# Free Objective Test 02 Practice Test - 11th and 12th

The coefficient of x3 in the expansion of  (1+3x)212x

will be

A.

8

B.

32

C.

50

D.

None of these

#### SOLUTION

Solution : C

= (1+3x)2(12x)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.

If 1(x2+1x)43 can be expanded by binomial theorem if

A.

x < 1

B.

|x| < 1

C.

x > 1

D.

|x| > 1

#### SOLUTION

Solution : D

The expansion is x83(1+1x3)43

Hence | 1x3| < 1  ⇒ |x| > 1

In the expansion of (1+x1x)2, the coefficient of xn

A.

4n

B.

4n-3

C.

4n+1

D.

None of these

#### SOLUTION

Solution : A

Given term can be written as (1+x)2(1x)2

=(1+2x+x2)[1+2x+3x2+.......+(n1)xn2+nxn1+(n+1)xn+...........]

= xn(n+1+2n+n1)+..........

Therefore coefficient of xn is 4n.

(r+1)th term in the expansion of (1x)4 will be

A.

xrr!

B.

(r+1)(r+2)(r+3)6xr

C.

(r+2)(r+3)2xr

D.

None of these

#### SOLUTION

Solution : B

(1x)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.

(aa+x)12(aax)12 =

A.

2 +  3x24a2 + .......

B.

1 +  3x28a2 + .....

C.

2 +  xa3x24a2 + ......

D.

2 -  xa3x24a2 + .......

#### SOLUTION

Solution : A

(a+xa)12(axa)12 =(1+xa)12+(1xa)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.

Cube root of 217 is

A.

6.01

B.

6.04

C.

6.02

D.

None of these

#### SOLUTION

Solution : A

(217)13 = (63+1)13 = 6(1+163)13

On expansion by binomial theorem

= 6(1+13×2161×23×3×2(1216)2+.........) = 6.01

The fourth term in the expansion of (12x)32 will be

A.

- 34x4

B.

x32

C.

-x32

D.

34x4

#### SOLUTION

Solution : B

Expansion of (12x)32

= 1+32(2x)+32.12.12(2x)2+32.12(12)16(2x)3+.........

Hence 4th term is  x32

C0C1+C2C3+........+(1)nCn is equal to

A.

2n

B.

2n1

C.

0

D.

2n1

#### SOLUTION

Solution : C

We know that

(1+x)n=nC0+nC1x+nC2x2+......+nCnxn

Putting x = -1, we get

(11)nnC0 -  nC1 +  nC2 - .......(1)n  nCn

Therefore C0C1+C2C3+......(1)nCn = 0

The value of  C12C34C56 + ...... is equal to

A.

2n1n+1

B.

n.2n

C.

2nn

D.

2n1

#### SOLUTION

Solution : A

We know that

(1+x)n(1x)n2 = C1x+C3x3+C5x5+........

Integrating from x = 0 to x = 1, we get

1210(1+x)n(1x)n dx

10(C1x+C3x3+C5x5+.......) dx

12{(1+x)n+1n+1+(1x)n+1n+1}10=C12 +  C34 + C56 + ....

or  C12 +  C34 + C56 + .......=  12  {2n+11n+1 +  01n+1}

=  12 2n+12n+1 =  2n1n+1

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

A.

a2n

B.

na

C.

0

D.

None of these

#### SOLUTION

Solution : C

We can write

aC0 - (a + d)C1 + (a + 2d)C2 - ........ upto (n+1) terms

=a(C0C1+C2........)+d(C1+2C23C3+......)            .............(i)

Again,(1x)n=C0C1x+C2x2.........+(1)nCnxn   ..........(ii)

Differentiating with respect to x,

n(1x)n1 = -C1 + 2C2x - .......... + (1)nCnnxn1     ......................(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.

C01C23C45C67 +..........=

A.

2n+1n+1

B.

2n+11n+1

C.

2nn+1

D.

None of these

#### SOLUTION

Solution : C

Putting the values of C0,C2,C4........., we get

= 1 +  n(n1)3.2!n(n1)(n2)(n3)5.4! + .........

= 1n+1[(n+1) + (n+1)n(n1)3!(n+1)n(n1)(n2)(n3)5! + ..........]

Put n+1 = N

1N[N +  N(N1)(N2)3!N(N1)(N2)(N3)(N4)5! + .........]

1N{NC1NC3 + NC5 +..........}

= 1N{2N1} =  2nn+1      { ∵ N = n+1}

Trick: Put n = 1, then S11C0111 = 1

At n = 2, S2 =   2C01 +   2C23 = 1 +   13 =   43

Also (c)  ⇒ S1 = 1, S243

If (1+x)n=C0+C1x+C2x2+.......+Cnxn, then C1C0+2C2C1+3C3C2+........+nCnCn1=

A.

n(n1)2

B.

n(n+2)2

C.

n(n+1)2

D.

(n1)(n2)2

#### SOLUTION

Solution : C

C1C0+2C2C1+3C3C2+........+nCnCn1

=n1+2n(n1)1.2n+3n(n1)(n2)3.2.1n(n1)1.2+......+n.1n

=n+(n1)+(n2)........+1=n=n(n+1)2

If (1+x)n=C0+C1x+C2x2+........+Cnx2, then

C20+C21+C22+C23+..........+C2n =

A.

n!n!n!

B.

(2n)!n!n!

C.

(2n)!n!

D.

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,

S22C20+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

In the expansion of ( ax+bx)12, the coefficient of x10 will be

A.

12a11

B.

12b11a

C.

12a11b

D.

12a11b11

#### SOLUTION

Solution : C

Tr+1=12cr(ax)12r(bx)r =12cr.a12r.br.x2r122x12=10r=1coefficient=12.a11.b

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

A.

n22np+4p2 = 0

B.

n2n(4p+1)+4p22 = 0

C.

n2n(4p+1)+4p2 = 0

D.

None of these

#### SOLUTION

Solution : B

2ncp=ncp1+pcp+12(np)!P!=1(np+1)!(p1)!+1(np1)!(p+1)!2(np)p=1(np+1)(np)+1(p+1)pon simplifing we getn2n(4p+1)+4p22=0