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Solved Objective Question on Binomial Expression Set 2

Posted on - 04-01-2017

Math

IIT JEE

Question.1  

The sum nC0 +nC1 +nC2+----+nCn is eqaul to

(A)

(B) nn

(C) n!

(D) 2n

Solution

(1 + x)n = nC0 + nC1 x + nC2 x2 + nC3 x3 + L nCn xn

Put x = 1,

2n = nC0 + nC1 + nC2 + L + nCn

Question.2

The digits at unit’s place in the number is

(A) 0

(B) 1

(C) 2

(D) 3

Solution

We have

=

= a multiple of 10 + 1

Thus, the unit’s place digits is 1.

Question.3

The coefficient of x5 in the expansion of (1 + x)21 +(1 + x)22 + …..+ (1 + x)30 is.

(A) 51C5

(B) 9C5

(C) 31C6 -21C6

(D) 30C5 + 20C5

Solution

(1 + x)21 +(1 + x)22 + …..+(1 + x)30.

=

\ coefficient of x5 in the given expression

= coefficient of x5 in

= coefficient of x6 in

Question.4

If the co-efficients of x7 and x8 in are equal, then n is

(A) 56

(B) 55

(C) 45

(D) 15

Solution


&⇒
n = 55

Question.5

If z = , then

(A) Re(z) =0

(B) Im(z) =0

(C) Re(z) >0, Im(z) >0

(D) Re(z) >0, Im(z) <0

Solution

z = 2

= Purely real number.

Hence Im(z) = 0

Question.6

If (1-x +x2)n = a0+a1x+a2x2 +------+a2nx2n then a0 +a2 +a4+---+a2n equals

(A)

(B)

(C)

(D)

Solution

Put x = 1
&⇒
1 = a0 + a1 + a2 + L + a2n

Put x = -1
&⇒
3n = a0 - a1 + a2 - a3 + L + a2n

Adding, 3n + 1 = 2 (a0 + a2 + a4 + L+ a2n)

Question.7

The positive integer which is just greater than (1+0.0001)1000 is .

(A) 3

(B) 4

(C) 5

(D) 2

Solution

Expression on expansion gives

1 + 1000 ´ 10-4 + < 1 +

=

So integer just greater than the given expression must be 2.

Question.8

If n is an even natural number and coefficient of xr in the expansion of is 2n, (|x| < 1), then

(A) r £ n/2

(B) r ≥

(C) r £

(D) r ≥ n

Solution

= (C0 + C1x +C2x2 + . . . + Cnxn) (1+ x+ x2 + ….)

The coefficient of xr = C0 + C1+C2 +C3 + . . . + Cr = 2n for r = n.

Moreover coefficient of xr is C0 + C1+C2 +C3 +. . .+ Cr if r > n. So r ≥ n.

Question.9

The number of terms in the expansion of (a+b+c)n, where n∈N, is

(A)

(B) n+1

(C) n+2

(D) (n+1)n

Solution

(a + (b+c))n = an +nC1 an-1(b + c)1+nC2an-2 (b + c)2 + L+ nCn (b + c)n

Further expanding each term of R.H.S.,.

First term on expansion gives one term

Second term on expansion gives two terms

Third term on expansion gives three terms and so on.

\ Total no. of terms = 1 + 2 + 3 + L + (n + 1) =

Question.10

If the co-efficient of the second, third and fourth terms in the expansion of (1+x)n are in A.P. then n is equal to .

(A) 2

(B) 7

(C) 9

(D) None of these

Solution

nC1, nC2, nC3 are in A.P.
&⇒
nC1 + nC3 = 2nC2.


&⇒
n2 - 9n + 14 = 0
&⇒
n = 7,2

But n = 2 is rejected as nC3 is not possible.

 
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