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1

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

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

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 $$x=2.$$
3

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 $$\alpha .$$
4

AIEEE 2005

The normal to the curve
$$x = a\left( {\cos \theta + \theta \sin \theta } \right),y = a\left( {\sin \theta - \theta \cos \theta } \right)$$ at any point
$$\theta\, '$$ is such that
A
it passes through the origin
B
it makes an angle $${\pi \over 2} + \theta$$ with the $$x$$-axis
C
it passes through $$\left( {a{\pi \over 2}, - a} \right)$$
D
it is at a constant distance from the origin

Explanation

$$x = a\left( {\cos \theta + \theta \sin \theta } \right)$$

$$\Rightarrow {{dx} \over {d\theta }} = a\left( { - \sin \theta + \sin \theta + \theta \cos \theta } \right)$$

$$\Rightarrow {{dx} \over {d\theta }} = a\theta \cos \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

$$y = a\left( {\sin \theta - \theta \cos \theta } \right)$$

$${{dy} \over {d\theta }} = a\left[ {\cos \theta - \cos \theta + \theta \sin \theta } \right]$$

$$\Rightarrow {{dy} \over {d\theta }} = a\theta \sin \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$

From equations $$(1)$$ and $$(2),$$ we get

$${{dy} \over {dx}} = \tan \theta \Rightarrow$$ Slope of normal $$= - \cot \,\theta$$

Equation of normal at

$$'\theta '$$ is $$y - a\left( {\sin \theta - \theta \cos \theta } \right)$$

$$= - \cot \theta \left( {x - a\left( {\cos \theta + \theta \sin \theta } \right)} \right.$$

$$\Rightarrow y\sin \theta - a{\sin ^2}\theta + a\,\,\theta \cos \theta \sin \theta$$

$$= - x\cos \theta + a{\cos ^2}\theta + a\,\theta \sin \theta \cos \theta$$

$$\Rightarrow x\cos \theta + y\sin \theta = a$$

Clearly this is an equation of straight line -

which is at a constant distance $$'a'$$ from origin.

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