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1

### AIEEE 2007

The solution for $$x$$ of the equation $$\int\limits_{\sqrt 2 }^x {{{dt} \over {t\sqrt {{t^2} - 1} }} = {\pi \over 2}}$$ is
A
$${{\sqrt 3 } \over 2}$$
B
$$2\sqrt 2$$
C
$$2$$
D
None

## Explanation

$$\int_{\sqrt 2 }^x {{{dt} \over {t\sqrt {{t^2} - 1} }}} = {\pi \over 2}$$

$$\therefore$$ $$\left[ {{{\sec }^{ - 1}}t} \right]_{\sqrt 2 }^x = {\pi \over 2}$$

$$\left[ {} \right.$$ As $$\int {{{dx} \over {x\sqrt {{x^2} - 1} }}} = {\sec ^{ - 1}}x$$ $$\left. {} \right]$$

$$\Rightarrow {\sec ^{ - 1}}x - {\sec ^{ - 1}}\sqrt 2 = {\pi \over 2}$$

$$\Rightarrow {\sec ^{ - 1}}x - {\pi \over 4} = {\pi \over 2}$$

$$\Rightarrow {\sec ^{ - 1}}x = {\pi \over 2} + {\pi \over 4}$$

$$\Rightarrow {\sec ^{ - 1}}x = {{3\pi } \over 4}$$

$$\Rightarrow x = \sec {{3\pi } \over 4}$$

$$\Rightarrow x = - \sqrt 2$$
2

### AIEEE 2007

Let $$F\left( x \right) = f\left( x \right) + f\left( {{1 \over x}} \right),$$ where $$f\left( x \right) = \int\limits_l^x {{{\log t} \over {1 + t}}dt,}$$ Then $$F(e)$$ equals
A
$$1$$
B
$$2$$
C
$$1/2$$
D
$$0$$

## Explanation

Given $$f\left( x \right) = f\left( x \right) + f\left( {{1 \over x}} \right),$$

where $$f\left( x \right) = \int_1^x {{{\log \,t} \over {1 + t}}} \,dt$$

$$\therefore$$ $$F\left( e \right) = f\left( e \right) + f\left( {{1 \over e}} \right)$$

$$\Rightarrow F\left( e \right)$$

$$= \int_1^e {{{\log \,t} \over {1 + t}}dt + \int_1^{1/e} {{{\log t} \over {1 + t}}} } dt\,\,\,....\left( A \right)$$

Now for solving, $$I = \int_1^{1/e} {{{\log t} \over {1 + t}}dt}$$

$$\therefore$$ Put $${1 \over t} = z \Rightarrow - {1 \over {{t^2}}}dt = dz$$

$$\Rightarrow dt = - {{dz} \over {{z^2}}}$$

and limit for $$t = 1 \Rightarrow z = 1$$ and for

$$t = 1/e \Rightarrow z = e$$

$$\therefore$$ $$I = \int_1^e {{{\log \left( {{1 \over z}} \right)} \over {1 + {1 \over z}}}} \left( { - {{dz} \over {{z^2}}}} \right)$$

$$= \int_1^e {{{\left( {\log 1 - \log z} \right).z} \over {z + 1}}\left( { - {{dz} \over {{z^2}}}} \right)}$$

$$= \int_1^e { - {{\log z} \over {\left( {z + 1} \right)}}\left( { - {{dz} \over z}} \right)}$$

[ as $$\log 1 = 0$$ ]

$$= \int_1^e {{{\log z} \over {z\left( {z + 1} \right)}}} dz$$

$$\therefore$$ $$I = \int_1^e {{{\log \,t} \over {t\left( {t + 1} \right)}}dt}$$

$$\left[ {} \right.$$ By property $$\int_a^b {f\left( t \right)dt}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$= \int_a^b {f\left( x \right)dx}$$ $$\left. {} \right]$$

