Free Objective Test 02 Practice Test - 11th and 12th
Question 1
The coefficient of x3 in the expansion of (1+3x)21−2x
will be
8
32
50
None of these
SOLUTION
Solution : C
= (1+3x)2(1−2x)−1
= (1+3x)2(1 + 2x + 1.22.1(−2x)2+.......)
= (1+6x+9x2)(1+2x+4x2+8x3+........)
Therefore coefficient of x3 is (8 + 24 + 18) = 50.
Question 2
If 1(x2+1x)43 can be expanded by binomial theorem if
x < 1
|x| < 1
x > 1
|x| > 1
SOLUTION
Solution : D
The expansion is x−83(1+1x3)−43
Hence | 1x3| < 1 ⇒ |x| > 1
Question 3
In the expansion of (1+x1−x)2, the coefficient of xn
4n
4n-3
4n+1
None of these
SOLUTION
Solution : A
Given term can be written as (1+x)2(1−x)−2
=(1+2x+x2)[1+2x+3x2+.......+(n−1)xn−2+nxn−1+(n+1)xn+...........]
= xn(n+1+2n+n−1)+..........
Therefore coefficient of xn is 4n.
Question 4
(r+1)th term in the expansion of (1−x)−4 will be
xrr!
(r+1)(r+2)(r+3)6xr
(r+2)(r+3)2xr
None of these
SOLUTION
Solution : B
(1−x)−4 = [1.2.36x0+2.3.46x+3.4.56x2+4.5.66x3+.......+(r+1)(r+2)(r+3)6xr+.........]
Therefore Tr+1 = (r+1)(r+2)(r+3)6xr.
Question 5
(aa+x)12 + (aa−x)12 =
2 + 3x24a2 + .......
1 + 3x28a2 + .....
2 + xa + 3x24a2 + ......
2 - xa + 3x24a2 + .......
SOLUTION
Solution : A
(a+xa)−12 + (a−xa)−12 =(1+xa)−12+(1−xa)−12
=[1 + (- 12)( xa) + (−12)(−32)2.1(xa)2 + ........]
+ [1+(−12)(−xa)+(−12)(32)2.1(−xa)2+........]
= 2 + 3x24a2 + ......
Here odd terms cancel each other.
Question 6
Cube root of 217 is
6.01
6.04
6.02
None of these
SOLUTION
Solution : A
(217)13 = (63+1)13 = 6(1+163)13
On expansion by binomial theorem
= 6(1+13×216−1×23×3×2(1216)2+.........) = 6.01
Question 7
The fourth term in the expansion of (1−2x)32 will be
- 34x4
x32
-x32
34x4
SOLUTION
Solution : B
Expansion of (1−2x)32
= 1+32(−2x)+32.12.12(−2x)2+32.12(−12)16(−2x)3+.........
Hence 4th term is x32
Question 8
C0−C1+C2−C3+........+(−1)nCn is equal to
2n
2n−1
0
2n−1
SOLUTION
Solution : C
We know that
(1+x)n=nC0+nC1x+nC2x2+......+nCnxn
Putting x = -1, we get
(1−1)n = nC0 - nC1 + nC2 - .......(−1)n nCn
Therefore C0−C1+C2−C3+......(−1)nCn = 0
Question 9
The value of C12 + C34 + C56 + ...... is equal to
2n−1n+1
n.2n
2nn
2n−1
SOLUTION
Solution : A
We know that
(1+x)n−(1−x)n2 = C1x+C3x3+C5x5+........
Integrating from x = 0 to x = 1, we get
12∫10(1+x)n−(1−x)n dx
= ∫10(C1x+C3x3+C5x5+.......) dx
⇒ 12{(1+x)n+1n+1+(1−x)n+1n+1}10=C12 + C34 + C56 + ....
or C12 + C34 + C56 + .......= 12 {2n+1−1n+1 + 0−1n+1}
= 12 2n+1−2n+1 = 2n−1n+1
Question 10
If a and d are two complex numbers, then the sum
to (n+1) terms of the following series
aC0 - (a + d)C1 + (a + 2d)C2 - ........... is
a2n
na
0
None of these
SOLUTION
Solution : C
We can write
aC0 - (a + d)C1 + (a + 2d)C2 - ........ upto (n+1) terms
=a(C0−C1+C2−........)+d(−C1+2C2−3C3+......) .............(i)
Again,(1−x)n=C0−C1x+C2x2−.........+(−1)nCnxn ..........(ii)
Differentiating with respect to x,
−n(1−x)n−1 = -C1 + 2C2x - .......... + (−1)nCnnxn−1 ......................(iii)
Putting x = 1 in (ii) and (iii), we get
C0 - C1 + C2 - ........ + (-1)nCn = 0
and -C1 + 2C2 - ........+(−1)nn.Cn = 0
Thus the required sum to (n+1) terms, by (i)
= a.0 + d.0 = 0.
