Q 21 :

The term independent of x in the expansion of (2x5-12xx)11 is

  • 5th term

     

  • 6th term

     

  • 11th term

     

  • no term

     

(4)

Let Tr+1 be general term of given expansion

  Tr+1=Cr11(2x5)11-r(-12xx)r

=Cr11(25)11-r×(-12)rx11-r2-3r2

  Tr+1=Cr11(25)11-r(-12)rx(11-4r)/2

For a term independent of x, put 11-4r2=0  or  r=114

But r can not be in fraction.

  There is no term independent of x.



Q 22 :

The ninth term of the expansion (3x-12x)8 is

  • 1512x9

     

  • -1512x9

     

  • -1256x8

     

  • 1256x8

     

(4)

Given expansion is (3x-12x)8

T9=C88(3x)0(-12x)8=1256x8



Q 23 :

The term independent of x in the expansion of (32x2-13x)6 is

  • 512

     

  • 125

     

  • 56

     

  • none of these

     

(1)

(32x2-13x)6

Tr+1=Cr6(32x2)6-r(-13x)r=Cr6(32)6-r(-13)rx12-3r

Put  12-3r=0    r=4

  Term independent of x is T5=C46(32)2(-13)4=512



Q 24 :

The term independent of x in the expansion of [x3+32x2]10 is:

  • 54

     

  • 74

     

  • 94

     

  • 45

     

(1)

Let (r+1)th term in the expansion of

[x3+32x2]10 is independent of x

So  Tr+1=Cr10(x3)10-r(32x2)r

=Cr10(13)10-r(32)rx10-5r2

Putting x(10-5r2) equal to x0    10-5r2=0    r=2

  T2+1=C210(13)10-2(32)2=54



Q 25 :

If xr occurs in the expansion of (x+1x)n, then its coefficient is

  • n!(r!)2

     

  • n!(r+1)!(r-1)!

     

  • n!(n+r2)!(n-r2)!

     

  • n![(r2)!]2

     

(3)

Tm+1=Cmn(x)n-m(1x)m=Cmnxn-2m

If xr occur in expansion, then

  n-2m=r    m=n-r2

Coefficient of xr=C(n-r2)n=n!(n-r2)!(n+r2)!



Q 26 :

If A and B are coefficients of xn in the expansions of (1+x)2n and (1+x)2n-1 respectively, then BA is equal to

  • 12

     

  • 2

     

  • 1

     

  • 1n

     

(1)

Now, A=Cn2n and B=Cn2n-1

Thus, BA=(2n-1)!n!(n-1)!×n!n!(2n)!=n2n=12



Q 27 :

The coefficient of the term independent of x in the expansion of (x+1x)10 is equal to

  • 10

     

  • 252

     

  • 20

     

  • 256

     

(2)

General term of expansion (x+1x)10 is

Tr+1=Cr10(x)10-r(1x)r=Cr10x5-r

Term independent of x, means power of x is 0

  5-r=0    r=5

Hence,  T5+1=C510=252



Q 28 :

The 13th term in the expansion of (x2+2x)n is independent of x, then the sum of the divisors of n is

  • 39

     

  • 36

     

  • 37

     

  • 38

     

(1)

T13=C12n(x2)n-12·(2x)12

  x2n-24-12=x0    n=18

  The sum of the divisors of n is 39



Q 29 :

If the coefficient of x8 in (ax2+1bx)13 is equal to the coefficient of x-8 in (ax-1bx2)13, then a and b will satisfy the relation

  • ab+1=0

     

  • ab=1

     

  • a=1-b

     

  • a+b=-1

     

(1)

     Tr+1 in (ax2+1bx)13

=Cr13(ax2)13-r(1bx)r=Cr13a13-rbrx26-3r

For coefficient of x8, we have 26-3r=8    r=6

  Coefficient of x8=C613a7b6

  Tr+1 in (ax-1bx2)13=Cr13(ax)13-r(-1bx2)r

=Cr13a13-rbrx13-3r(-1)r

For coefficient of x-8, we have 13-3r=-8    r=7

  Coefficient of x-8=-C713a6b7

Now,  C613a7b6=-C713a6b7    ab=-1    ab+1=0



Q 30 :

The value of  (C121-C110)+(C221-C210)+(C321-C310)+(C421-C410)++(C1021-C1010)  is

  • 221-210

     

  • 220-29

     

  • 220-210

     

  • 221-211

     

(3)

(C121-C110)+(C221-C210)++(C1021-C1010)

=(C121+C221++C1021)-(C110+C210++C1010)

=2212-210=220-210



Q 31 :

Let X=(C110)2+2(C210)2+3(C310)2++10(C1010)2, where Cr10, r{1,2,,10}, denote binomial coefficients. Then, the value of 11430X is ______ .



