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1

WB JEE 2010

MCQ (Single Correct Answer)

If in the expansion (a $$-$$ 2b)n, the sum of the 5th and 6th term is zero, then the value of $${a \over b}$$ is

A
$${{n - 4} \over 5}$$
B
$${{2(n - 4)} \over 5}$$
C
$${5 \over {n - 4}}$$
D
$${5 \over {2(n - 4)}}$$

Explanation

$$\because$$ $${t_5} + {t_6} = 0$$

$$\therefore$$ $${}^n{C_4}{a^{n - 4}}{( - 2b)^4} + {}^n{C_5}{a^{n - 5}}{( - 2b)^5} = 0$$

$$ \Rightarrow {{\left| \!{\underline {\, n \,}} \right. } \over {\left| \!{\underline {\, 4 \,}} \right. \left| \!{\underline {\, {n - 4} \,}} \right. }}a = {{\left| \!{\underline {\, n \,}} \right. } \over {\left| \!{\underline {\, 5 \,}} \right. \left| \!{\underline {\, {n - 5} \,}} \right. }}\,.\,(2b) \Rightarrow {a \over {n - 4}} = {{2b} \over 5}$$

$$ \Rightarrow {a \over b} = {{2(n - 4)} \over 5}$$

2

WB JEE 2010

MCQ (Single Correct Answer)

$$({2^{3n}} - 1)$$ will be divisible by $$(\forall n \in N)$$

A
25
B
8
C
7
D
3

Explanation

$$n \in N$$ $$\therefore$$ n = 1, 2, 3, ......

$$\therefore$$ Putting n = 1, $${2^{3n}} - 1 = {2^3} - 1 = 7$$

$$\therefore$$ $${2^{3n}} - 1$$ will be divisible by $$7\forall n \in N$$

3

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.

4

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 $${x^3} = {}^9{C_3} \times {3^6} \times {a^3}$$

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

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

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