Question 11
C01 + C23 + C45 + C67 +..........=
2n+1n+1
2n+1−1n+1
2nn+1
None of these
SOLUTION
Solution : C
Putting the values of C0,C2,C4........., we get
= 1 + n(n−1)3.2! + n(n−1)(n−2)(n−3)5.4! + .........
= 1n+1[(n+1) + (n+1)n(n−1)3! + (n+1)n(n−1)(n−2)(n−3)5! + ..........]
Put n+1 = N
= 1N[N + N(N−1)(N−2)3! + N(N−1)(N−2)(N−3)(N−4)5! + .........]
= 1N{NC1 + NC3 + NC5 +..........}
= 1N{2N−1} = 2nn+1 { ∵ N = n+1}
Trick: Put n = 1, then S1 = 1C01 = 11 = 1
At n = 2, S2 = 2C01 + 2C23 = 1 + 13 = 43
Also (c) ⇒ S1 = 1, S2 = 43
Question 12
If (1+x)n=C0+C1x+C2x2+.......+Cnxn, then C1C0+2C2C1+3C3C2+........+nCnCn−1=
n(n−1)2
n(n+2)2
n(n+1)2
(n−1)(n−2)2
SOLUTION
Solution : C
C1C0+2⋅C2C1+3⋅C3C2+........+n⋅CnCn−1
=n1+2n(n−1)1.2n+3n(n−1)(n−2)3.2.1n(n−1)1.2+......+n.1n
=n+(n−1)+(n−2)........+1=∑n=n(n+1)2
Question 13
If (1+x)n=C0+C1x+C2x2+........+Cnx2, then
C20+C21+C22+C23+..........+C2n =
n!n!n!
(2n)!n!n!
(2n)!n!
None of these
SOLUTION
Solution : B
(1+x)n=C0+C1x+C2x2+.........+CnXn ...........(i)
and (1+1x)n =C0+C11x+C2(1x)2+........+Cn(1x)n ..........(ii)
If we multiply (i) and (ii), we get
C20+C21+C22+........+C2n
is the term independent of x and hence it is equal to the term independent of x in the product (1+x)n(1+1x)n or in 1xn(1+x)2n or term
containing xn in (1+x)2n. Clearly the coefficient of xn in (1+x)2x is Tn+1 and equal to 2nCn = (2n)!n!n!
Trick: Solving conversely.
Put n = 1, n = 2,.........then we get S1 = 1C20+1C21= 2,
S2 = 2C20+2C21+2C22=12+22+12 = 6
Now check the options
(a) Does not hold given condition,
(b) (i) Put n = 1, then 2!1!1!=2
(ii) Put n = 2, then 4!2!2!=4×3×2×12×1×2×1=6
Note: Students should remember this question as identity
Question 14
In the expansion of ( ax+bx)12, the coefficient of x−10 will be
12a11
12b11a
12a11b
12a11b11
SOLUTION
Solution : C
Tr+1=12cr(ax)12−r(bx)r =12cr.a12−r.br.x2r−12∴2x−12=−10⇒r=1∴coefficient=12.a11.b
Question 15
If the coefficients of pth, (p+1)th and (p+2)th terms in the expansion of (1+x)n are in A.P., then
n2−2np+4p2 = 0
n2−n(4p+1)+4p2−2 = 0
n2−n(4p+1)+4p2 = 0
None of these
SOLUTION
Solution : B
2ncp=ncp−1+pcp+1⇒2(n−p)!P!=1(n−p+1)!(p−1)!+1(n−p−1)!(p+1)!⇒2(n−p)p=1(n−p+1)(n−p)+1(p+1)pon simplifing we getn2−n(4p+1)+4p2−2=0