(646)

X=r=110r(Cr10)2=r=110r·Cr10·Cr10

=10r=110Cr-19·C10-r10=10·C919

  X1430=10·C9191430=C919143=646



Q 32 :

What is the value of the expression (1+2)4-(1-2)4?

  • 34

     

  • 102

     

  • 12+82

     

  • 242

     

(4)

Consider,   (1+2)4-(1-2)4

=2[C142+C34(2)3]                       [ (a+b)4-(a-b)4=2(C14a3b+C34ab3)]

=2[42+4×22]=2(122)=242



Q 33 :

The number 81 is the coefficient of xk in the binomial expansion of (x2+3x)4, x0. Then the value of k equals

  • - 2

     

  • 2

     

  • - 4

     

  • 4

     

(3)

(x2+3x)4=C04(3x)4+C14x2(3x)3+C24x4(3x)2+C34x6(3x)1+C44x8(3x)0

=81x4+4×x2×27x3+6×9×x2+12x5+x8

=81x4+108x+54x2+12x5+x8

Here, 81 is coefficient of x-4       k=-4



Q 34 :

If α and β be the coefficients of x4 and x2 respectively in the expansion of (x+x2-1)6+(x-x2-1)6, then

  • α+β=-30

     

  • α-β=-132

     

  • α+β=60

     

  • α-β=60

     

(2)

(x+x2-1)6+(x-x2-1)6

=2[C06x6+C26x4(x2-1)+C46x2(x2-1)2+C66(x2-1)3]

=2[x6+15(x6-x4)+15x2(x4-2x2+1)+(-1+3x2-3x4+x6)]

=2(32x6-48x4+18x2-1)

  α=-96 and β=36

  α-β=-132



Q 35 :

Let m,nN and gcd(2,n)=1. If 30(300)+29(301)++2(3028)+1(3029)=n·2m, then n+m is equal to _________.          (Here, (nk)=Ckn)



(45)

We have,

30(300)+29(301)++2(3028)+1(3029)

a=30(3030)+29(3029)++2(302)+1(301)

=r=130r(30r)=r=130r·30r(29r-1)

=30[(290)+(291)+(292)++(2929)]

=30·229=15×230=n·2m  (given)

  n+m=15+30=45



Q 36 :

k=020(Ck20)2 is equal to

  • C2140

     

  • C2041

     

  • C2040

     

  • C1940

     

(3)

We know that

(1+x)n=C0n+C1nx+C2nx2++Cnnxn                  ...(i)

Also, (x+1)n=C0nxn+C1nxn-1+C2nxn-2++Cnn          ...(ii)

Multiply (i) & (ii), we get

(1+x)2n=(C0n+C1nx++Cnnxn)(C0nxn+C1nxn-1++Cnn)

Comparing coefficient of xn on both sides, we get

            Cn2n=(C0n)2+(C1n)2+(C2n)2++(Cnn)2

r=0n(Crn)2=Cn2n

   k=020(Ck20)2=C2040       [ n=20]



Q 37 :

Fractional part of the number 4202215 is equal to

  • 115

     

  • 415

     

  • 1415

     

  • 815

     

(1)

We have,   {4202215}={(42)101115}={(1+15)101115}

={115+15101015+}={115+0+}={115}



Q 38 :

The remainder, when 7103 is divided by 17, is ______.



(12)

7103=7·7102

7·7102=7(72)51=7(49)51=7(51-2)51

Remainder=7(-2)51=-7(2)51=-7(248+3)=-7·8(248)

=-56(24)12=-56(17-1)12

Again Remainder=-56×(-1)12=(-56)

Remainder=12



Q 39 :

The coefficient of x17 in (1-x)13(1+x+x2)12 is

  • C612

     

  • C79

     

  • 0

     

  • 1

     

(3)

We have,   (1-x)13(1+x+x2)12

=(1-x)[(1-x)(1+x+x2)]12=(1-x)(1-x3)12

In the expansion, all the powers of x is either in the form of 3k or (3k+1) for some k. But 17 has the form (3k+2).

   Coefficient of x17 in the given expansion is 0.



Q 40 :

The remainder when 4282024 is divided by 21 is _______.



