Let f : R $$\to$$ R be a continuous function. Then $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{{\pi \over 4}\int\limits_2^{{{\sec }^2}x} {f(x)\,dx} } \over {{x^2} - {{{\pi ^2}} \over {16}}}}$$ is equal to :
Options:
[{"identifier": "A", "content": "f (2)"}, {"identifier": "B", "content": "2f (2)"}, {"identifier... | ["B"]
Explanation:
$$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{{\pi \over 4}\int\limits_2^{{{\sec }^2}x} {f(x)\,dx} } \over {{x^2} - {{{\pi ^2}} \over {16}}}}$$<br><br>= $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {\pi \over 4}.{{\left[ {f({{\sec }^2}x).2\sec x.\sec x\tan x} \right]} \over {2x}}$$<br><br>=... |
<p>If m and n respectively are the number of local maximum and local minimum points of the function $$f(x) = \int\limits_0^{{x^2}} {{{{t^2} - 5t + 4} \over {2 + {e^t}}}dt} $$, then the ordered pair (m, n) is equal to</p>
Options:
[{"identifier": "A", "content": "(3, 2)"}, {"identifier": "B", "content": "(2, 3)"}, {"id... | ["B"]
Explanation:
<p>$$f(x) = \int_0^{{x^2}} {{{{t^2} - 5t + 4} \over {2 + {e^t}}}dt} $$</p>
<p>$$f'(x) = 2x\left( {{{{x^4} - 5{x^2} + 4} \over {2 + {e^{{x^2}}}}}} \right) = 0$$</p>
<p>$$x = 0$$, or $$({x^2} - 4)({x^2} - 1) = 0$$</p>
<p>$$x = 0,$$ $$x = \pm 2,\, \pm 1$$</p>
<p>Now, $$f'(x) = {{2x(x + 1)(x - 1)(x + 2... |
<p>Let f be a differentiable function in $$\left( {0,{\pi \over 2}} \right)$$. If $$\int\limits_{\cos x}^1 {{t^2}\,f(t)dt = {{\sin }^3}x + \cos x} $$, then $${1 \over {\sqrt 3 }}f'\left( {{1 \over {\sqrt 3 }}} \right)$$ is equal to</p>
Options:
[{"identifier": "A", "content": "$$6 - 9\\sqrt 2 $$"}, {"identifier": "B"... | ["B"]
Explanation:
<p>$$\int\limits_{\cos x}^1 {{t^2}f(t)dt = {{\sin }^3}x + \cos x} $$</p>
<p>$$ \Rightarrow \sin x{\cos ^2}x\,f(\cos x) = 3{\sin ^2}x\cos x - \sin x$$</p>
<p>$$ \Rightarrow f(\cos x) = 3\tan x - {\sec ^2}x$$</p>
<p>$$ \Rightarrow f'(\cos x).\,( - \sin x) = 3{\sec ^2}x - 2{\sec ^2}x\tan x$$</p>
<p><im... |
<p>Let $$f$$ be a twice differentiable function on $$\mathbb{R}$$. If $$f^{\prime}(0)=4$$ and $$f(x) + \int\limits_0^x {(x - t)f'(t)dt = \left( {{e^{2x}} + {e^{ - 2x}}} \right)\cos 2x + {2 \over a}x} $$, then $$(2 a+1)^{5}\, a^{2}$$ is equal to _______________.</p>
Options:
[] | 8
Explanation:
$$
\begin{aligned}
\because f(x)+\int_0^x(x-t) f^{\prime}(t) d t & =\left(e^{2 x}+e^{-2 x}\right) \cos 2 x+\frac{2 x}{a} ~~...(i)
\end{aligned}
$$<br/><br/>
Here $f(0)=2 \hspace{0.5cm} ...(ii)$<br/><br/>
On differentiating equation (i) w.r.t. $x$ we get :<br/><br/>
$$
\begin{aligned}
& f^{\prime}(x)+\i... |
<p>Let f be a differentiable function satisfying $$f(x)=\frac{2}{\sqrt{3}} \int\limits_{0}^{\sqrt{3}} f\left(\frac{\lambda^{2} x}{3}\right) \mathrm{d} \lambda, x>0$$ and $$f(1)=\sqrt{3}$$. If $$y=f(x)$$ passes through the point $$(\alpha, 6)$$, then $$\alpha$$ is equal to _____________.</p>
Options:
[] | 12
Explanation:
<p>$$\because$$ $$f(x) = {2 \over {\sqrt 3 }}\int\limits_0^{\sqrt 3 } {f\left( {{{{\lambda ^2}x} \over 3}} \right)d\lambda ,\,x > 0} $$</p>
<p>On differentiating both sides w.r.t., x, we get</p>
<p>$$f'(x) = {2 \over {\sqrt 3 }}\int\limits_0^{\sqrt 3 } {{{{\lambda ^2}} \over 3}f'\left( {{{{\lambda ^2}x... |
<p>The minimum value of the twice differentiable function $$f(x)=\int\limits_{0}^{x} \mathrm{e}^{x-\mathrm{t}} f^{\prime}(\mathrm{t}) \mathrm{dt}-\left(x^{2}-x+1\right) \mathrm{e}^{x}$$, $$x \in \mathbf{R}$$, is :</p>
Options:
[{"identifier": "A", "content": "$$-\\frac{2}{\\sqrt{\\mathrm{e}}}$$"}, {"identifier": "B", ... | ["A"]
Explanation:
<p>$$f(x) = \int\limits_0^x {{e^{x - t}}f'(t)dt - ({x^2} - x + 1){e^x}} $$</p>
<p>$$f(x) = {e^x}\int\limits_0^x {{e^{ - t}}f'(t)dt - ({x^2} - x + 1){e^x}} $$</p>
<p>$${e^{ - x}}f(x) = \int\limits_0^x {{e^{ - t}}f'(t)dt - ({x^2} - x + 1)} $$</p>
<p>Differentiate on both side</p>
<p>$${e^{ - x}}f'(x) ... |
If $\phi(x)=\frac{1}{\sqrt{x}} \int\limits_{\frac{\pi}{4}}^x\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t, x>0$,
