# Free Objective Test 03 Practice Test - 11th and 12th

The first 3 terms in the expansion of (1+ax)n (n ≠ 0) are 1, 6x and 16x2. Then the value of a and n are respectively

A.

2 and 9

B.

3 and 2

C.

23 and 9

D.

32 and 6

#### SOLUTION

Solution : C

nc1.ax=6x;nc2(ax)2=16x2an=6;n(n1)2a2=16a=23n=9

If the third term in the binomial expansion of (1+x)m is -18x2, then the rational value of m is

A.

2

B.

12

C.

3

D.

4

#### SOLUTION

Solution : B

mc2=18m(m1)2=184m24m+1=0(2m1)2=0m=12

If the (r+1)th term in the expansion of (3ab+b3a)21 has the same power of a and b, then value of r is

A.

9

B.

10

C.

8

D.

6

#### SOLUTION

Solution : A

We have Tr+121Cr(3ab)21r(b3a)r

21Cra7(r2)b23r(72)

Since the powers of a and b are the same,

therefore

7 -  r223r -  72  ⇒ r = 9

If x4 occurs in the rth term in the expansion of (x4+1x3)16, then r =

A.

7

B.

8

C.

9

D.

10

#### SOLUTION

Solution : C

Tr16Cr1 (x4)16r(1x3)r116Cr1x677r

677r=4r=9

If coefficient of (2r+3)th and (r1)th terms in the expansion of (1+x)15 are equal, then value of r is

A.

5

B.

6

C.

4

D.

3

#### SOLUTION

Solution : A

15C2r+215Cr2

But 15C2r+215C15(2r+2)15C132r

⇒ 15C132r15Cr2  ⇒  r = 5.

In (32+133)n if the ratio of 7th term from the beginning to the 7th term from the end is  16, then n =

A.

7

B.

8

C.

9

D.

None of these

#### SOLUTION

Solution : C

16=nC6(213)n6(313)6nCn6(213)6(313)n6

61=613(n12)

n12=3

n=9

rth term in the expansion of (a+2x)n is

A.

n(n+1).....(nr+1)r!anr+1(2x)r

B.

n(n1).....(nr+2)(r1)!anr+1(2x)r1

C.

n(n+1).....(nr)(r+1)!anr(x)r

D.

None of these

#### SOLUTION

Solution : B

rth term of (a+2x)n is nCr1(a)nr+1(2x)r1

n!(nr+1)!(r1)!anr+1(2x)r1

n(n1).....(nr+2)(r1)!anr+1(2x)r1

6th term in expansion of (2x213x2)10 is

A.

458017

B.

- 89627

C.

558017

D.

None of these

#### SOLUTION

Solution : B

Applying Tr+1nCrxnrd for (x+a)n

Hence T610C5(2x2)5(- 13x2)5

= - (10!5!5!)(32) (1243) = - 89627

If the ratio of the coefficient of third and fourth term in the expansion of  (x12x)n is 1:2, then the value of n will be

A.

18

B.

16

C.

12

D.

-10

#### SOLUTION

Solution : D

T3nC2(x)n2(12x)2 and T4nC3(x)n3(12x)3

But according to the condition,

n(n1)×3×2×1×8n(n1)(n2)×2×1×4 = 12 ⇒ n = -10

Find the value of
(183+73+318725)36+62432+15814+20278+15916+6332+64

A.

1

B.

5

C.

25

D.

100

#### SOLUTION

Solution : A

The numerator is of the form

a3 + b3+3ab(a+b)=(a+b)3

∴ N' = (18+7)3 = 253

∴ D' = 36+6C13521+6C23422+6C33323+6C43224+6C53125+6C626

This is clearly the expansion of (3+2)6=56=(25)3

ND=(25)3(25)3=1

If number of terms in the expansion of (x2y+3z)n are 45, then n =

A.

7

B.

8

C.

9

D.

None of these

#### SOLUTION

Solution : B

(n+1)(n+2)2 = 45 or n2 + 3n - 88 = 0   n = 8 .

The greatest integer less than or equal to (2+1)6 is

A.

196

B.

197

C.

198

D.

199

#### SOLUTION

Solution : B

Let (2+1)6 = k + f, where k is integral part and f

the fraction (0 ≤ f < 1).

Let  (21)6 = f, (0 < f < 1),

Since 0 < (2 - 1) < 1

Now, k + f + f' =  (2+1)6 +  (21)6

= 2{6C0.23+6C2.22+6C4.2+6C6} = 198 ..........(i)

∴ f + f' = 198 - k = an integer

But, 0 ≤ f < 1 and 0 < f' < 1  ⇒ 0 < (f + f') < 2

⇒ f + f' = 1, ( ∵ f + f' is an integer)

∴ By (i), l = 198 - (f + f') = 198 - 1 = 197.

The number of terms which are free from radical signs in the expansion of (y15+x110)55 is

A.

5

B.

6

C.

7

D.

None of these

#### SOLUTION

Solution : B

In the expansion of (y15+x110)55, the general

term is

Tr+155Cr(y15)55r(x110)r55Cry11r5xr10.

This Tr+1 will be independent of radicals if the exponents  r5 and  r10 are integers, for

0 ≤ r ≤ 55 which is possible only when

r  = 0, 10, 20, 30, 40, 50.

∴ There are six terms viz. T1,T11,T21,T31,T41,T51

In the expansion of (512+718)1024, the number of integral terms is

A.

128

B.

129

C.

130

D.

131

#### SOLUTION

Solution : B

Here n = 1024 = 210, a power of 2, where as the

power of 7 is  18 = 23

Now first term 1024C0(512)1024 = 5512 (integer)

And after 8 terms, the 9th term =  1024C8(512)1016 (718)8  = an integer

Again, 17th term =  1024C16(512)1008 (718)16

= An integer.
Continuing like this, we get an A.P. 1, 9, 17,..........., 1025,

because 1025th term = the last term in the expansion

=  1024C1024(718)1024 = 7128 (an integer)

If n is the number of terms of above A.P. we have

1025 = Tn = 1 + (n - 1)8  ⇒ n = 129.

If n is positive integer and three consecutive coefficients in the expansion of (1+x)n are in the ratio 6 : 33 : 110, then n =

A.

4

B.

6

C.

12

D.

16

#### SOLUTION

Solution : C

Let the consecutive coefficient of (1+x)n are

nCr1nCrnCr+1

By condition, nCr1nCrnCr+1 = 6 : 33 : 110

Now nCr1nCr = 6 : 33

2n - 13r + 2 = 0                             ..............(i)

and nCrnCr+1 = 33: 110

3n - 13r - 10 = 0                            ...............(ii)

Solving (i) and (ii), we get n = 12 and r = 2.

Aliter: We take first n = 4 [By alternate (a)] but

(a) does not hold. Similarity (b).

So alternate (c), n = 12 gives

(1+x)12 =[1+12x+12.112.1 x212.11.103.2.1x3+.........]

So coefficient of II, III and IV terms are

12,6×11,2×11×10.

So, 12:6×11:2×11×10=6:33:110.