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1

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 $${{3\sqrt 2 } \over 8}$$
2

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 $$x = 2\pi$$
3

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 $$c \in R$$

$$f\left( c \right) = {1 \over 3}$$
4

AIEEE 2010

The equation of the tangent to the curve $$y = x + {4 \over {{x^2}}}$$, that
is parallel to the $$x$$-axis, is
A
$$y=1$$
B
$$y=2$$
C
$$y=3$$
D
$$y=0$$

Explanation

Since tangent is parallel to $$x$$-axis,

$$\therefore$$ $${{dy} \over {dx}} = 0 \Rightarrow 1 - {8 \over {{x^3}}} = 0 \Rightarrow x = 2 \Rightarrow y = 3$$

Equation of tangent is $$y - 3 = 0\left( {x - 2} \right) \Rightarrow y = 3$$

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