<br/><br/>then $\emptyset^{\prime}\left(\frac{\pi}{4}\right)$ is equal to :
Options:
[{"identifier": "A", "content": "$\\frac{4}{6+\\sqrt{\\pi}}$"}, {"identifier": "B", "content": "$\\frac{4}{6-... | ["D"]
Explanation:
$\phi(x)=\frac{1}{\sqrt{x}} \int_{\pi / 4}^{x}\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t$
<br/><br/>$\Rightarrow \phi^{\prime}(x)=\frac{-1}{2 x^{3 / 2}} \int_{\pi / 4}^{x}\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t$
<br/><br/>$+\frac{1}{\sqrt{x}}\left(4 \sqrt{2} \sin (x)-3 \ph... |
<p>$$\lim_\limits{x \rightarrow 0} \frac{48}{x^{4}} \int_\limits{0}^{x} \frac{t^{3}}{t^{6}+1} \mathrm{~d} t$$ is equal to ___________.</p>
Options:
[] | 12
Explanation:
$$
48 \lim\limits_{x \rightarrow 0} \frac{\int_0^x \frac{t^3}{t^6+1} d t}{x^4}\left(\frac{0}{0}\right)
$$
<br/><br/>Applying L' Hospitals Rule
<br/><br/>$$
\begin{aligned}
48 \lim _{x \rightarrow 0} \frac{x^3}{x^6+1} \times \frac{1}{4 x^3}\\\\
\end{aligned}
$$
<p>= $${{48} \over 4}$$$$\mathop {\lim }\l... |
<p>Let $$f$$ be $$a$$ differentiable function defined on $$\left[ {0,{\pi \over 2}} \right]$$ such that $$f(x) > 0$$ and $$f(x) + \int_0^x {f(t)\sqrt {1 - {{({{\log }_e}f(t))}^2}} dt = e,\forall x \in \left[ {0,{\pi \over 2}} \right]}$$. Then $$\left( {6{{\log }_e}f\left( {{\pi \over 6}} \right)} \right)^2$$ is e... | 27
Explanation:
$f(x)+\int_{0}^{x} f(t) \sqrt{1-\left(\log _{e} f(t)\right)^{2}} d t=e\quad...(1)$
<br/><br/>
So, $f(0)=e$
<br/><br/>
Now differentiate w.r. to $x$
<br/><br/>
$$
\begin{gathered}
f^{\prime}(x)+f(x) \sqrt{1-\left(\log _{e} f(x)^{2}\right.}=0 \\\\
\frac{f^{\prime}(x)}{f(x) \sqrt{1-\left(\log _{e} f(x)\ri... |
<p>Let $$f$$ be a continuous function satisfying $$\int_\limits{0}^{t^{2}}\left(f(x)+x^{2}\right) d x=\frac{4}{3} t^{3}, \forall t > 0$$. Then
$$f\left(\frac{\pi^{2}}{4}\right)$$ is equal to :</p>
Options:
[{"identifier": "A", "content": "$$-\\pi\\left(1+\\frac{\\pi^{3}}{16}\\right)$$"}, {"identifier": "B", "conte... | ["B"]
Explanation:
Given that
<br/><br/>$$
\int\limits_0^{t^2}\left(f(x)+x^2\right) d x=\frac{4}{3} t^3, \forall t>0
$$
<br/><br/>On differentiating using Leibnitz rule, we get
<br/><br/>$$
\begin{aligned}
& \left(f\left(t^2\right)+t^4\right) \times 2 t=\frac{4}{3} \times 3 t^2 \\\\
& \Rightarrow f\left(t^2\right)+t^4... |
Let $f:(0, \infty) \rightarrow \mathbf{R}$ and $\mathrm{F}(x)=\int\limits_0^x \mathrm{t} f(\mathrm{t}) \mathrm{dt}$. If $\mathrm{F}\left(x^2\right)=x^4+x^5$, then $\sum\limits_{\mathrm{r}=1}^{12} f\left(\mathrm{r}^2\right)$ is equal to ____________.
Options:
[] | 219
Explanation:
$F(x)=\int\limits_0^x t \cdot f(t) d t$
<br/><br/>$\begin{aligned} & F'(x)=x f(x) \\\\ & F\left(x^2\right)=x^4+x^5, \quad \text { let } x^2=t \\\\ & F(t)=t^2+t^{5 / 2} \\\\ & F^{\prime}(t)=2 t+5 / 2 t^{3 / 2} \\\\ & t \cdot f(t)=2 t+5 / 2 t^{3 / 2} \\\\ & f(t)=2+5 / 2 r^{1 / 2}\end{aligned}$
<br/><br/... |
<p>Let $$S=(-1, \infty)$$ and $$f: S \rightarrow \mathbb{R}$$ be defined as</p>
<p>$$f(x)=\int_\limits{-1}^x\left(e^t-1\right)^{11}(2 t-1)^5(t-2)^7(t-3)^{12}(2 t-10)^{61} d t \text {, }$$</p>
<p>Let $$\mathrm{p}=$$ Sum of squares of the values of $$x$$, where $$f(x)$$ attains local maxima on $$S$$, and $$\mathrm{q}=$$ ... | 27
Explanation:
<p>$$\mathrm{f}^{\prime}(\mathrm{x})=\left(\mathrm{e}^{\mathrm{x}}-1\right)^{11}(2 \mathrm{x}-1)^5(\mathrm{x}-2)^7(\mathrm{x}-3)^{12}(2 \mathrm{x}-10)^{61}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsnjdht7/4e50bf09-665e-42ec-a15b-ac37fbbf6eff/783c33b0-cc2d-11ee-b2... |
<p>$$\mathop {\lim }\limits_{x \to {\pi \over 2}} \left( {{1 \over {{{\left( {x - {\pi \over 2}} \right)}^2}}}\int\limits_{{x^3}}^{{{\left( {{\pi \over 2}} \right)}^3}} {\cos \left( {{t^{{1 \over 3}}}} \right)dt} } \right)$$ is equal to</p>
Options:
[{"identifier": "A", "content": "$$\\frac{3 \\pi^2}{4}$$\n"}, {"id... | ["B"]
Explanation:
<p>Using L'hospital rule</p>
