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1

### AIEEE 2007

A value of $$c$$ for which conclusion of Mean Value Theorem holds for the function $$f\left( x \right) = {\log _e}x$$ on the interval $$\left[ {1,3} \right]$$ is
A
$${\log _3}e$$
B
$${\log _e}3$$
C
$$2\,\,{\log _3}e$$
D
$${1 \over 2}{\log _3}e$$

## Explanation

Using Lagrange's Mean Value Theorem

Let $$f(x)$$ be a function defined on $$\left[ {a,b} \right]$$

then, $$f'\left( c \right) = {{f\left( b \right) - f\left( a \right)} \over {b - a}}\,\,\,\,\,\,\,\,\,\,\,\,....\left( i \right)$$

$$c\,\, \in \left[ {a,b} \right]$$

$$\therefore$$ Given $$f\left( x \right) = {\log _e}x$$

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

$$\therefore$$ equation $$(i)$$ become $${1 \over c} = {{f\left( 3 \right) - f\left( 1 \right)} \over {3 - 1}}$$

$$\Rightarrow {1 \over c} = {{{{\log }_e}3 - {{\log }_e}1} \over 2} = {{{{\log }_e}3} \over 2}$$

$$\Rightarrow c = {2 \over {{{\log }_e}3}} \Rightarrow c = 2\,{\log _3}e$$
2

### AIEEE 2007

The function $$f\left( x \right) = {\tan ^{ - 1}}\left( {\sin x + \cos x} \right)$$ is an incresing function in
A
$$\left( {0,{\pi \over 2}} \right)$$
B
$$\left( { - {\pi \over 2},{\pi \over 2}} \right)$$
C
$$\left( { {\pi \over 4},{\pi \over 2}} \right)$$
D
$$\left( { - {\pi \over 2},{\pi \over 4}} \right)$$

## Explanation

Given $$f\left( x \right) = {\tan ^{ - 1}}\left( {\sin x + \cos x} \right)$$

$$f'\left( x \right) = {1 \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}.\left( {\cos x - \sin x} \right)$$

$$= {{\sqrt 2 .\left( {{1 \over {\sqrt 2 }}\cos x - {1 \over {\sqrt 2 }}\sin x} \right)} \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}$$

$$= {{\left( {\cos {\pi \over 4}.\cos x - \sin {\pi \over 4}.\sin x} \right)} \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}$$

$$\therefore$$ $$f'\left( x \right) = {{\sqrt 2 \cos \left( {x + {\pi \over 4}} \right)} \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}$$

if $$f'\left( x \right) > O$$ then $$f\left( x \right)$$ is increasing function.

Hence $$f(x)$$ is increasing, if $$- {\pi \over 2} < x + {\pi \over 4} < {\pi \over 2}$$

$$\Rightarrow - {{3\pi } \over 4} < x < {\pi \over 4}$$

Hence, $$f(x)$$ is increasing when $$n \in \left( { - {\pi \over 2},{\pi \over 4}} \right)$$
3

### AIEEE 2006

A triangular park is enclosed on two sides by a fence and on the third side by a straight river bank. The two sides having fence are of same length $$x$$. The maximum area enclosed by the park is
A
$${3 \over 2}{x^2}$$
B
$$\sqrt {{{{x^3}} \over 8}}$$
C
$${1 \over 2}{x^2}$$
D
$$\pi {x^2}$$

## Explanation

Area $$= {1 \over 2}{x^2}\,\sin \,\theta$$

Maximum value of $$\sin \theta$$ is $$1$$ at $$\theta = {\pi \over 2}$$

$${A_{\max }} = {1 \over 2}{x^2}$$
4

### AIEEE 2006

The function $$f\left( x \right) = {x \over 2} + {2 \over x}$$ has a local minimum at
A
$$x=2$$
B
$$x=-2$$
C
$$x=0$$
D
$$x=1$$

## Explanation

$$f\left( x \right) = {x \over 2} + {2 \over x} \Rightarrow f'\left( x \right) = {1 \over 2} - {2 \over {{x^2}}} = 0$$

$$\Rightarrow {x^2} = 4$$ or $$x=2,-2;$$ $$\,\,\,\,\,f''\left( x \right) = {4 \over {{x^3}}}$$

$$f''{\left. {\left( x \right)} \right]_{x = 2}} = + ve \Rightarrow f\left( x \right)$$

has local min at $$x=2.$$

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