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1

### AIEEE 2005

The value of $$\int\limits_{ - \pi }^\pi {{{{{\cos }^2}} \over {1 + {a^x}}}dx,\,\,a > 0,}$$ is
A
$$a\,\pi$$
B
$${\pi \over 2}$$
C
$${\pi \over a}$$
D
$${2\pi }$$

## Explanation

Let $$I = \int\limits_{ - \pi }^\pi {{{{{\cos }^2}x} \over {1 + {a^x}}}} dx\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

$$= \int\limits_{ - \pi }^\pi {{{{{\cos }^2}\left( { - x} \right)} \over {1 + {a^{ - x}}}}} dx$$

$$\left[ \, \right.$$ Using $$\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)dx}$$ $$\left. \, \right]$$

$$= \int\limits_{ - \pi }^\pi {{{{{\cos }^2}x} \over {1 + {\alpha ^x}}}} dx\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$

Adding equations $$(1)$$ and $$(2)$$ we get

$$2I = \int\limits_{ - \pi }^\pi {{{\cos }^2}} x\left( {{{1 + {a^x}} \over {1 + {a^x}}}} \right)dx$$

$$= \int\limits_{ - \pi }^\pi {{{\cos }^2}} x\,dx$$

$$= 2\int\limits_0^\pi {{{\cos }^2}} x\,dx$$

$$= 2 \times 2\int\limits_0^{{\pi \over 2}} {{{\cos }^2}} x\,dx$$

$$= 4\int\limits_0^{{\pi \over 2}} {{{\sin }^2}} x\,dx$$

$$\Rightarrow I = 2\int\limits_0^{{\pi \over 2}} {{{\sin }^2}} \,x\,dx$$

$$= 2\int\limits_0^{{\pi \over 2}} {\left( {1 - {{\cos }^2}x\,dx} \right)}$$

$$\Rightarrow I\,\, = 2\int\limits_0^{{\pi \over 2}} {dx - 2\int\limits_0^{{\pi \over 2}} {{{\cos }^2}} } \,x\,dx$$

$$\Rightarrow I + I = 2\left( {{\pi \over 2}} \right) = \pi$$

$$\Rightarrow I = {\pi \over 2}$$
2

### AIEEE 2005

Let $$f(x)$$ be a non - negative continuous function such that the area bounded by the curve $$y=f(x),$$ $$x$$-axis and the ordinates $$x = {\pi \over 4}$$ and $$x = \beta > {\pi \over 4}$$ is $$\left( {\beta \sin \beta + {\pi \over 4}\cos \beta + \sqrt 2 \beta } \right).$$ Then $$f\left( {{\pi \over 2}} \right)$$ is
A
$$\left( {{\pi \over 4} + \sqrt 2 - 1} \right)$$
B
$$\left( {{\pi \over 4} - \sqrt 2 + 1} \right)$$
C
$$\left( {1 - {\pi \over 4} - \sqrt 2 } \right)$$
D
$$\left( {1 - {\pi \over 4} + \sqrt 2 } \right)$$

## Explanation

Given that

$$\int\limits_{\pi /4}^\beta {f\left( x \right)} dx = \beta \sin \beta + {\pi \over 4}\cos \,\beta + \sqrt 2 \beta$$

Differentiating $$w.r.t$$ $$\beta$$

$$f\left( \beta \right) = \beta \cos \beta + \sin \beta - {\pi \over 4}\sin \beta + \sqrt 2$$

$$f\left( {{\pi \over 2}} \right) = \left( {1 - {\pi \over 4}} \right)\sin {\pi \over 2} + \sqrt 2$$

$$= 1 - {\pi \over 4} + \sqrt 2$$
3

### AIEEE 2005

The parabolas $${y^2} = 4x$$ and $${x^2} = 4y$$ divide the square region bounded by the lines $$x=4,$$ $$y=4$$ and the coordinate axes. If $${S_1},{S_2},{S_3}$$ are respectively the areas of these parts numbered from top to bottom ; then $${S_1},{S_2},{S_3}$$ is
A
$$1:2:1$$
B
$$1:2:3$$
C
$$2:1:2$$
D
$$1:1:1$$

## Explanation

Intersection points of $${x^2} = 4y$$ and $${y^2} = 4x$$ are $$\left( {0,0} \right)$$ and $$\left( {4,4} \right).$$ The graph is as shown in the figure. By symmetry, we observe

$${S_1} = {S_3} = \int\limits_0^4 {ydx = \int\limits_0^4 {{{{x^2}} \over 4}dx} }$$

$$= \left[ {{{{x^3}} \over {12}}} \right]_0^4 = {{16} \over 3}$$ sq. unit

Also $${S_2} = \int\limits_0^4 {\left( {2\sqrt x - {{{x^2}} \over 4}} \right)} dx$$

$$\,\,\,\,\,\,\,\,\,\,\,\,$$ $$= \left[ {{{2{x^{{3 \over 2}}}} \over {{3 \over 2}}} - {{{x^3}} \over {12}}} \right]_0^4$$

$$\,\,\,\,\,\,\,\,\,\,\,\,$$ $$= {4 \over 3} \times 8 - {{16} \over 3} = {{16} \over 3}$$

$$\therefore$$ $${S_1}:{S_2}:{S_3} = 1:1:1$$
4

### AIEEE 2005

The area enclosed between the curve $$y = {\log _e}\left( {x + e} \right)$$ and the coordinate axes is
A
$$1$$
B
$$2$$
C
$$3$$
D
$$4$$

## Explanation

The graph of the curve $$y = {\log _e}\left( {x + e} \right)$$ is as shown in the fig. Required area

$$A = \int\limits_{1 - e}^0 {ydx} = \int\limits_{1 - e}^0 {{{\log }_e}} \left( {x + e} \right)dx$$

put $$x + e = t \Rightarrow dx = dt$$

also At $$x = 1 - e,t = 1$$

At $$x = 0,\,\,t = e$$

$$\therefore$$ $$A = \int\limits_1^e {{{\log }_e}} \,tdt = \left[ {t\,{{\log }_e}t - t_1^e} \right]$$

$$e - e - 0 + 1 = 1$$

Hence the required area is $$1$$ square unit.

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