<p>$$\begin{aligned}
& =\lim _\limits{x \rightarrow \frac{\pi^{-}}{2}} \frac{0-\cos x \times 3 x^2}{2\left(x-\frac{\pi}{2}\right)} \\
& =\lim _\limits{x \rightarrow \frac{\pi^{-}}{2}} \frac{\sin \left(x-\frac{\pi}{2}\right)}{2\left(x-\frac{\pi}{2}\right)} \times \frac{3 ... |
<p>Let the slope of the line $$45 x+5 y+3=0$$ be $$27 r_1+\frac{9 r_2}{2}$$ for some $$r_1, r_2 \in \mathbb{R}$$. Then $$\lim _\limits{x \rightarrow 3}\left(\int_3^x \frac{8 t^2}{\frac{3 r_2 x}{2}-r_2 x^2-r_1 x^3-3 x} d t\right)$$ is equal to _________.</p>
Options:
[] | 12
Explanation:
<p>According to the question,</p>
<p>$$\begin{aligned}
& 27 r_1+\frac{9 r_2}{2}=-9 \\
& \lim _\limits{x \rightarrow 3} \frac{\int_\limits3^x 8 t^2 d t}{\frac{3 r_2 x}{2}-r_2 x^2-r_1 x^3-3 x} \\
& =\lim _\limits{x \rightarrow 3} \frac{8 x^2}{\frac{3 r_2^2}{2}-2 r_2 x-3 r_1 x^2-3} \text { (using LH' Rule... |
<p>Let $$\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R}$$ be defined as $$f(x)=a e^{2 x}+b e^x+c x$$. If $$f(0)=-1, f^{\prime}\left(\log _e 2\right)=21$$ and $$\int_0^{\log _e 4}(f(x)-c x) d x=\frac{39}{2}$$, then the value of $$|a+b+c|$$ equals</p>
Options:
[{"identifier": "A", "content": "16"}, {"identifier": "B", "c... | ["C"]
Explanation:
<p>$$\begin{array}{ll}
\mathrm{f}(\mathrm{x})=a \mathrm{e}^{2 \mathrm{x}}+b \mathrm{e}^{\mathrm{x}}+\mathrm{cx} & \mathrm{f}(0)=-1 \\\\
& \mathrm{a}+\mathrm{b}=-1 \\\\
\mathrm{f}^{\prime}(\mathrm{x})=2 a \mathrm{e}^{2 \mathrm{x}}+b \mathrm{e}^{\mathrm{x}}+\mathrm{c} & \mathrm{f}^{\prime}(\ln 2)=21 \... |
<p>$$\lim _\limits{x \rightarrow \frac{\pi}{2}}\left(\frac{\int_{x^3}^{(\pi / 2)^3}\left(\sin \left(2 t^{1 / 3}\right)+\cos \left(t^{1 / 3}\right)\right) d t}{\left(x-\frac{\pi}{2}\right)^2}\right)$$ is equal to</p>
Options:
[{"identifier": "A", "content": "$$\\frac{3 \\pi^2}{2}$$\n"}, {"identifier": "B", "content": "... | ["B"]
Explanation:
<p>$$\lim _\limits{x \rightarrow \frac{\pi}{2}}\left(\frac{\int_{x^3}^{(\pi / 2)^3}\left(\sin \left(2 t^{1 / 3}\right)+\cos \left(t^{1 / 3}\right)\right) d t}{\left(x-\frac{\pi}{2}\right)^2}\right)$$</p>
<p>Using Newton Leibniz theorem</p>
<p>$$\begin{aligned}
& =\lim _\limits{x \rightarrow \frac{\p... |
<p>Let $$f(x)=\int_0^x\left(t+\sin \left(1-e^t\right)\right) d t, x \in \mathbb{R}$$. Then, $$\lim _\limits{x \rightarrow 0} \frac{f(x)}{x^3}$$ is equal to</p>
Options:
[{"identifier": "A", "content": "$$\\frac{1}{6}$$\n"}, {"identifier": "B", "content": "$$-\\frac{1}{6}$$\n"}, {"identifier": "C", "content": "$$\\frac... | ["B"]
Explanation:
<p>Given $$f(x)=\int_\limits0^x\left(t+\sin \left(1-e^t\right)\right) d t$$</p>
<p>Now, $$\lim _\limits{x \rightarrow 0} \frac{f(x)}{x^3}\left(\frac{0}{0} \text { form }\right)$$</p>
<p>$$\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{\int_\limits0^x\left(t+\sin \left(1-e^t\right)\right) d t}{x^3}... |
$${I_n} = \int\limits_0^{\pi /4} {{{\tan }^n}x\,dx} $$ then $$\,\mathop {\lim }\limits_{n \to \infty } \,n\left[ {{I_n} + {I_{n + 2}}} \right]$$ equals
Options:
[{"identifier": "A", "content": "$${1 \\over 2}$$ "}, {"identifier": "B", "content": "$$1$$"}, {"identifier": "C", "content": "$$\\infty $$ "}, {"identifier"... | ["B"]
Explanation:
$${I_n} + {I_{n + 2}}$$
<br><br>$$ = \int\limits_0^{\pi /4} {{{\tan }^n}} \,x\left( {1 + {{\tan }^2}x} \right)dx$$
<br><br>$$ = \int\limits_0^{\pi /4} {{{\tan }^n}} x\,{\sec ^2}x\,dx$$
<br><br>$$ = \left[ {{{{{\tan }^{n + 1}}x} \over {n + 1}}} \right]_0^{\pi /4}$$
<br><br>$$ = {{1 - 0} \over {n + 1... |
$$\int\limits_0^{10\pi } {\left| {\sin x} \right|dx} $$ is
Options:
[{"identifier": "A", "content": "$$20$$"}, {"identifier": "B", "content": "$$8$$"}, {"identifier": "C", "content": "$$10$$"}, {"identifier": "D", "content": "$$18$$"}] | ["A"]
Explanation:
$$I = \int\limits_0^{10\pi } {\left| {\sin x} \right|} dx$$
<br><br>$$ = 10\int\limits_0^\pi {\left| {\sin x\,} \right|} \,dx$$
<br><br>$$ = 10\int\limits_0^\pi {\sin \,x\,dx} $$
