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1

### AIEEE 2012

A spherical balloon is filled with $$4500\pi$$ cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of $$72\pi$$ cubic meters per minute, then the rate (in meters per minute) at which the radius of the balloon decreases $$49$$ minutes after the leakage began is :
A
$${{9 \over 7}}$$
B
$${{7 \over 9}}$$
C
$${{2 \over 9}}$$
D
$${{9 \over 2}}$$

## Explanation

Volume of spherical balloon $$= V = {4 \over 3}\pi {r^3}$$

$$\Rightarrow 4500\pi = {{4\pi {r^3}} \over 3}$$

( as Given, volume $$= 4500\pi {m^3}$$ )

Differentiating both the sides, $$w.r.t't'$$ we get,

$${{dV} \over {dt}} = 4\pi {r^2}\left( {{{dr} \over {dt}}} \right)$$

Now, it is given that $${{dV} \over {dt}} = 72\pi$$

$$\therefore$$ After $$49$$ min, Volume -

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {4500 - 49 \times 72} \right)\pi$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {4500 - 3528} \right)\pi$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 972\pi {m^3}$$

$$\Rightarrow V = 972\,\,\pi {m^3}$$

$$\therefore$$ $$972\pi = {4 \over 3}\pi r{}^3$$

$$\Rightarrow {r^3} = 3 \times 243 = 3 \times {3^5} = {3^6} = {\left( {{3^2}} \right)^3} \Rightarrow r = 9$$

Also, we have $${{dV} \over {dt}} = 72\pi$$

$$\therefore$$ $$72\pi = 4\pi \times 9 \times 9\left( {{{dr} \over {dt}}} \right) \Rightarrow {{dr} \over {dt}} = \left( {{2 \over 9}} \right)$$
2

### AIEEE 2011

The shortest distance between line $$y-x=1$$ and curve $$x = {y^2}$$ is
A
$${{3\sqrt 2 } \over 8}$$
B
$${8 \over {3\sqrt 2 }}$$
C
$${4 \over {\sqrt 3 }}$$
D
$${{\sqrt 3 } \over 4}$$

## Explanation

Shortest distance between two curve occurred along -

the common normal

Slope of normal to $${y^2} = x$$ at point

$$P\left( {{t^2},t} \right)$$ is $$-2t$$ and

slope of line $$y - x = 1$$ is $$1.$$

As they are perpendicular to each other

$$\therefore$$ $$\left( { - 2t} \right) = - 1 \Rightarrow t = {1 \over 2}$$

$$\therefore$$ $$P\left( {{1 \over 4},{1 \over 2}} \right)$$

and shortest distance $$= \left| {{{{1 \over 2} - {1 \over 4} - 1} \over {\sqrt 2 }}} \right|$$

So shortest distance between them is $${{3\sqrt 2 } \over 8}$$
3

### AIEEE 2011

For $$x \in \left( {0,{{5\pi } \over 2}} \right),$$ define $$f\left( x \right) = \int\limits_0^x {\sqrt t \sin t\,dt.}$$ Then $$f$$ has
A
local minimum at $$\pi$$ and $$2\pi$$
B
local minimum at $$\pi$$ and local maximum at $$2\pi$$
C
local maximum at $$\pi$$ and local minimum at $$2\pi$$
D
local maximum at $$\pi$$ and $$2\pi$$

## Explanation

$$f'\left( x \right) = \sqrt x \sin x$$

At local maxima or minima, $$f'\left( x \right) = 0$$

$$\Rightarrow x = 0$$ or $$sin$$ $$x=0$$

$$\Rightarrow x = 2\pi ,\,\,\pi \in \left( {0,{{5\pi } \over 2}} \right)$$

$$f''\left( x \right) = \sqrt x \cos \,x + {1 \over {2\sqrt x }}\sin \,x$$

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

At $$x = \pi ,$$ $$f''\left( x \right) < 0$$

Hence, local maxima at $$x = \pi$$

At $$x = 2\pi ,\,\,\,f''\left( x \right) > 0$$

Hence local minima at $$x = 2\pi$$
4

### AIEEE 2010

Let $$f:R \to R$$ be a continuous function defined by $$f\left( x \right) = {1 \over {{e^x} + 2{e^{ - x}}}}$$\$

Statement - 1 : $$f\left( c \right) = {1 \over 3},$$ for some $$c \in R$$.

Statement - 2 : $$0 < f\left( x \right) \le {1 \over {2\sqrt 2 }},$$ for all $$x \in R$$

A
Statement - 1 is true, Statement -2 is true; Statement - 2 is not a correct explanation for Statement - 1.
B
Statement - 1 is true, Statement - 2 is false.
C
Statement - 1 is false, Statement - 2 is true.
D
Statement - 1 is true, Statement -2 is true; Statement -2 is a correct explanation for Statement - 1.

## Explanation

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

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

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

$${e^{2x}} = 2 \Rightarrow {e^x} = \sqrt 2$$

maximum $$f\left( x \right) = {{\sqrt 2 } \over 4} = {1 \over {2\sqrt 2 }}$$

$$0 < f\left( x \right) \le {1 \over {2\sqrt 2 }}\,\,\forall x \in R$$

Since $$0 < {1 \over 3} < {1 \over {2\sqrt 2 }} \Rightarrow$$ for some $$c \in R$$

$$f\left( c \right) = {1 \over 3}$$

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