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1

WB JEE 2009

MCQ (Single Correct Answer)

Product of any r consecutive natural numbers is always divisible by

A
r!
B
(r + 4)!
C
(r + 1)!
D
(r + 2)!

Explanation

Product of r consecutive natural numbers

= 1 . 2 . 3 . 4 . ...... . r = r! which is divisible by r! .

2

WB JEE 2008

MCQ (Single Correct Answer)

The coefficient of x$$-$$10 in $${\left( {{x^2} - {1 \over {{x^3}}}} \right)^{10}}$$ is

A
$$-$$ 252
B
$$-$$ 210
C
$$-$$ (5!)
D
$$-$$ 120

Explanation

Let $${T_{r + 1}}$$ contains coefficient of x$$-$$10 in $${\left( {{x^2} - {1 \over {{x^3}}}} \right)^{10}}$$

$$\therefore$$ $${T_{r + 1}} = {}^{10}{C_r}{({x^2})^{10 - r}}{\left( { - {1 \over {{x^3}}}} \right)^r}$$

$$ = {}^{10}{C_r}{( - 1)^r}{x^{20 - 5r}}$$

Comparing $${x^{ - 10}} = {x^{20 - 5r}}$$

$$ \Rightarrow - 10 = 20 - 5r \Rightarrow r = 6$$

$$\therefore$$ Coefficient $$ = {}^{10}{C_6}{( - 1)^6} = 210$$

$$\therefore$$ no option is given.

3

WB JEE 2008

MCQ (Single Correct Answer)

If $${}^{16}{C_r} = {}^{16}{C_{r + 1}}$$, then the value of $${}^r{P_{r - 3}}$$ is

A
31
B
120
C
210
D
840

Explanation

$$\because$$ $${}^{16}{C_r} = {}^{16}{C_{r + 1}}$$

$$ \Rightarrow {}^{16}{C_r} = {}^{16}{C_{15 - r}} \Rightarrow r = 15 - r$$

$$ \Rightarrow r = 7.5$$ (not possible since r is whole number)

So no option is given.

4

WB JEE 2008

MCQ (Single Correct Answer)

If the magnitude of the coefficient of x7 in the expansion of $${\left( {a{x^2} + {1 \over {bx}}} \right)^8}$$, where a, b are positive numbers, is equal to the magnitude of the coefficient of x7 in the expansion of $${\left( {ax + {1 \over {b{x^2}}}} \right)^8}$$, then a and b are connected by the relation

A
ab = 1
B
ab = 2
C
a2b = 1
D
ab2 = 2

Explanation

Let (r + 1)th term contains x7 in the expansion of $${\left( {a{x^2} + {1 \over {bx}}} \right)^8}$$

$$\therefore$$ $${T_{r + 1}} = {}^8{C_r}{(a{x^2})^{8 - r}}{\left( {{1 \over {bx}}} \right)^r}$$

$$\therefore$$ $${x^{16 - 3r}} = {x^7} \Rightarrow r = 3$$

$$\therefore$$ Coefficient of $${x^7} = {}^8{C_3}{a^5}{b^{ - 3}}$$ ..... (i)

Let (r + 1)th term contains x$$-$$7 in the expansion of $${\left( {ax + {1 \over {b{x^2}}}} \right)^8}$$

$$\therefore$$ $${T_{r + 1}} = {}^8{C_r}{(ax)^{8 - r}}{\left( {{1 \over {b{x^2}}}} \right)^r} = {}^8{C_r}{a^{8 - r}}{(b)^{ - r}}{x^{8 - 3r}}$$

$$\therefore$$ $${x^{8 - 3r}} = {x^{ - 7}} \Rightarrow r = 5$$

$$\therefore$$ coefficient of $${x^{ - 7}} = {}^8{C_5}{a^3}{(b)^{ - 5}} = {}^8{C_5}{a^3}{b^{ - 5}}$$ ..... (ii)

from (i) and (ii)

$${}^8{C_3}{a^5}{b^{ - 3}} = {}^8{C_5}{a^3}{b^{ - 5}}$$

$$ \Rightarrow {a^2}{b^2} = 1 \Rightarrow ab = 1$$ or $$-$$1.

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