<br><br>$$\left[ \, \right.$$ as $$\left| {\sin x} \right|$$ is periodic with period $$\pi $$
<br><br>and $$sin$$ $$x &... |
$$\int\limits_0^2 {\left[ {{x^2}} \right]dx} $$ is
Options:
[{"identifier": "A", "content": "$$2 - \\sqrt 2 $$ "}, {"identifier": "B", "content": "$$2 + \\sqrt 2 $$"}, {"identifier": "C", "content": "$$\\,\\sqrt 2 - 1$$ "}, {"identifier": "D", "content": "$$ - \\sqrt 2 - \\sqrt 3 + 5$$ "}] | ["D"]
Explanation:
$$\int\limits_0^2 {\left[ {{x^2}} \right]} dx = \int\limits_0^1 {\left[ {{x^2}} \right]dx} + \int\limits_1^{\sqrt 2 } {\left[ {{x^2}} \right]} dx + $$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$$$\int\limits_{\sqrt 2 }^{\sqrt 3 } {\left[ {{x^2}} \right]} + \int\limits_{\sqrt 3 }^2 {\left[ {{x^2... |
$$\int_{ - \pi }^\pi {{{2x\left( {1 + \sin x} \right)} \over {1 + {{\cos }^2}x}}} dx$$ is
Options:
[{"identifier": "A", "content": "$${{{\\pi ^2}} \\over 4}$$ "}, {"identifier": "B", "content": "$${{\\pi ^2}}$$ "}, {"identifier": "C", "content": "zero "}, {"identifier": "D", "content": "$${\\pi \\over 2}$$ "}] | ["B"]
Explanation:
$$\int_{ - \pi }^\pi {{{2x\left( {1 + \sin \,x} \right)} \over {1 + {{\cos }^2}x}}} dx$$
<br><br>$$ = \int_{ - \pi }^\pi {{{2x\,dx} \over {1 + {{\cos }^2}x}} + 2\int_{ - \pi }^\pi {{{x\,\sin x} \over {1 + {{\cos }^2}x}}} } dx$$
<br><br>$$ = 0 + 4\int_0^\pi {{{x\sin x\,dx} \over {1 + {{\cos }^2}x... |
If $$y=f(x)$$ makes +$$ve$$ intercept of $$2$$ and $$0$$ unit on $$x$$ and $$y$$ axes and encloses an area of $$3/4$$ square unit with the axes then $$\int\limits_0^2 {xf'\left( x \right)dx} $$ is
Options:
[{"identifier": "A", "content": "$$3/2$$ "}, {"identifier": "B", "content": "$$1$$ "}, {"identifier": "C", "cont... | ["D"]
Explanation:
We have $$\int\limits_0^2 {f\left( x \right)} dx = {3 \over 4};Now,$$
<br><br>$$\int\limits_0^2 {xf'\left( x \right)} dx$$
<br><br>$$ = x\int\limits_0^2 {f'\left( x \right)dx} - \int\limits_0^2 {f\left( x \right)} dx$$
<br><br>$$ = \left[ {x\,f\left( x \right)} \right]_0^2 - {3 \over 4}$$
<br><br>$... |
If $$f\left( y \right) = {e^y},$$ $$g\left( y \right) = y;y > 0$$ and
<br/><br>$$F\left( t \right) = \int\limits_0^t {f\left( {t - y} \right)g\left( y \right)dy,} $$ then :</br>
Options:
[{"identifier": "A", "content": "$$F\\left( t \\right) = t{e^{ - t}}$$ "}, {"identifier": "B", "content": "$$F\\left( t \\right... | ["C"]
Explanation:
$$F\left( t \right) = \int\limits_0^t {f\left( {t - y} \right)g\left( y \right)} dy$$
<br><br>$$ = \int\limits_0^t {{e^{t - y}}} ydy = {e^t}\int\limits_0^t {{e^{ - y}}} \,ydy$$
<br><br>$$ = {e^t}\left[ { - y{e^{ - y}} - {e^{ - y}}} \right]_0^t$$
<br><br>$$ = - {e^t}\left[ {y{e^{ - y}} + {e^{ - y}}}... |
The value of the integral $$I = \int\limits_0^1 {x{{\left( {1 - x} \right)}^n}dx} $$ is
Options:
[{"identifier": "A", "content": "$${1 \\over {n + 1}} + {1 \\over {n + 2}}$$ "}, {"identifier": "B", "content": "$${1 \\over {n + 1}}$$ "}, {"identifier": "C", "content": "$${1 \\over {n + 2}}$$ "}, {"identifier": "D", "c... | ["D"]
Explanation:
$$I = \int\limits_0^1 {x{{\left( {1 - x} \right)}^n}} dx$$
<br><br>$$ = \int\limits_0^1 {\left( {1 - x} \right){{\left( {1 - 1 + x} \right)}^n}dx} $$
<br><br>$$ = \int\limits_0^1 {\left( {1 - x} \right)} {x^n}dx$$
<br><br>$$ = \left[ {{{{x^{n + 1}}} \over {n + 1}} - {{{x^{n + 2}}} \over {n + 2}}} \... |
Let $$f(x)$$ be a function satisfying $$f'(x)=f(x)$$ with $$f(0)=1$$ and $$g(x)$$ be a function that satisfies $$f\left( x \right) + g\left( x \right) = {x^2}$$. Then the value of the integral $$\int\limits_0^1 {f\left( x \right)g\left( x \right)dx,} $$ is
Options:
[{"identifier": "A", "content": "$$e + {{{e^2}} \\ove... | ["D"]
Explanation:
Given $$f'\left( x \right) = f\left( x \right) \Rightarrow {{f'\left( x \right)} \over {f\left( x \right)}} = 1$$
<br><br>Integrating log
<br><br>$$f\left( x \right) = x + c \Rightarrow f\left( x \right) = {e^{x + c}}$$
<br><br>$$f\left( 0 \right) = 1 \Rightarrow f\left( x \right) = {e^x}$$
