NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

JEE Main 2014 (Offline)

MCQ (Single Correct Answer)
If $$x=-1$$ and $$x=2$$ are extreme points of $$f\left( x \right) = \alpha \,\log \left| x \right|+\beta {x^2} + x$$ then
A
$$\alpha = 2,\beta = - {1 \over 2}$$
B
$$\alpha = 2,\beta = {1 \over 2}$$
C
$$\alpha = - 6,\beta = {1 \over 2}$$
D
$$\alpha = - 6,\beta = -{1 \over 2}$$

Explanation

Let $$f\left( x \right) = \alpha \log \left| x \right| + \beta {x^2} + x$$

Differentiating both sides,

$$f'\left( x \right) = {\alpha \over x} + 2\beta x + 1$$

Since $$x=-1$$ and $$x=2$$ are extreme points therefore

$$f'\left( x \right) = 0$$ at these points.

Put $$x = - 1$$ and $$x = 2$$ in $$f'\left( x \right),$$

we get $$ - \alpha - 2\beta + 1 = 0$$

$$ \Rightarrow \alpha + 2\beta = 1\,\,...\left( i \right)$$

$${\alpha \over 2} + 4\beta + 1 = 0$$

$$ \Rightarrow \alpha + 8\beta = - 2\,\,...\left( {ii} \right)$$

On solving $$(i)$$ and $$(ii)$$, we get

$$6\beta = - 3 \Rightarrow \beta = - {1 \over 2}$$

$$\therefore$$ $$\,\,\,\,\alpha = 2$$
2

JEE Main 2013 (Offline)

MCQ (Single Correct Answer)
If $$y = \sec \left( {{{\tan }^{ - 1}}x} \right),$$ then $${{{dy} \over {dx}}}$$ at $$x=1$$ is equal to :
A
$${1 \over {\sqrt 2 }}$$
B
$${1 \over 2}$$
C
$$1$$
D
$$\sqrt 2 $$

Explanation

Let $$y = \sec \left( {{{\tan }^{ - 1}}x} \right)$$

and $${\tan ^{ - 1}}\,\,x = \theta .$$

$$ \Rightarrow x = \tan \theta $$



Thus, we have $$y = \sec \,\theta $$

$$ \Rightarrow y = \sqrt {1 + {x^2}} $$

$$\left( {\,\,} \right.$$ As $$\,\,\,\,\,\,{\sec ^2}\theta = 1 + {\tan ^2}\theta $$ $$\left. {\,\,} \right)$$

$$ \Rightarrow {{dy} \over {dx}} = {1 \over {2\sqrt {1 + {x^2}} }}.2x$$

At $$x = 1,\,\,{{dy} \over {dx}} = {1 \over {\sqrt 2 }}.$$
3

AIEEE 2011

MCQ (Single Correct Answer)
$${{{d^2}x} \over {d{y^2}}}$$ equals:
A
$$ - {\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{ - 1}}{\left( {{{dy} \over {dx}}} \right)^{ - 3}}$$
B
$${\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{}}{\left( {{{dy} \over {dx}}} \right)^{ - 2}}$$
C
$$ - \left( {{{{d^2}y} \over {d{x^2}}}} \right){\left( {{{dy} \over {dx}}} \right)^{ - 3}}$$
D
$${\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{ - 1}}$$

Explanation

$${{{d^2}x} \over {d{y^2}}} = {d \over {dy}}\left( {{{dx} \over {dy}}} \right)$$

$$ = {d \over {dx}}\left( {{{dx} \over {dy}}} \right){{dx} \over {dy}}$$

$$ = {d \over {dx}}\left( {{1 \over {dy/dx}}} \right){{dx} \over {dy}}$$

$$ = - {1 \over {{{\left( {{{dy} \over {dx}}} \right)}^2}}}.{{{d^2}y} \over {d{x^2}}}.{1 \over {{{dy} \over {dx}}}}$$

$$ = - {1 \over {{{\left( {{{dy} \over {dx}}} \right)}^3}}}{{{d^2}y} \over {d{x^2}}}$$
4

AIEEE 2010

MCQ (Single Correct Answer)
Let $$f:\left( { - 1,1} \right) \to R$$ be a differentiable function with $$f\left( 0 \right) = - 1$$ and $$f'\left( 0 \right) = 1$$. Let $$g\left( x \right) = {\left[ {f\left( {2f\left( x \right) + 2} \right)} \right]^2}$$. Then $$g'\left( 0 \right) = $$
A
$$-4$$
B
$$0$$
C
$$-2$$
D
$$4$$

Explanation

$$g'\left( x \right) = 2\left( {f\left( {2f\left( x \right) + 2} \right)} \right)$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{d \over {dx}}\left( {f\left( {2f\left( x \right) + 2} \right)} \right)} \right)$$

$$ = 2f\left( {2f\left( x \right) + 2} \right)f'\left( {2f\left( x \right)} \right)$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \left. 2 \right).\left( {2f'\left( x \right)} \right)$$

$$ \Rightarrow g'\left( 0 \right) = 2f\left( {2f\left( 0 \right) + 2} \right).$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f'\left( {2f\left( 0 \right) + 2} \right).2f'\left( 0 \right)$$

$$ = 4f\left( 0 \right){\left( {f'\left( 0 \right)} \right)^2}$$

$$ = 4\left( { - 1} \right){\left( 1 \right)^2} = - 4$$

Joint Entrance Examination

JEE Main JEE Advanced WB JEE

Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

Medical

NEET

CBSE

Class 12