Question.1
The ratio of the
co-efficient of x15 to the term independent of x in [x2+2/x]15
is
(A) 12: 32
(B) 1:32
(C) 32 :12
(D) 32:1
Solution
General term in the expansion is 15Cr(x2)15-r 
i.e. 15Cr
x30-3r×2r
Coefficient of x15
is 15C5 25 (r = 5)
Coefficient of
constant term is 15C10 210 (r =
10)
Ratio is 1 : 32. 
.
Question.2
If f(x) = xn, then the value of
f(1) + 
, where fr (x) denotes
the rth order derivative of f(x) with respect to x, is
(A) n
(B) 2n
(C) 2n –1
(D) none of these
Solution
We have f (x) = xn. So,.
Now,
Question.3
The value of 
(A) 1
(B) 2
(C) 3
(D) none of these
Solution
The numerator is of the form
a3 + b3 + 3ab (a+b) =
(a+b)3
Where a = 18, and b = 7
\
Nr = (18+7)3 = (25)3
Denominator can be written as
36 + 6C1.35.21
+ 6C2.34.22 +6C3
33.23+ 6C4 32.24+
6C5 3.25 + 6C626.
= (3+2)6 = 56 = (25)3
\
= 
Question.4
The term independent of x in 
is
(A) 1
(B) 5/12
(C) 10C1
(D) None of these
Solution
General term in the expansion is 
= 
For constant term, 
&⇒ 
which is not an integer.
Therefore, there will be no constant term .
Question.5
If the sum of the
coefficients in the expansion of (1 +2x)n is 6561, the greatest term
in the expansion for x = 
is
(A) 4th
(B) 5th
(C) 6th
(D) none of these
Solution
sum of the coefficient in the expansion of
(1 +2x)n = 6561
&⇒ (1 +2x)n =
6561, when x = 1
&⇒ 3n = 6561
&⇒ 3n = 38
&⇒ n
= 8
Now, 
&⇒ 
[
x = ½]
\ 
Hence, 5th term
is the greatest term.
Question.6
Given the integers r>1, n> 2, and
co-efficients of (3r) th and (r+2)nd term in the
binomial expansion of (1+x)2n are equal, then
(A) n = 2r
(B) n =3r
(C) n = 2r+1
(D) None of these
Solution
Coefficients of (3r)th and (r +
2)th terms will be 2nC3r-1 and 2nCr+1
These are equal
&⇒ (3r - 1) + (r + 1) = 2n
&⇒
n = 2r
Question.7
If xm
occurs in the expansion of 
, then the
co-efficient of xm is
(A) 
(B) 
(C) 
(D) None of these
Solution
General term in the
expansion is 
= 
For xm,
2n - 3r = m
&⇒ 
So coefficient of xm
is 
Question.8
The co-efficients of
xp and xq (p and q are positive integers) in the
expansion of (1+x)p+q are
(A) equal
(B) equal with
opposite signs
(C) reciprocals to
each other
(D) None of these
Solution
The coefficients of xp and xq
are 
Both of which will
be equal.
Question.9
The value of the
expression
is
(A) 2
(B) 1
(C) 3
(D) 0
Solution
The expression can
be divided into two parts.
+ 
= 
= 
=
=
0 .
Question.10
In the usual
notations C1+2C2x +3C3x2+-----+nCnxn-1
is equal to
(A) n(1+x)n-1
(B) n(1+x)n
(C) (n-1)(1+x)n-1
(D) (n-1)(1+x)n
Solution
(1 +x)n = C0 +C1x
+ C2x2 +C3x3 + . . . . + Cn
xn.
Differentiating,
n(1+ x)n-1
= C1 + 2C2x + 3C3x2 + . . .nCn
xn-1.