<br><br... |
If $$f\left( {a + b - x} \right) = f\left( x \right)$$ then $$\int\limits_a^b {xf\left( x \right)dx} $$ is equal to
Options:
[{"identifier": "A", "content": "$${{a + b} \\over 2}\\int\\limits_a^b {f\\left( {a + b + x} \\right)dx} $$ "}, {"identifier": "B", "content": "$${{a + b} \\over 2}\\int\\limits_a^b {f\\left( {... | ["C"]
Explanation:
$$I = \int\limits_a^b {xf\left( x \right)} dx$$
<br><br>$$ = \int\limits_a^b {\left( {a + b - x} \right)} f\left( {a + b - x} \right)dx$$
<br><br>$$ = \left( {a + b} \right)\int\limits_a^b {f\left( {a + b - x} \right)} dx - \int\limits_a^b {xf} \left( {a + b - x} \right)dx$$
<br><br>$$ = \left( {a +... |
The value of $$I = \int\limits_0^{\pi /2} {{{{{\left( {\sin x + \cos x} \right)}^2}} \over {\sqrt {1 + \sin 2x} }}dx} $$ is
Options:
[{"identifier": "A", "content": "$$3$$ "}, {"identifier": "B", "content": "$$1$$"}, {"identifier": "C", "content": "$$2$$"}, {"identifier": "D", "content": "$$0$$"}] | ["C"]
Explanation:
$$I = \int\limits_0^{{\pi \over 2}} {{{{{\left( {\sin \,x + \cos x} \right)}^2}} \over {\sqrt {1 + \sin 2x} }}} dx$$
<br><br>We know $$\left[ {{{\left( {\sin x + \cos x} \right)}^2} = 1 + \sin 2x} \right],\,$$ So
<br><br>$$I = \int\limits_0^{{\pi \over 2}} {{{{{\left( {\sin x + \cos x} \right)}^2}... |
The value of $$\int\limits_{ - 2}^3 {\left| {1 - {x^2}} \right|dx} $$ is
Options:
[{"identifier": "A", "content": "$${1 \\over 3}$$ "}, {"identifier": "B", "content": "$${14 \\over 3}$$"}, {"identifier": "C", "content": "$${7 \\over 3}$$"}, {"identifier": "D", "content": "$${28 \\over 3}$$"}] | ["D"]
Explanation:
$$\int\limits_{ - 2}^3 {\left| {1 - {x^2}} \right|} dx = \int\limits_{ - 2}^3 {\left| {{x^2} - 1} \right|} dx$$
<br><br>Now $$\left| {{x^2} - 1} \right| = \left\{ {\matrix{
{{x^2} - 1} & {if} & {x \le - 1} \cr
{1 - {x^2}} & {if} & { - 1 \le x \le 1} \cr
{{x^2} - 1} &... |
If $$\int\limits_0^\pi {xf\left( {\sin x} \right)dx = A\int\limits_0^{\pi /2} {f\left( {\sin x} \right)dx,} } $$ then $$A$$ is
Options:
[{"identifier": "A", "content": "$$2\\pi $$ "}, {"identifier": "B", "content": "$$\\pi $$ "}, {"identifier": "C", "content": "$${\\pi \\over 4}$$ "}, {"identifier": "D", "content... | ["B"]
Explanation:
Let $$I = \int\limits_0^\pi {xf\left( {\sin x} \right)} dx$$
<br><br>$$ = \int\limits_0^\pi {\left( {\pi - x} \right)} f\left( {\sin x} \right)dx$$
<br><br>$$\therefore$$ $$2I = \pi \int\limits_2^\pi {f\left( {\sin x} \right)} dx$$
<br><br>$$ = \pi .2\int\limits_0^{{\pi \over 2}} {f\left( {\sin... |
If $$f\left( x \right) = {{{e^x}} \over {1 + {e^x}}},{I_1} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {xg\left\{ {x\left( {1 - x} \right)} \right\}dx} $$
<br/>and $${I_2} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {g\left\{ {x\left( {1 - x} \right)} \right\}dx} ,$$ then the value of $${... | ["D"]
Explanation:
$$f\left( x \right) = {{{e^x}} \over {1 + {e^x}}}$$
<br><br>$$ \Rightarrow f\left( { - x} \right) = {{{e^{ - x}}} \over {1 + {e^{ - x}}}} = {1 \over {{e^x} + 1}}$$
<br><br>$$\therefore$$ $$f\left( x \right) + f\left( { - x} \right) = 1\forall x$$
<br><br>Now $${I_1} = \int\limits_{f\left( { - a} \ri... |
If $${I_1} = \int\limits_0^1 {{2^{{x^2}}}dx,{I_2} = \int\limits_0^1 {{2^{{x^3}}}dx,\,{I_3} = \int\limits_1^2 {{2^{{x^2}}}dx} } } $$ and $${I_4} = \int\limits_1^2 {{2^{{x^3}}}dx} $$ then
Options:
[{"identifier": "A", "content": "$${I_2} > {I_1}$$ "}, {"identifier": "B", "content": "$${I_1} > {I_2}$$"}, {"identif... | ["B"]
Explanation:
$${I_1} = \int\limits_0^1 {{2^{{x^2}}}} dx,\,{I_2} = \int\limits_0^1 {{2^{{x^3}}}} dx,$$
<br><br>$$ = {I_3} = \int\limits_0^1 {{2^{{x^2}}}} dx,\,$$
<br><br>$${I_4} = \int\limits_0^1 {{2^{{x^3}}}} dx\,\,$$
<br><br>$$\forall 0 < x < 1,\,{x^2} > {x^3}$$
<br><br>$$ \Rightarrow \int\limits_0^1 {... |
The value of $$\int\limits_{ - \pi }^\pi {{{{{\cos }^2}} \over {1 + {a^x}}}dx,\,\,a > 0,} $$ is
Options:
