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Integrals · Question 25

JEE Main 2026 (Online) 23rd January Evening ShiftMCQMedium📈 Topic in 23% papers+4−1
1

Let I(x) = 3 dx(4x + 6) ( 4x2 + 8x + 3 ) and I(0) = 34 + 20. If I ( 12 ) = a 2b + c, where a, b, c N, (a, b) = 1, then a + b + c is equal to :

  1. 30
  2. 29
  3. 28
  4. 31

Answer

(4) 31

Solution

4x2 + 8x + 3 = (2x + 1)(2x + 3) and 4x + 6 = 2(2x + 3)

Put t = 2x + 3 dx = dt2, 2x + 1 = t - 2

I = 3 dt22t t(t - 2) = 34 dtt2 1 - 2t

Now put 1 - 2t = u2 2t2 dt = 2u du dtt2 = u du

I = 34 u duu = 34u + C

I(x) = 34 2x + 12x + 3 + C

I(0) = 34 1 3 + C = 34 + C

I(0) = 34 + 20 C = 20

I ( 12 ) = 34 24 + 20 = 34 22 + 20 = 3 28 + 20

a = 3, b = 8, c = 20 (and (3, 8) = 1)

a + b + c = 3 + 8 + 20 = 31

Answer: (4)

JEE Main 2026 (Online) 23rd January Evening Shift

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