NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

### 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 $$0 \le h < 1$$

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

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

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

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

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

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

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

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

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

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

$$\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\left. { + f\left( 3 \right) + ........f\left( {\left[ a \right]} \right)} \right\}$$
2

### 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 $$\left. {} \right]$$

$$= \int\limits_0^{{\pi \over 2}} {\left( {1 + \cos 2t} \right)} dt = {\pi \over 2} + 0$$
3

### AIEEE 2006

$$\int\limits_0^\pi {xf\left( {\sin x} \right)dx}$$ is equal to
A
$$\pi \int\limits_0^\pi {f\left( {\cos x} \right)dx}$$
B
$$\,\pi \int\limits_0^\pi {f\left( {sinx} \right)dx}$$
C
$${\pi \over 2}\int\limits_0^{\pi /2} {f\left( {sinx} \right)dx}$$
D
$$\pi \int\limits_0^{\pi /2} {f\left( {\cos x} \right)dx}$$

## Explanation

$$I = \int\limits_0^\pi {xf\left( {\sin \,x} \right)dx}$$

$$= \int\limits_0^\pi {\left( {\pi - x} \right)f\left( {\sin x} \right)dx}$$

$$= \pi \int\limits_0^\pi {f\left( {\sin x} \right)dx - 1}$$

$$\Rightarrow 2I = \pi {\pi \over 0}f\left( {\sin x} \right)dx$$

$$I = {\pi \over 2}\int\limits_0^\pi {f\left( {\sin x} \right)dx}$$

$$= \pi \int\limits_0^{\pi /2} {f\left( {\sin x} \right)dx}$$

$$= \pi \int\limits_0^{\pi /2} {f\left( {\cos x} \right)dx}$$
4

### AIEEE 2005

The value of integral, $$\int\limits_3^6 {{{\sqrt x } \over {\sqrt {9 - x} + \sqrt x }}} dx$$ is
A
$${1 \over 2}$$
B
$${3 \over 2}$$
C
$$2$$
D
$$1$$

## Explanation

$$I = \int\limits_3^6 {{{\sqrt x } \over {\sqrt {9 - x} + \sqrt x }}} dx\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

$$I = \int\limits_3^6 {{{\sqrt {9 - x} } \over {\sqrt {9 - x} + \sqrt x }}} dx\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$

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

Adding equation $$(1)$$ and $$(2)$$

$$2I = \int\limits_3^6 {dx} = 3 \Rightarrow I = {3 \over 2}$$

### Joint Entrance Examination

JEE Main JEE Advanced WB JEE

### Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

NEET

Class 12