[{"identifier": "A", "content": "$$a\\,\\pi $$ "}, {"identifier": "B", "content": "$${\\pi \\over 2}$$ "}, {"identifier": "C", "content": "$${\\pi \\over a}$$"}, {"identifier": "D", "content": "$${2\\pi }$$... | ["B"]
Explanation:
Let $$I = \int\limits_{ - \pi }^\pi {{{{{\cos }^2}x} \over {1 + {a^x}}}} dx\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
<br><br>$$ = \int\limits_{ - \pi }^\pi {{{{{\cos }^2}\left( { - x} \right)} \over {1 + {a^{ - x}}}}} dx$$
<br><br>$$\left[ \, \right.$$ Using $$\int\limits_a^b {f\left( x \rig... |
The value of integral, $$\int\limits_3^6 {{{\sqrt x } \over {\sqrt {9 - x} + \sqrt x }}} dx $$ is
Options:
[{"identifier": "A", "content": "$${1 \\over 2}$$ "}, {"identifier": "B", "content": "$${3 \\over 2}$$"}, {"identifier": "C", "content": "$$2$$"}, {"identifier": "D", "content": "$$1$$"}] | ["B"]
Explanation:
$$I = \int\limits_3^6 {{{\sqrt x } \over {\sqrt {9 - x} + \sqrt x }}} dx\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
<br><br>$$I = \int\limits_3^6 {{{\sqrt {9 - x} } \over {\sqrt {9 - x} + \sqrt x }}} dx\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$
<br><br>$$\left[ \, \right.$$ using $$\int\limits_a^b {f\lef... |
$$\int\limits_0^\pi {xf\left( {\sin x} \right)dx} $$ is equal to
Options:
[{"identifier": "A", "content": "$$\\pi \\int\\limits_0^\\pi {f\\left( {\\cos x} \\right)dx} $$ "}, {"identifier": "B", "content": "$$\\,\\pi \\int\\limits_0^\\pi {f\\left( {sinx} \\right)dx} $$ "}, {"identifier": "C", "content": "$${\\pi ... | ["D"]
Explanation:
$$I = \int\limits_0^\pi {xf\left( {\sin \,x} \right)dx} $$
<br><br>$$ = \int\limits_0^\pi {\left( {\pi - x} \right)f\left( {\sin x} \right)dx} $$
<br><br>$$ = \pi \int\limits_0^\pi {f\left( {\sin x} \right)dx - 1} $$
<br><br>$$ \Rightarrow 2I = \pi {\pi \over 0}f\left( {\sin x} \right)dx$$
<br>... |
$$\int\limits_{ - {{3\pi } \over 2}}^{ - {\pi \over 2}} {\left[ {{{\left( {x + \pi } \right)}^3} + {{\cos }^2}\left( {x + 3\pi } \right)} \right]} dx$$ is equal to
Options:
[{"identifier": "A", "content": "$${{{\\pi ^4}} \\over {32}}$$ "}, {"identifier": "B", "content": "$${{{\\pi ^4}} \\over {32}} + {\\pi \\over ... | ["C"]
Explanation:
$$I = \int\limits_{ - {{3\pi } \over 2}}^{ - {\pi \over 2}} {\left[ {{{\left( {x + \pi } \right)}^3} + {{\cos }^2}\left( {x + 3\pi } \right)} \right]} \,dx$$
<br><br>Put $$x + \pi = t$$
<br><br>$$I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\left( {{t^3} + {{\cos }^2}t} \right)dt} $$
<br>... |
The value of $$\int\limits_1^a {\left[ x \right]} f'\left( x \right)dx,a > 1$$ where $${\left[ x \right]}$$ denotes the greatest integer not exceeding $$x$$ is
Options:
[{"identifier": "A", "content": "$$af\\left( a \\right) - \\left\\{ {f\\left( 1 \\right) + f\\left( 2 \\right) + .............f\\left( {\\left[ a ... | ["B"]
Explanation:
Let $$a = k + h$$ where $$k$$ is an integer such that
<br><br>$$\left[ a \right] = k$$ and $$0 \le h < 1$$
<br><br>$$\therefore$$ $$\int\limits_1^a {\left[ x \right]f'\left( x \right)dx = \int\limits_1^2 {1f'\left( x \right)} } \,dx$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\, + \int\limits_2^3 {2f'\lef... |
The solution for $$x$$ of the equation $$\int\limits_{\sqrt 2 }^x {{{dt} \over {t\sqrt {{t^2} - 1} }} = {\pi \over 2}} $$ is
Options:
[{"identifier": "A", "content": "$${{\\sqrt 3 } \\over 2}$$ "}, {"identifier": "B", "content": "$$2\\sqrt 2 $$ "}, {"identifier": "C", "content": "$$2$$ "}, {"identifier": "D", "conte... | ["D"]
Explanation:
$$\int_{\sqrt 2 }^x {{{dt} \over {t\sqrt {{t^2} - 1} }}} = {\pi \over 2}$$
<br><br>$$\therefore$$ $$\left[ {{{\sec }^{ - 1}}t} \right]_{\sqrt 2 }^x = {\pi \over 2}$$
<br><br>$$\left[ {} \right.$$ As $$\int {{{dx} \over {x\sqrt {{x^2} - 1} }}} = {\sec ^{ - 1}}x$$ $$\left. {} \right]$$
<br><br>$$... |
Let $$I = \int\limits_0^1 {{{\sin x} \over {\sqrt x }}dx} $$ and $$J = \int\limits_0^1 {{{\cos x} \over {\sqrt x }}dx} .$$ Then which one of the following is true?
