Free Objective Test 01 Practice Test - 11th and 12th
Question 1
The value of (√2+1)6 + (√2−1)6 will be
-198
198
99
-99
SOLUTION
Solution : B
(x+a)n + (x−a)n = 2[xn+nC2xn−2a2+nC4xn−4a4+nC6xn−6a6+.......]
Here, n = 6, x = √2, a = 1;
6C2 = 15, 6C4 = 15, 6C6 = 1
∴ (√2+1)6 + (√2−1)6 = 2[(√2)6+15.(√2)4.1+15(√2)2.1+1.1]
=2[8+15×4+15×2+1]=198
Question 2
Middle term in the expansion of (1+3x+3x2+x3)6 is
4th
3rd
10th
None of these
SOLUTION
Solution : C
(1+3x+3x2+x3)6 = ((1+x)3)6 = (1+x)18
Hence the middle term is 10th.
Question 3
The middle term in the expansion of (1+x)2n is
1.3.5.......(5n−1)n!xn
2.4.6.......2nn!x2n+1
1.3.5.......(2n−1)n!xn
1.3.5.......(2n−1)n!2nxn
SOLUTION
Solution : D
Middle term of (1+2x)2n is Tn+1 = 2nCnxn
= (2n)!n!n!xn = 1.3.5.......(2n−1)n! 2nxn.
Question 4
If the coefficient of (2r+4)th and (r−2)th terms in the expansion of (1+x)18 are equal, then r =
12
10
8
6
SOLUTION
Solution : D
18C2r+3 = 18Cr−3 ⇒ 2r + 3 + r - 3 = 18 ⇒ r = 6
Question 5
If the coefficient of x in the expansion of (x2+kx)5 is 270, then k =
1
2
3
4
SOLUTION
Solution : C
Tr+1 = 5Cr(x2)5−r(kx)r
For coefficient of x, 10 - 2r - r = 1 ⇒ r = 3
Hence, T3+1 = 5C3(x2)5−3(kx)3
According to question, 10k3 = 270 ⇒ k = 3.
Question 6
If the coefficient of x7 in (ax2+1bx)11 is equal to the coefficient of x−7 in (ax−1bx2)11, then ab =
1
12
2
3
SOLUTION
Solution : A
In the expansion of (ax2+1bx)11, the general
term is
Tr+1 = 11Cr(ax2)11−r(1bx)r = 11Cr(a)11−r(1br)x22−3r
For x7, we must have 22 - 3r = 7 ⇒ r = 5, and
the coefficient of x7 = 11C5.a11−515) = 11C5 a6b5
Similarly, in the expansion of (ax−1bx2)11, the
general term is Tr+1 = 11Cr(−1)r a11−rbr.x11−3r
For x−7 we must have, 11 - 3r = -7 ⇒ r = 6, and
the coefficient of x−7 is 11C6 a5b6 = 11C5 a5b6.
As given, 11C5 a6b5 = 11C5 a5b6 ⇒ ab = 1.
Question 7
In the polynomial (x - 1)(x - 2)(x - 3)............... .........(x - 100), the coefficient of x99 is
5050
-5050
100
99
SOLUTION
Solution : B
(x - 1)(x - 2)(x - 3)..............(x - 100)
Number of terms = 100;
∴ Coefficient of x99
= (x - 1)(x - 2)(x - 3)........(x -100)
= (-1 - 2 - 3 - .......... - 100) = -(1 + 2 +.........+ 100)
= - 100×1012 = - 5050.
Question 8
If the coefficient of the middle term in the expansion of (1+x)2n+2 is p and the coefficients of middle terms in the expansion of (1+x)2n+1 are q and r, then
p + q = r
p + r = q
p = q + r
p + q + r = 0
SOLUTION
Solution : C
Since (n+2)th term is the middle term in the
expansion of (1+x)2n+2, therefore p = 2n+2Cn+1.