Equation $$(A)$$ becomes

$$F\left( e \right) = \int_1^e {{{\log t} \over {1 + t}}dt + \int_1^e {{{\log t} \over {t\left( {1 + t} \right)}}} } dt$$

$$= \int_1^e {{{t.\log t + \log t} \over {t\left( {1 + t} \right)}}} dt$$

$$= \int_1^e {{{\left( {\log t} \right)\left( {t + 1} \right)} \over {t\left( {1 + t} \right)}}}$$

$$\Rightarrow F\left( e \right) = \int_1^e {{{\log t} \over t}dt}$$

Let $$\log t = x$$

$$\therefore$$ $${1 \over t}dt = dx$$

$$\left[ {} \right.$$ for limit $$t = 1,x = 0$$

and $$t = e,x = \log \,e = 1$$ $$\left. {} \right]$$

$$\therefore$$ $$F\left( e \right) = \int_0^1 {x\,dx}$$

$$\Rightarrow F\left( e \right) = \left[ {{{{x^2}} \over 2}} \right]_0^1$$

$$\Rightarrow F\left( e \right) = {1 \over 2}$$
3

### AIEEE 2006

The value of $$\int\limits_1^a {\left[ x \right]} f'\left( x \right)dx,a > 1$$ where $${\left[ x \right]}$$ denotes the greatest integer not exceeding $$x$$ is
A
$$af\left( a \right) - \left\{ {f\left( 1 \right) + f\left( 2 \right) + .............f\left( {\left[ a \right]} \right)} \right\}$$
B
$$\left[ a \right]f\left( a \right) - \left\{ {f\left( 1 \right) + f\left( 2 \right) + ...........f\left( {\left[ a \right]} \right)} \right\}$$
C
$$\left[ a \right]f\left( {\left[ a \right]} \right) - \left\{ {f\left( 1 \right) + f\left( 2 \right) + ...........f\left( a \right)} \right\}$$
D
$$af\left( {\left[ a \right]} \right) - \left\{ {f\left( 1 \right) + f\left( 2 \right) + .............f\left( a \right)} \right\}$$

## Explanation

Let $$a = k + h$$ where $$k$$ is an integer such that

$$\left[ a \right] = k$$ and $$0 \le h < 1$$

$$\therefore$$ $$\int\limits_1^a {\left[ x \right]f'\left( x \right)dx = \int\limits_1^2 {1f'\left( x \right)} } \,dx$$

$$\,\,\,\,\,\,\,\,\,\,\,\, + \int\limits_2^3 {2f'\left( x \right)dx + } ....$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\int\limits_{k - 1}^k {\left( {k - 1} \right)dx + \int\limits_k^{k + h} {kf'\left( x \right)} } dx$$

$$\left\{ {f\left( 2 \right) - f\left( 1 \right)} \right\} + 2\left\{ {f\left( 3 \right) - f\left( 2 \right)} \right\}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$+ 3\left\{ {f\left( 4 \right) - f\left( 3 \right)} \right\} + \,........\,$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$+ \left( {k - 1} \right)\left\{ {f\left( k \right) - f\left( {k - 1} \right)} \right\}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$+ k\left\{ {f\left( {k + h} \right) - f\left( k \right)} \right\}$$

$$= - f\left( 1 \right) - f\left( 2 \right) - f\left( 3 \right).......$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$- f\left( k \right) + kf\left( {k + h} \right)$$

$$= \left[ a \right]f\left( a \right) - \left\{ {f\left( 1 \right) + f\left( 2 \right)} \right.$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\left. { + f\left( 3 \right) + ........f\left( {\left[ a \right]} \right)} \right\}$$
4

### AIEEE 2006

$$\int\limits_{ - {{3\pi } \over 2}}^{ - {\pi \over 2}} {\left[ {{{\left( {x + \pi } \right)}^3} + {{\cos }^2}\left( {x + 3\pi } \right)} \right]} dx$$ is equal to
A
$${{{\pi ^4}} \over {32}}$$
B
$${{{\pi ^4}} \over {32}} + {\pi \over 2}$$
C
$${\pi \over 2}$$
D
$${\pi \over 4} - 1$$

## Explanation

$$I = \int\limits_{ - {{3\pi } \over 2}}^{ - {\pi \over 2}} {\left[ {{{\left( {x + \pi } \right)}^3} + {{\cos }^2}\left( {x + 3\pi } \right)} \right]} \,dx$$

Put $$x + \pi = t$$

$$I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\left( {{t^3} + {{\cos }^2}t} \right)dt}$$

$$= 2\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\cos }^2}} tdt$$

$$\left[ {} \right.$$ using the property of even and odd function $$\left. {} \right]$$

$$= \int\limits_0^{{\pi \over 2}} {\left( {1 + \cos 2t} \right)} dt = {\pi \over 2} + 0$$

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