Options:
[{"identifier": "A", "content": "$$1 > {2 \\over 3}$$ and $$J > 2$$ "}, {"identifier": "B", "content": "$$1 < {2 \\over 3}$$ and $$J <... | ["B"]
Explanation:
We know that $${{\sin x} \over x} < 1,$$ for $$x \in \left( {0,1} \right)$$
<br><br>$$ \Rightarrow {{\sin x} \over {\sqrt x }} < \sqrt x $$ on $$x \in \left( {0,1} \right)$$
<br><br>$$ \Rightarrow \int\limits_0^1 {{{\sin x} \over {\sqrt x }}dx < \int\limits_0^1 {\sqrt x dx} = \left[ {{{2{x... |
Let $$F\left( x \right) = f\left( x \right) + f\left( {{1 \over x}} \right),$$ where $$f\left( x \right) = \int\limits_l^x {{{\log t} \over {1 + t}}dt,} $$ Then $$F(e)$$ equals
Options:
[{"identifier": "A", "content": "$$1$$"}, {"identifier": "B", "content": "$$2$$"}, {"identifier": "C", "content": "$$1/2$$ "}, {"ide... | ["C"]
Explanation:
Given $$f\left( x \right) = f\left( x \right) + f\left( {{1 \over x}} \right),$$
<br><br>where $$f\left( x \right) = \int_1^x {{{\log \,t} \over {1 + t}}} \,dt$$
<br><br>$$\therefore$$ $$F\left( e \right) = f\left( e \right) + f\left( {{1 \over e}} \right)$$
<br><br>$$ \Rightarrow F\left( e \right)... |
$$\int\limits_0^\pi {\left[ {\cot x} \right]dx,} $$ where $$\left[ . \right]$$ denotes the greatest integer function, is equal to:
Options:
[{"identifier": "A", "content": "$$1$$"}, {"identifier": "B", "content": "$$-1$$ "}, {"identifier": "C", "content": "$$ - {\\pi \\over 2}$$ "}, {"identifier": "D", "content": "$... | ["C"]
Explanation:
Let $$I = \int_0^\pi {\left[ {\cot x} \right]dx\,\,\,\,\,\,...\left( 1 \right)} $$
<br><br>$$ = \int_0^\pi {\left[ {\cot \left( {\pi - x} \right)} \right]} dx$$
<br><br>$$ = \int_0^\pi {\left[ { - \cot x} \right]dx\,\,\,\,\,\,...\left( 2 \right)} $$
<br><br>Adding two values of $$I$$ in $$e{q^n}... |
Let $$p(x)$$ be a function defined on $$R$$ such that $$p'(x)=p'(1-x),$$ for all $$x \in \left[ {0,1} \right],p\left( 0 \right) = 1$$ and $$p(1)=41.$$ Then $$\int\limits_0^1 {p\left( x \right)dx} $$ equals :
Options:
[{"identifier": "A", "content": "$$21$$"}, {"identifier": "B", "content": "$$41$$ "}, {"identifier": ... | ["A"]
Explanation:
$$p'\left( x \right) = p'\left( {1 - x} \right)$$
<br><br>$$ \Rightarrow p\left( x \right) = - p\left( {1 - x} \right) + c$$
<br><br>at $$x=0$$
<br><br>$$p\left( 0 \right) = - p\left( 1 \right) + c \Rightarrow 42 = c$$
<br><br>Now, $$p\left( x \right) = - p\left( {1 - x} \right) + 42$$
<br><br>... |
The value of $$\int\limits_0^1 {{{8\log \left( {1 + x} \right)} \over {1 + {x^2}}}} dx$$ is
Options:
[{"identifier": "A", "content": "$${\\pi \\over 8}\\log 2$$ "}, {"identifier": "B", "content": "$${\\pi \\over 2}\\log 2$$"}, {"identifier": "C", "content": "$$\\log 2$$ "}, {"identifier": "D", "content": "$$\\pi \\... | ["D"]
Explanation:
$$I = \int\limits_0^1 {{{8\log \left( {1 + x} \right)} \over {1 + {x^2}}}} dx$$
<br><br>put $$x = \tan \,\theta ,$$
<br><br>$$\therefore$$ $${{dx} \over {d\theta }} = {\sec ^2}\theta \Rightarrow dx = {\sec ^2}\theta d\theta $$
<br><br>$$\therefore$$ $$I = 8\int\limits_0^{\pi /4} {{{\log \left( {1 ... |
If $$g\left( x \right) = \int\limits_0^x {\cos 4t\,dt,} $$ then $$g\left( {x + \pi } \right)$$ equals
Options:
[{"identifier": "A", "content": "$${{g\\left( x \\right)} \\over {8\\left( \\pi \\right)}}$$ "}, {"identifier": "B", "content": "$$g\\left( x \\right) + g\\left( \\pi \\right)$$ "}, {"identifier": "C", "con... | Explanation:
$$\left( {b,c} \right)g\left( {x + \pi } \right) = \int\limits_0^{x + \pi } {\cos \,4t\,dt} $$
<br><br>$$ = \int\limits_0^\pi {\cos 4tdt + \int\limits_\pi ^{\pi + x} {\cos 4t\,dt} } $$
<br><br>$$ = g\left( \pi \right) + \int\limits_0^x {\cos \,4t\,dt} $$
<br><br>Putting $$t = \pi + y$$ in second inte... |
<b>Statement-1 :</b> The value of the integral
<br/>$$\int\limits_{\pi /6}^{\pi /3} {{{dx} \over {1 + \sqrt {\tan \,x} }}} $$ is equal to $$\pi /6$$
<p><b>Statement-2 :</b> $$\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)} dx.$$</p>
Options:
[{"identifier": "A", "content": "Stat... | ["D"]
Explanation:
Let $$I = \int\limits_{\pi /6}^{\pi /3} {{{dx} \over {1 + \sqrt {\tan \,x} }}} $$
<br><br>$$ = \int\limits_{\pi /6}^{\pi /3} {{{dx} \over {\sqrt {\tan \left( {{\pi \over 2} - x} \right)} }}} $$
<br><br>$$ = \int\limits_{\pi /6}^{\pi /3} {{{\sqrt {\tan \,x} \,dx} \over {1 + \sqrt {\tan \,x} }}} \,\,... |
