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1

### AIEEE 2011

MCQ (Single Correct Answer)
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 $$x = 2\pi$$
2

### AIEEE 2010

MCQ (Single Correct Answer)
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 $$c \in R$$ $$f\left( c \right) = {1 \over 3}$$ 3 ### AIEEE 2010 MCQ (Single Correct Answer) 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 $$y - 3 = 0\left( {x - 2} \right) \Rightarrow y = 3$$ 4 ### AIEEE 2010 MCQ (Single Correct Answer) Let $$f:R \to R$$ be defined by $$f\left( x \right) = \left\{ {\matrix{ {k - 2x,\,\,if} & {x \le - 1} \cr {2x + 3,\,\,if} & {x > - 1} \cr } } \right.$$$

If $$f$$has a local minimum at $$x=-1$$, then a possible value of $$k$$ is

A
$$0$$
B
$$- {1 \over 2}$$
C
$$-1$$
D
$$1$$

## Explanation

$$f\left( x \right) = \left\{ {\matrix{ {k - 2x,\,\,\,\,if\,\,\,\,x \le - 1} \cr {2x + 3,\,\,\,\,if\,\,\,\,x > - 1} \cr } } \right.$$

This is true where $$k=-1$$

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