(1)

We have,   4282024=(420+8)2024

=C02024(420)2024+C12024(420)2023(8)1++C20232024(420)82023+C2024202482024

=420[Some integer]+641012

=21×20[Some integer]+(63+1)1012

=21×20[Some integer]+63×Some integer+1

=21×20[Some integer]+21×3×Some integer+1

Hence, remainder=1



Q 41 :

The middle term of expansion (10x+x10)10 is

  • C58

     

  • C510

     

  • C57

     

  • C59

     

(2)

Given expansion is (10x+x10)10

Since, n=10 is even

  Middle term=tn2+1=t6

  t6=C510(10x)5(x10)5=C510



Q 42 :

The constant term in the expansion of (x2-1x2)16 is

  • C816

     

  • C716

     

  • C916

     

  • C1016

     

(1)

Let (r+1)th term be the constant term in the expansion of (x2-1x2)16

  Tr+1=Cr16x2r(-1x2)16-r

=(-1)16-rCr16x2r·x-32+2r=(-1)16-rCr16x4r-32

For constant term, 4r-32=0r=8

  T9=(-1)8C816=C816



Q 43 :

If the fourth term in the binomial expansion of (2x+xlog8x)6 (x>0) is 20×87, then a value of x is

  • 8

     

  • 8-2

     

  • 83

     

  • 82

     

(4)

 



Q 44 :

The positive value of λ for which the coefficient of x2 in the expression x2(x+λx2)10 is 720, is

  • 5

     

  • 4

     

  • 22

     

  • 3

     

(2)

Tr+1 term of (x+λx2)10 is

Cr10(x)10-r(λx2)r=Cr10x5-5r2(λ)r

For this to be independent of x, 5-5r2=0r=2

  C210λ2=72045λ2=720

  λ2=16λ=±4

  The positive value of λ for which the coefficient of x2 is 720 is 4.



Q 45 :

If (3644)k  is the term, independent of x, in the binomial expansion of (x4-12x2)12, then k is equal to ________.



(55)

 



Q 46 :

The term independent of 'x' in the expansion of (x+1x2/3-x1/3+1-x-1x-x1/2)10, where x0,1 is equal to ________.



(210)

The term independent of x in the expansion of

(x+1x2/3-x1/3+1-x-1x-x1/2)10

=[(x1/3+1)-(x1/2+1)x1/2]10=(x1/3-x-1/2)10

Tr+1=Cr10(x1/3)r·(-x-1/2)10-r

=Cr10(-1)10-r·xr/3+r/2-5

For this term to be independent, r3+r2=5

  r=6

  Term independent is T7=C610=210



Q 47 :

Let the coefficients of third, fourth and fifth terms in the expansion of (x+ax2)n, x0, be in the ratio 12 : 8 : 3. Then the term independent of x in the expansion, is equal to _______.



(4)

Tr+1=Crn(x)n-r(ax2)r=Crnarxn-3r

We have,  C2na2C3na3=128    2=a(n-2)                   ...(i)

and  C3na3C4na4=83    3=2a(n-3)                              ...(ii)

Dividing (i) by (ii), we get

23=n-22(n-3)    n=6

From (i),  2=a(6-2)    a=12

For the term independent of x, n-3r=0

  6-3r=0    r=2

So,  T3=C26a2=1544  (Approx.)



Q 48 :

If the constant term, in binomial expansion of (2xr+1x2)10 is 180, then r is equal to ________.



(8)

 



Q 49 :

Let the ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of  (24+134)n, in the increasing powers of 134 be 64:1. If the sixth term from the beginning is α34, then α is equal to _______.



(84)

We have,   (21/4+131/4)n

Given that  T5Tn-3=61/41

  C4n(21/4)n-4(3-1/4)4Cn-4n(21/4)4(3-1/4)n-4=61/41

  (21/4)n-8(3-1/4)8-n=61/4

  (61/4)n-8=61/4n-8=1n=9

Now,  T6=C59(21/4)4(3-1/4)5

=126×2×(3-1)(3-1/4)

=2523131/4=8434=α34

Thus,  α=84



Q 50 :

Let a and b be two nonzero real numbers. If the coefficient of x5 in the expansion of (ax2+7027bx)4 is equal to the coefficient of x-5 in the expansion of  (ax-1bx2)7, then the value of 2b is __________.



(3)

Given, (ax2+7027bx)4

The general term is given by

Tr+1=Cr4(ax2)4-r(7027bx)r                      ...(1)

For coefficient of x5, put 8-2r-r=5r=1

  The coefficient of x5 in (ax2+7027bx)4=C14·a3(70)27·b

General term of (ax-1bx2)7 is given by

Tr+1=Cr7(ax)7-r(-1bx2)r

For coefficient of x-5, put 7-r-2r=-5r=4

  The coefficient of x-5 in (ax-1bx2)7=C47a3(-1)4b4

Now,  C14a3(70)27(b)=C47a3b4

  4×7027(b)=35b4b3=278b=32

  2b=3