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Binomial Theorem · Question 28

JEE Main 2025 (Online) 3rd April Evening ShiftNumericalMedium📈 Topic in 52% papers+4
1

Let (1 + x + x2 )10 = a0 + a1x + a2x2 + + a20x20. If (a1 + a3 + a5 + + a19) - 11a2 = 121k, then k is equal to .

Answer

239

Solution

Put x = 1: a0 + a1 + a2 + + a20 = 310

Put x = -1: a0 - a1 + a2 - + a20 = 1

Subtracting:

a1 + a3 + + a19 = 310 - 12 = 59049 - 12 = 29524

a2 = coefficient of x2: take x2 from one factor or x from two factors

⇒ a2 = 10C1 + 10C2 = 10 + 45 = 55

⇒ 29524 - 11 55 = 29524 - 605 = 28919

⇒ 121k = 28919

∴ k = 239

Answer: 239

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2

The term independent of x in the expansion of ( (x + 1)(x2/3 + 1 - x1/3) - (x - 1)(x - x1/2) )10, x > 1, is :

  1. 240
  2. 120
  3. 150
  4. 210
JEE Main 2025 (Online) 2nd April Morning Shift
3

If r=010 ( 10r+1 - 110r ) 11Cr+1 = 11 - 11111010, then is equal to :

  1. 11
  2. 20
  3. 24
  4. 15
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4

The least value of n for which the number of integral terms in the Binomial expansion of ( [3]7 + [12]11)n is 183, is :

  1. 2184
  2. 2172
  3. 2196
  4. 2148
JEE Main 2025 (Online) 29th January Morning Shift
5

If = 1 + r=16(-3)r-1 12C2r-1, then the distance of the point (12, 3) from the line x - 3y + 1 = 0 is .

JEE Main 2025 (Online) 28th January Morning Shift
6

Let the coefficients of three consecutive terms Tr, Tr+1 and Tr+2 in the binomial expansion of (a + b)12 be in a G.P. and let p be the number of all possible values of r. Let q be the sum of all rational terms in the binomial expansion of ( [4]3 + [3]4)12. Then p + q is equal to:

  1. 295
  2. 283
  3. 299
  4. 287
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