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1

### WB JEE 2009

MCQ (Single Correct Answer)

using binomial theorem, the value of (0.999)3 correct to 3 decimal places is

A
0.999
B
0.998
C
0.997
D
0.995

## Explanation

(0.999)3 = (1 $$-$$ 0.001)3

= 1 $$-$$ 3 $$\times$$ (0.001) = 0.997.

2

### WB JEE 2009

MCQ (Single Correct Answer)

If the coefficients of x2 and x3 in the expansion of (3 + ax)9 be same, then the value of a is

A
3/7
B
7/3
C
7/9
D
9/7

## Explanation

$${T_3} = {}^9{C_2}{(3)^{9 - 2}}.\,{(ax)^2} = {}^9{C_2}{3^7}\,.\,{a^2}\,.\,{x^2}$$

$$\therefore$$ Coeff. of $${x^2} = {}^9{C_2} \times {3^7} \times {a^2}$$

$${T_4} = {}^9{C_3}{(3)^{9 - 3}}\,.\,{(ax)^3} = {}^9{C_3}\,.\,{3^6}\,.\,{a^3}\,.\,{x^3}$$

$$\therefore$$ Coeff. of $${x^3} = {}^9{C_3} \times {3^6} \times {a^3}$$

$$\therefore$$ $${}^9{C_2} \times {3^7} \times {a^2} = {}^9{C_3} \times {3^6} \times {a^3}$$

$$\therefore$$ $$a = {{{}^9{C_2} \times {3^7}} \over {{}^9{C_3} \times {3^6}}} = {{{{9!} \over {2! \times 7!}} \times 3} \over {{{9!} \over {3! \times 6!}}}} = {{3! \times 6! \times 3} \over {2! \times 7!}} = {9 \over 7}$$

3

### WB JEE 2009

MCQ (Single Correct Answer)

If C0, C1, C2, ......, Cn denote the coefficients in the expansion of (1 + x)n then the value of C1 + 2C2 + 3C3 + ..... + nCn is

A
n . 2n $$-$$ 1
B
(n + 1)2n $$-$$ 1
C
(n + 1)2n
D
(n + 2)2n $$-$$ 1

## Explanation

$${(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + ..... + {C_n}{x^n}$$ (Binomial theorem)

Differentiating both sides w.r. to x, we get

$$n{(1 + x)^{n - 1}} = {C_1} + 2{C_2}x + 3{C_3}{x^2} + ..... + n{C_n}{x^{n - 1}}$$

Keeping x = 1 we get

$$n\,.\,{2^{n - 1}} = {C_1} + 2{C_2} + 2{C_3} + .... + n{C_n}$$

4

### WB JEE 2009

MCQ (Single Correct Answer)

For each n $$\in$$ N, 23n $$-$$ 1 is divisible by

here N is a set of natural numbers.

A
7
B
8
C
6
D
16

## Explanation

Let P(n) = 23n $$-$$ 1

Putting n = 1

P(1) = 23 . 1 $$-$$ 1 = 7 which is divisible by 7

Putting n = 2

P(2) = 23 . 2 $$-$$ 1 = 26 $$-$$ 1 = 63 which is divisible by 7 and so on

Let P(k) = 23k $$-$$ 1 is also divisible by 7

$$\therefore$$ 23k $$-$$ 1 = 7P $$\Rightarrow$$ 23k = 7P + 1

P(k + 1) = 23(k + 1) $$-$$ 1 = 23k . 23 $$-$$ 1

= (7P + 1)8 $$-$$ 1 = 7 . 8P + 8 $$-$$ 1 = 7(8P + 1)

$$\therefore$$ By process of M.I. P(k) is divisible by 7.

So, 23n $$-$$ 1 is divisible by 7.

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