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Mere Teacher · Previous Year Questions
JEE Main Mathematics · Binomial Theorem · Set 2 (Q31-60)
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31.
If n is an odd positive integer and (1+x+x2+x3)n=∑r=03narxr, then a0−a1+a2−a3+…..−a3n is
(1)4n
(2)1
(3)-1
(4)0
32.
If the term independent of x in the expansion of (ax2+2x31)10 is 105 , then a2 is equal to :
(1)2
(2)4
(3)6
(4)9
33.
For n,p∈N−1, the coefficient of x3 in (1−x)n(1−x)−1/p is equal to
(1)p3×3!(np+1)(np+p+1)(np+2p+1)
(2)3!p3(np+1)(np+p)(np+2p)
(3)3!p3(np+p)(np+2p)(np+3p)
(4)3!p3(np+1)(np+2)(np+3)
34.
If 21st and 22nd terms in the expansion of (1+x)44 are equal, then x is equal to
(1)78
(2)2221
(3)87
(4)2423
35.
The largest n∈N such that 3n divides 50! is:
(1)22
(2)20
(3)21
(4)23
36.
Let x∈R be so small that the powers of x beyond two are insignificant and negligibly small. For such x, if (1−x)3(2+x)6 is approximated by a+bx+cx2, then a+b+c=
(1)Statement -1 is true, Statement-2 is true ; Statement -2 is not a correct explanation for Statement -1.
(2)Statement-1 is true, Statement-2 is false.
(3)Statement -1 is false, Statement -2 is true.
(4)Statement -1 is true, Statement -2 is true; Statement-2 is a correct explanation for Statement-1.
38.
Let (1+x+2x2)20=a0+a1x+a2x2+…+a40x40 then a1+a3+a5+…+a37 is equal to
(1)220(220−21)
(2)219(220−21)
(3)219(220+21)
(4)220(220+21)
39.
For r = 0, 1, ...., 10, let Ar, Br and Cr denote, respectively, the coefficient of xr in the expansions of (1 + x)10 , (1 + x)20 and (1 + x)30 . Then ∑Ar(B10Br−C10Ar) is equal to
(1)B10 – C10
(2)A10 (B210 – C10 A10)
(3)0
(4)C10 – B10
40.
If in the expansion of (1+x)p(1−x)q, the coefficients of x and x2 are 1 and -2 , respectively, then p2+q2 is equal to:
(1)18
(2)13
(3)8
(4)20
41.
Suppose l,m,n respectively represent the coefficient of x10, the constant term and the coefficient of x−10 in the expansion of (ax2+x3b)15. If ml+nm=1126, then a2:b2=
(1)16:9
(2)9:4
(3)4:1.
(4)1:25
42.
If (1−qx)(1−px)−1=a0+a1x+a2x2+a3x3+……….. then an=
(1)q−ppn+1−qn+1
(2)p−qpn+1−qn+1
(3)q−ppn−qn
(4)p−qpn−qn
43.
If the greatest divisor of 30⋅52n+4⋅23n is p,∀n∈N and that of 22n+1−6n−2 is q,∀n∈N, then p+q=
(1)26
(2)52
(3)104
(4)13
44.
The maximum value of the term independent of ’t’in the expansion of (tx51+t(1−x)101)10 where x∈(0,1) is:
(1)3(5!)210!
(2)3(5!)22.10!
(3)3(5!)210!
(4)33(5!)22.10!
45.
When 32023 is divided by 16 , the remainder obtained is
(1)15
(2)11
(3)9
(4)7
46.
The coefficient of x301 in (1+x)500+x(1+x)499+x2(1+x)498+…+x500 is
(1)501C302
(2)500C301
(3)500C300
(4)501C200
47.
For all positive integers k, if the greatest divisor of 25k+12k−1 is d, then 4d=
(1)36
(2)8
(3)20
(4)24
48.
The constant term in the expansion of (x2−x21)16 is
(1)16C8
(2)16C7
(3)16C9
(4)16C10
49.
The coefficient of x7 in the expansion of (1 – x – x2 + x3)6 is :
(1)144
(2)– 132
(3)– 144
(4)132
50.
A ratio the 5th term from the beginning to the 5th term from the end in the binomial expansion of (231+2(3)311)10 is:
(1)1:2(6)31
(2)1:4(16)31
(3)4(36)31:1
(4)2(36)31:1
51.
If for some m,n;6Cm+2(6Cm+1)+6Cm+2>8C3 and n−1P3:nP4=1:8, then nPm+1+n+1Cm is equal to
(1)380
(2)376
(3)384
(4)372
52.
If the coefficients of (2r+1)th term and (r+1)th term in the expansion of (1+x)42 are equal then r can be
(1)12
(2)14
(3)16
(4)20
53.
If x is positive, the first negative term in the expansion of (1+x)527 is :
(1)7th term
(2)5th term
(3)8th term
(4)6th term.
54.
The value of r for which 20Cr20C0+20Cr−120C1+20Cr−220C2+…+20C020Cr is maximum is :
(1)15
(2)20
(3)11
(4)10
55.
The number of positive integers k such that the constant term in the binomial expansion of (2x3+xk3)12,x=0is 28.l, where lis an odd integer, is________.
56.
The number of rational terms in the binomial expansion of (441+561)120is ___
57.
Let the ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of (42+431)n, in the increasing powers of 431 be 46 : 1. If the sixth term from the beginning is 43α, then a is equal to_______.
58.
The coefficient of x9 in the expansion of (1 + x) (1 + x2) (1 + x3) ......(1 + x100) is
59.
If (1 + x + x² +... + xp)6 = a0 + a1x + a2x²+...+a6p x6p, then the value of : p(p+1)61 [a1 + 2a2 + 3a3 +.... + 6p a6p] is :
60.
Find the value of 6C0. 12C6.– 6C1 11C6+ 6C2 10C6 – 6C3 . 9C6 + 6C4 . 8C6 – 6C5 . 7C6 + 6C6 . 6C6