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1

### JEE Main 2014 (Offline)

If $$g$$ is the inverse of a function $$f$$ and $$f'\left( x \right) = {1 \over {1 + {x^5}}},$$ then $$g'\left( x \right)$$ is equal to:
A
$${1 \over {1 + {{\left\{ {g\left( x \right)} \right\}}^5}}}$$
B
$$1 + {\left\{ {g\left( x \right)} \right\}^5}$$
C
$$1 + {x^5}$$
D
$$5{x^4}$$

## Explanation

Since $$f(x)$$ and $$g(x)$$ are inverse of each other

$$\therefore$$ $$g'\left( {f\left( x \right)} \right) = {1 \over {f'\left( x \right)}}$$

$$\Rightarrow g'\left( {f\left( x \right)} \right) = 1 + {x^5}$$

$$\left( \, \right.$$ As $$\,f'\left( x \right) = {1 \over {1 + {x^5}}}$$ $$\left. \, \right)$$

Here $$x=g(y)$$

$$\therefore$$ $$g'\left( y \right) = 1 + \left\{ {g\left( y \right)} \right\}$$

$$\Rightarrow g'\left( x \right) = 1 + \left\{ {g\left( x \right)} \right\}$$
2

### JEE Main 2014 (Offline)

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

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

$$\therefore$$ $$\,\,\,\,\alpha = 2$$
3

### JEE Main 2013 (Offline)

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 $$x = 1,\,\,{{dy} \over {dx}} = {1 \over {\sqrt 2 }}.$$
4

### AIEEE 2011

$${{{d^2}x} \over {d{y^2}}}$$ 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

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

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

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

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

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

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