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Binomial Theorem

JEE Main 2025 (Online) 23rd January Evening ShiftMCQMedium📈 Topic in 5% papers+4−1
Ques. 1JEE Main 2025, 23 January (Evening Shift)

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. 8
  2. 20
  3. 13
  4. 18

Answer

(3) 13

Solution

Coefficient of x: p-q=1. Coefficient of x2: p(p-1)2-pq+ q(q-1)2=-2 (p-q)2-(p+q)=-4

1-(p+q)=-4 p+q=5 p=3, q=2

p2+q2=13

Answer: (3)

Keep practising — Binomial Theorem

Ques. 2JEE Main 2025, 22 January (Morning Shift)

If r=05 11C2r+12r+2= mn, (m,n)=1, then m-n is equal to .

Ques. 3JEE Main 2025, 22 January (Evening Shift)

Let , , and be the coefficients of x7,x5,x3 and x respectively in the expansion of (x+ x3-1 )5+ (x- x3-1 )5, x>1. If u and v satisfy the equations u+ v=18, u+ v=20, then u+v equals :

  1. 4
  2. 3
  3. 5
  4. 8
Ques. 4JEE Main 2025, 22 January (Evening Shift)

If r=130 r2 (30Cr )230Cr-1= 229, then is equal to .

Ques. 5JEE Main 2024, 9 April (Morning Shift)

The coefficient of x70 in x2(1 + x)98 + x3(1 + x)97 + x4(1 + x)96 + + x54(1 + x)46 is 99Cp - 46Cq. Then a possible value of p + q is :

  1. 61
  2. 83
  3. 55
  4. 68
Ques. 6JEE Main 2024, 9 April (Morning Shift)

The remainder when 4282024 is divided by 21 is .

Ques. 7JEE Main 2024, 9 April (Evening Shift)

The sum of the coefficients of x2/3 and x-2/5 in the binomial expansion of (x2/3 + 12x-2/5 )9 is

  1. 19/4
  2. 69/16
  3. 63/16
  4. 21/4
Ques. 8JEE Main 2024, 8 April (Morning Shift)

Let = r=0n (4r2 + 2r + 1 )nCr and = ( r=0n nCrr + 1 ) + 1n + 1. If 140 < 2 < 281, then the value of n is .

Ques. 9JEE Main 2024, 8 April (Evening Shift)

If the term independent of x in the expansion of ( ax2 + 12x3 )10 is 105, then a2 is equal to :

  1. 6
  2. 4
  3. 2
  4. 9
Ques. 10JEE Main 2024, 6 April (Morning Shift)

If the second, third and fourth terms in the expansion of (x + y)n are 135, 30 and 103, respectively, then 6 (n3 + x2 + y ) is equal to .

Ques. 11JEE Main 2024, 5 April (Morning Shift)

If the constant term in the expansion of (1 + 2x - 3x3 ) ( 32x2 - 13x )9 is p, then 108p is equal to .

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