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1

WB JEE 2009

MCQ (Single Correct Answer)

The value of $$\int\limits_0^\infty {{{dx} \over {({x^2} + 4)({x^2} + 9)}}} $$ is

A
$${\pi \over {60}}$$
B
$${\pi \over {20}}$$
C
$${\pi \over {40}}$$
D
$${\pi \over {80}}$$

Explanation

$$\int\limits_0^\infty {{{dx} \over {({x^2} + 4)({x^2} + 9)}} = {1 \over 5}\int\limits_0^\infty {\left[ {{1 \over {{x^2} + 4}} - {1 \over {{x^2} + 9}}} \right]dx} } $$

$$ = {1 \over 5}\left[ {\left[ {{1 \over 2}{{\tan }^{ - 1}}{x \over 2}} \right]_0^\infty - \left[ {{1 \over 3}{{\tan }^{ - 1}}{x \over 3}} \right]_0^\infty } \right]$$

$$ = {1 \over 5}\left[ {{1 \over 2}{{\tan }^{ - 1}}\infty - {1 \over 3}{{\tan }^{ - 1}}\infty } \right] = {1 \over 5}\left[ {{\pi \over 4} - {\pi \over 6}} \right] = {\pi \over {60}}$$

2

WB JEE 2009

MCQ (Single Correct Answer)

If $$f(x) = f(a - x)$$, then $$\int\limits_0^a {xf(x)dx} $$ is equal to

A
$$\int\limits_0^a {f(x)dx} $$
B
$${{{a^2}} \over 2}\int\limits_0^a {f(x)dx} $$
C
$${a \over 2}\int\limits_0^a {f(x)dx} $$
D
$$ - {a \over 2}\int\limits_0^a {f(x)dx} $$

Explanation

Let $$I = \int\limits_0^a {xf(x)dx = \int\limits_0^a {(a - x)f(a - x)dx} } $$ ($$\because$$ $$\int\limits_0^a {f(x)dx = \int\limits_0^a {f(a - x)dx} } $$)

$$ = a\int\limits_0^a {f(a - x)dx - \int\limits_0^a {xf(a - x)dx} } $$

$$ = a\int\limits_0^a {f(x)dx - \int\limits_0^a {xf(x)dx} } $$ ($$\because$$ $$f(a - x) = f(x)$$)

$$ = a\int\limits_0^a {f(x)dx - I \Rightarrow 2I = a\int\limits_0^a {f(x)dx \Rightarrow I = {a \over 2}\int\limits_0^a {f(x)dx} } } $$

3

WB JEE 2008

MCQ (Single Correct Answer)

The value of the $$\mathop {\lim }\limits_{n \to \infty } \left( {{1 \over {n + 1}} + {1 \over {n + 2}} + ... + {1 \over {6n}}} \right)$$ is

A
log 2
B
log 6
C
1
D
log 3

Explanation

Let $$I = \mathop {\lim }\limits_{n \to \infty } \left( {{1 \over {n + 1}} + {1 \over {n + 2}} + ... + {1 \over {6n}}} \right)$$

$$ = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\left( {{1 \over {\left( {1 + {1 \over n}} \right)}} + {1 \over {\left( {1 + {2 \over n}} \right)}} + .... + {1 \over 6}} \right)$$

$$ = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 1}^{5n} {{1 \over {\left( {1 + {r \over n}} \right)}}} $$

$$\therefore$$ $$I = \int\limits_0^5 {{1 \over {1 + x}}dx = \left| {\log (1 + x)} \right|_0^5 = \log 6} $$.

4

WB JEE 2008

MCQ (Single Correct Answer)

The value of the integral $$\int\limits_{ - a}^a {{{x{e^{{x^2}}}} \over {1 + {x^2}}}dx} $$ is

A
$${{e^{{a^2}}}}$$
B
0
C
$${e^{ - {a^2}}}$$
D
a

Explanation

Let $$I = \int\limits_{ - a}^a {{{x{e^{{x^2}}}} \over {1 + {x^2}}}dx} $$

$$f(x) = {{x{e^{{x^2}}}} \over {1 + {x^2}}}$$ is an odd function

$$\therefore$$ $$I = 0$$. ($$\because$$ $$f( - x) = {{ - x{e^{{{( - x)}^2}}}} \over {(1 + {{( - x)}^2}}} = - {{x{e^{{x^2}}}} \over {1 + {x^2}}} = - f(x)$$)

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