Since (n+1)th and (n+2)th terms are middle terms in the expansion of (1+x)2n+1,
therefore q = 2n+1Cn and r =2n+1C2n+1.But ,2n+1Cn +2n+1Cn+1 = 2n+2Cn+1
∴ q + r = p
Question 9
The value of x in the expression [x+xlog10(x)]5, if the third term in the expansion is 10,00,000
10
11
12
None of these
SOLUTION
Solution : A
T3 = 5C2. x2(xlog10x)3 = 106
Put 5C2 = 10 [ ∵log1010=1].
If x = 10, then 103.102.1 = 105 is satisfied.
Hence x = 10.
Question 10
If (1−3x)12+(1−x)53√4−x is approximately equal to a + bx for small values of x, then (a, b) =
(1, 3524)
(1, - 3524)
(2, 3512)
(2, - 3512)
SOLUTION
Solution : B
(1−3x)1/2+(1−x)5/32[1−x4]1/212[(1−32x)+(1−5x3)](1+x8)12[2−196x]](1+x8)=12[2−(−14+196)x]=1−3524x∴(a,b)=(1,−3524)
Question 11
If (1+ax)n = 1 + 8x + 24 x2 + ......, then the value of a and n is
2, 4
2, 3
3, 6
1, 2
SOLUTION
Solution : A
As given (1+ax)n = 1 + 8x + 24x2 + .........
⇒ 1 + n1ax + n(n−1)1.2a2x2 + ......... = 1 + 8x + 24x2 + ...........
⇒na = 8, n(n−1)1.2a2 = 24 ⇒ na(n-1)a = 48
⇒8(8 - a) = 48 ⇒ 8 - a = 6 ⇒ a = 2 ⇒ n = 4.
Question 12
If P is a prime number, then np - n is divisible by p when n is a
Natural number greater than 1
Irrational number
Complex number
Odd number
SOLUTION
Solution : A
np - n is divisible by p for any natural number
greater than 1. It is Fermet's theorem.
Trick: Let n = 4 and p = 2
Then (4)2 - 4 = 16 - 4 = 12, it is divisible by 2.
So,it is true for any natural number greater than 1.
Question 13
The value of the expression (2+√2)4 lies between
134 and 135
135 and 136
136 and 137
None of these
SOLUTION
Solution : B
(2+√(2))4 = (√2)4(√2+1)4
= 4[4C0 + 4C1(√2) + 4C2(√2)2 + 4C3(√2)3 +4C4 (√2)4]
= 4 [1 + 4 √2 + 4.32.2 + 4.3.21.2.3.2 √2 + 4]
= 4[1 + 4 √2 + 12 + 8 √2 + 4] = 4[17 + 12 √2]
= 4[17 + 17] = 4[34] = 136
Question 14
The number of integral terms in the expansion of (512+716)642 is
106
108
103
109
SOLUTION
Solution : B
Tr+1 = 642Cr(512)642−r.(716)r
Obviously, r should be a multiple of 6.
Total number of terms = 6426 = 107,
but first term for r = 0 is also an integer. Hence total terms are 107 + 1 = 108.
Question 15
If a1,a2,a3,a4 are the coefficients of any four consecutive terms in the expansion of (1+x)n, then a1a1+a2+a3a3+a4=
a2a2+a3
12 a2a2+a3
2a2a2+a3
2a3a2+a3
SOLUTION
Solution : C
Let a1, a2, a3, a4 be respectively the coefficients of (r+1)th, (r+2)th, (r+3)th and (r+4)th terms in the expansion of (1+x)n.
Then a1=nCr, a2=nCr+1, a3=nCr+2,a4=nCr+3
Now,
a1a1+a2+a3a3+a4=nCrnCr+nCr+1+nCr+2nCr+2+nCr+3=nCrn+1Cr+1+nCr+2n+1Cr+3=nCrn+1r+1nCr+nCrn+1r+3nCr+2=r+1n+1+r+3n+3=2(r+2)n+1
Also, solving the R.H.S, we get
2a2a2+a3=2nCr+1nCr+1+nCr+2=2nCr+1n+1Cr+2=2(r+2)n+1