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1

WB JEE 2008

MCQ (Single Correct Answer)

If $$h(x) = \int\limits_0^x {{{\sin }^4}t\,dt} $$, then $$h(x + \pi )$$ equals

A
$${{h(x)} \over {h(\pi )}}$$
B
$$h(x)h(\pi )$$
C
$$h(x) - h(\pi )$$
D
$$h(x) + h(\pi )$$

Explanation

$$h(x) = \int\limits_0^x {{{\sin }^4}t\,dt} $$

Taking $$h(x + \pi ) = \int\limits_0^{x + \pi } {{{\sin }^4}t\,dt} $$

$$ = \int\limits_0^x {{{\sin }^4}t\,dt + \int\limits_x^{x + \pi } {{{\sin }^4}t\,dt = h(x) + \int\limits_x^{x + \pi } {{{\sin }^4}t\,dt} } } $$

Since $$f(t) = {\sin ^4}t$$ is a periodic function of period $$\pi$$

$$\therefore$$ $$h(x + \pi ) = h(x) + \int\limits_0^\pi {{{\sin }^4}t\,dt = h(x) + h(\pi )} $$.

($$\int\limits_a^{a + nT} {f(x)dx = n\int\limits_0^T {f(x)dx,\,f(x)} } $$ is periodic function of period T)

2

WB JEE 2008

MCQ (Single Correct Answer)

If $$I = \int\limits_{ - \pi }^\pi {{{{e^{\sin x}}} \over {{e^{\sin x}} + {e^{ - \sin x}}}}dx} $$, then I equals

A
$$\pi$$/2
B
2$$\pi$$
C
$$\pi$$
D
$$\pi$$/4

Explanation

$$I = \int\limits_{ - \pi }^\pi {{{{e^{\sin x}}} \over {{e^{\sin x}} + {e^{ - \sin x}}}}dx} $$

$$I = \int\limits_{ - \pi }^0 {{{{e^{\sin x}}} \over {{e^{\sin x}} + {e^{ - \sin x}}}}dx + \int\limits_0^\pi {{{{e^{\sin x}}} \over {{e^{\sin x}} + {e^{ - \sin x}}}}dx} } $$

Put x = $$-$$y, dx = $$-$$ dy (in Ist integral)

$$ = - \int\limits_\pi ^0 {{{{e^{\sin ( - y)}}} \over {{e^{\sin ( - y)}} + {e^{ - \sin ( - y)}}}}dy + \int\limits_0^\pi {{{{e^{\sin x}}} \over {{e^{\sin x}} + {e^{ - \sin x}}}}dx} } $$

$$ = \int\limits_0^\pi {{{{e^{ - \sin y}}} \over {{e^{\sin y}} + {e^{ - \sin y}}}}dy + \int\limits_0^\pi {{{{e^{\sin x}}} \over {{e^{\sin x}} + {e^{ - \sin x}}}}dx} } $$

(using $$\int\limits_a^b {f(x)dx = - \int\limits_b^a {f(x)dx} } $$ and $$\int\limits_a^b {f(x)dx = \int\limits_a^b {f(y)dy} } $$)

$$ = \int\limits_0^\pi {{{({e^{ - \sin x}} + {e^{\sin x}})} \over {({e^{\sin x}} + {e^{ - \sin x}})}} = \int\limits_0^\pi {1\,dx = \pi } } $$.

3

WB JEE 2008

MCQ (Single Correct Answer)

$$\int\limits_{ - \pi /2}^{\pi /2} {{{\sin }^9}x{{\cos }^5}x\,dx} $$ equals

A
$${1 \over {20}}$$
B
20
C
0
D
$${1 \over {330}}$$

Explanation

Let $$I = \int\limits_{ - \pi /2}^{\pi /2} {{{\sin }^9}x{{\cos }^5}x\,dx} $$

Let $$f(x) = {\sin ^9}x{\cos ^5}x$$

$$f( - x) = {\sin ^9}( - x){\cos ^5}( - x) = {( - \sin x)^9}{\cos ^5}x$$

$$ = - {\sin ^9}x{\cos ^5}x$$

since $$f(x) = - f(x)$$

$$\therefore$$ f(x) is an odd function $$\therefore$$ I = 0.

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