The integral $$\int\limits_0^\pi {\sqrt {1 + 4{{\sin }^2}{x \over 2} - 4\sin {x \over 2}{\mkern 1mu} } } dx$$ equals:
Options:
[{"identifier": "A", "content": "$$4\\sqrt 3 - 4$$ "}, {"identifier": "B", "content": "$$4\\sqrt 3 - 4 - {\\pi \\over 3}$$ "}, {"identifier": "C", "content": "$$\\pi - 4$$ "}, {"identif... | ["B"]
Explanation:
Let $$I = \int\limits_0^\pi {\sqrt {1 + 4{{\sin }^2}{x \over 2} - 4\sin {x \over 2}} } dx$$
<br><br>$$ = \int\limits_0^\pi {\left| {2\sin {x \over 2} - 1} \right|} dx$$
<br><br>$$ = \int\limits_0^{\pi /3} {\left( {1 - 2\sin {x \over 2}} \right)} dx + \int\limits_{\pi /3}^\pi {\left( {2\sin {x \ov... |
The integral
<br/>$$\int\limits_2^4 {{{\log \,{x^2}} \over {\log {x^2} + \log \left( {36 - 12x + {x^2}} \right)}}dx} $$ is equal to :
Options:
[{"identifier": "A", "content": "$$1$$"}, {"identifier": "B", "content": "$$6$$"}, {"identifier": "C", "content": "$$2$$"}, {"identifier": "D", "content": "$$4$$"}] | ["A"]
Explanation:
$$I = \int\limits_2^4 {{{\log {x^2}} \over {\log {x^2} + \log \left( {36 - 12x + {x^2}} \right)}}} $$
<br><br>$$I = \int\limits_2^4 {{{\log {x^2}} \over {\log {x^2} + \log {{\left( {6 - x} \right)}^2}}}} \,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
<br><br>$$I = \int\limits_2^4 {{{\log {{\left( {6 - x}... |
The value of the integral
<br/><br/>$$\int\limits_4^{10} {{{\left[ {{x^2}} \right]dx} \over {\left[ {{x^2} - 28x + 196} \right] + \left[ {{x^2}} \right]}}} ,$$
<br/><br/>where [x] denotes the greatest integer less than or
equal to x, is :
Options:
[{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "... | ["B"]
Explanation:
Let I = $$\int\limits_4^{10} {{{\left[ {{x^2}} \right]\,dx} \over {\left[ {{x^2} - 28x + 196} \right] + \left[ {{x^2}} \right]}}} $$
<br><br> = $$\int\limits_4^{10} {{{\left[ {{x^2}} \right]dx} \over {\left[ {{{\left( {x - 14} \right)}^2}} ... |
If $$2\int\limits_0^1 {{{\tan }^{ - 1}}xdx = \int\limits_0^1 {{{\cot }^{ - 1}}} } \left( {1 - x + {x^2}} \right)dx,$$
<br/><br/>then $$\int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx$$ is equalto :
Options:
[{"identifier": "A", "content": "log4"}, {"identifier": "B", "content": "$${\\pi \\over ... | ["C"]
Explanation:
Given,
<br><br>$$2\int\limits_0^1 {{{\tan }^{ - {1_x}\,{d_x}}}} = \int\limits_0^1 {{{\cot }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx$$
<br><br>= $$\int\limits_0^1 {\left( {{\pi \over 2} - {{\tan }^{ - 1}}\left( {1 - x + {x^2}} \right)} \right)} dx$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$$... |
The integral $$\int_{{\pi \over {12}}}^{{\pi \over 4}} {\,\,{{8\cos 2x} \over {{{\left( {\tan x + \cot x} \right)}^3}}}} \,dx$$ equals :
Options:
[{"identifier": "A", "content": "$${{15} \\over {128}}$$"}, {"identifier": "B", "content": "$${{15} \\over {64}}$$"}, {"identifier": "C", "content": "$${{13} \\over {32}}... | ["A"]
Explanation:
tan x + cot x
<br><br>= $${{\sin x} \over {\cos x}}$$ + $${{\cos x} \over {\sin x}}$$
<br><br>= $${{{{\sin }^2}x + {{\cos }^2}x} \over {\sin x\,\,\cos x}}$$
<br><br>= $${1 \over {\sin x\,\,\cos x}}$$
<br><br>= $${2 \over {\sin 2x}}$$
<br><br>$$\therefore\,\,\,$$ $$\int\limits_{{\pi \ove... |
If $$\int\limits_1^2 {{{dx} \over {{{\left( {{x^2} - 2x + 4} \right)}^{{3 \over 2}}}}}} = {k \over {k + 5}},$$ then k is equal to :
Options:
[{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "4"}] | ["A"]
Explanation:
Given, I = $$\int\limits_1^2 {{{dx} \over {{{\left( {{x^2} - 2x + 4} \right)}^{{3 \over 2}}}}}} $$
<br><br>= $$\int\limits_1^2 {{{dx} \over {{{\left[ {{{\left( {x - 1} \right)}^2} + 3} \right]}^{{3 \over 2}}}}}} $$
<br><br>Let x $$-$$ 1 = $$\sqrt 3 $$ tan$$\theta $$
<br><br>$$ \Rightarrow $$$$\,\,\,... |
The integral $$\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{dx} \over {1 + \cos x}}} $$ is equal to
Options:
[{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "$$-$$ 1"}, {"identifier": "D", "content": "$$-$$ 2"}] | ["A"]
Explanation:
$$\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{dx} \over {1 + \cos x}}} $$
<br><br>= $$\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{dx} \over {2{{\cos }^2}{x \over 2}}}} $$
<br><br>= $${1 \over 2}\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{\sec }^2}} {x \over 2}\,dx$$
<br><br>... |
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