# Free Objective Test 01 Practice Test - 11th and 12th

The value of (2+1)6 + (21)6 will be

A.

-198

B.

198

C.

99

D.

-99

#### SOLUTION

Solution : B

(x+a)n + (xa)n = 2[xn+nC2xn2a2+nC4xn4a4+nC6xn6a6+.......]

Here, n = 6, x =  2, a = 1;

6C2 = 15, 6C4 = 15, 6C6 = 1

∴ (2+1)6 + (21)6 = 2[(2)6+15.(2)4.1+15(2)2.1+1.1]

=2[8+15×4+15×2+1]=198

Middle term in the expansion of  (1+3x+3x2+x3)6 is

A.

4th

B.

3rd

C.

10th

D.

None of these

#### SOLUTION

Solution : C

(1+3x+3x2+x3)6 = ((1+x)3)6 = (1+x)18

Hence the middle term is 10th.

The middle term in the expansion of (1+x)2n is

A.

1.3.5.......(5n1)n!xn

B.

2.4.6.......2nn!x2n+1

C.

1.3.5.......(2n1)n!xn

D.

1.3.5.......(2n1)n!2nxn

#### SOLUTION

Solution : D

Middle term of (1+2x)2n is Tn+12nCnxn

(2n)!n!n!xn1.3.5.......(2n1)n! 2nxn.

If the coefficient of (2r+4)th and (r2)th terms in the  expansion of (1+x)18 are equal, then r =

A.

12

B.

10

C.

8

D.

6

#### SOLUTION

Solution : D

18C2r+318Cr3   ⇒ 2r + 3 + r - 3 = 18   ⇒ r = 6

If the coefficient of x in the expansion of (x2+kx)5 is 270, then k =

A.

1

B.

2

C.

3

D.

4

#### SOLUTION

Solution : C

Tr+15Cr(x2)5r(kx)r

For coefficient of x, 10 - 2r - r = 1  ⇒ r = 3

Hence, T3+15C3(x2)53(kx)3

According to question, 10k3 = 270  ⇒ k = 3.

If the coefficient of x7 in (ax2+1bx)11 is equal to the coefficient of  x7 in (ax1bx2)11, then ab =

A.

1

B.

12

C.

2

D.

3

#### SOLUTION

Solution : A

In the expansion of (ax2+1bx)11, the general

term is

Tr+111Cr(ax2)11r(1bx)r =  11Cr(a)11r(1br)x223r

For x7, we must have 22 - 3r = 7  ⇒ r = 5, and

the coefficient of x711C5.a11515) = 11C5  a6b5

Similarly, in the expansion of (ax1bx2)11, the

general term is Tr+1 =  11Cr(1)r a11rbr.x113r

For x7 we must have, 11 - 3r = -7  ⇒ r = 6, and

the coefficient of x7 is 11C6 a5b6 =  11C5 a5b6.

As given,  11C5 a6b5 =  11C5 a5b6 ⇒ ab = 1.

In the polynomial (x - 1)(x - 2)(x - 3)............... .........(x - 100), the coefficient of x99 is

A.

5050

B.

-5050

C.

100

D.

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.

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

A.

p + q = r

B.

p + r = q

C.

p = q + r

D.

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+12n+2Cn+1

∴ q + r = p

The value of x in the expression [x+xlog10(x)]5, if the third term in the expansion is 10,00,000

A.

10

B.

11

C.

12

D.

None of these

#### SOLUTION

Solution : A

T35C2. x2(xlog10x)3 = 106

Put 5C2 = 10 [ ∵log1010=1].

If x = 10, then 103.102.1 = 105 is satisfied.

Hence x = 10.

If  (13x)12+(1x)534x is approximately equal to a + bx for small values of x, then (a, b) =

A.

(1, 3524)

B.

(1, - 3524)

C.

(2, 3512)

D.

(2, - 3512)

#### SOLUTION

Solution : B

(13x)1/2+(1x)5/32[1x4]1/212[(132x)+(15x3)](1+x8)12[2196x]](1+x8)=12[2(14+196)x]=13524x(a,b)=(1,3524)

If (1+ax)n = 1 + 8x + 24 x2 + ......, then the value of a and n is

A.

2, 4

B.

2, 3

C.

3, 6

D.

1, 2

#### SOLUTION

Solution : A

As given (1+ax)n = 1 + 8x + 24x2 + .........

⇒ 1 +  n1ax +  n(n1)1.2a2x2 + ......... = 1 + 8x + 24x2 + ...........

⇒na = 8, n(n1)1.2a2 = 24 ⇒ na(n-1)a = 48

⇒8(8 - a) = 48 ⇒ 8 - a = 6 ⇒ a = 2 ⇒ n = 4.

If P is a prime number, then np - n is divisible by p when n is a

A.

Natural number greater than 1

B.

Irrational number

C.

Complex number

D.

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.

The value of the expression (2+2)4  lies between

A.

134 and 135

B.

135 and 136

C.

136 and 137

D.

None of these

#### SOLUTION

Solution : B

(2+(2))4(2)4(2+1)4

= 4[4C04C1(2) + 4C2(2)24C3(2)3 +4C4 (2)4]

= 4 [1 + 4 24.32.2 +  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

The number of integral terms in the expansion of (512+716)642 is

A.

106

B.

108

C.

103

D.

109

#### SOLUTION

Solution : B

Tr+1642Cr(512)642r.(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.

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=

A.

a2a2+a3

B.

12 a2a2+a3

C.

2a2a2+a3

D.

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