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1

### 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 $$n \in \left( { - {\pi \over 2},{\pi \over 4}} \right)$$
2

### 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 $$\theta = {\pi \over 2}$$

$${A_{\max }} = {1 \over 2}{x^2}$$
3

### 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 $$x=2.$$
4

### AIEEE 2005

If the equation $${a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + ........... + {a_1}x = 0$$
$${a_1} \ne 0,n \ge 2,$$ has a positive root $$x = \alpha$$, then the equation
$$n{a_n}{x^{n - 1}} + \left( {n - 1} \right){a_{n - 1}}{x^{n - 2}} + ........... + {a_1} = 0$$ has a positive root, which is
A
greater than $$\alpha$$
B
smaller than $$\alpha$$
C
greater than or equal to smaller than $$\alpha$$
D
equal to smaller than $$\alpha$$

## Explanation

Let $$f\left( x \right) = {a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + ........... + {a_1}x = 0$$

The other given equation,

$$na{}_n{x^{n - 1}} + \left( {n - 1} \right){a_{n - 1}}{x^{n - 2}} + .... + {a_1} = 0 = f'\left( x \right)$$

Given $${a_1} \ne 0 \Rightarrow f\left( 0 \right) = 0$$

Again $$f(x)$$ has root $$\alpha ,\,\,\, \Rightarrow f\left( \alpha \right) = 0$$

$$\therefore$$ $$\,\,\,\,\,f\left( 0 \right) = f\left( \alpha \right)$$

$$\therefore$$ By Roll's theorem $$f'(x)=0$$ has root be-

tween $$\left( {0,\alpha } \right)$$

Hence $$f'(x)$$ has a positive root smaller than $$\alpha .$$

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