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JEE Main Mathematics Indefinite Integration 2026 JEE Main 2026 (28 January Shift 2)

JEE Main Mathematics Question (2026) — Solution

Question

Let f(x)= d x x^ ( 2 3 ) +2 x^ ( 1 2 ) be such that f(0)=-26+24 _ e (2). If f(1)= a + b _ e (3), where a , b Z , then a + b is equal to :

Options

  1. A. -26
  2. B. -11
  3. C. -5
  4. D. -18

Answer

B. -11

Step-by-step solution

Let u = x^ 1/6 , so x = u^6 and dx = 6u^5 du. Then x^ 2/3 = u^4, x^ 1/2 = u^3. f(x) = 6u^5 du u^4 + 2u^3 = 6u^2 du u+2 . By polynomial division: u^2 u+2 = u - 2 + 4 u+2 . f(x) = 6 ( u^2 2 - 2u + 4 |u+2| ) + C = 3u^2 - 12u + 24 (u+2) + C. Substituting back: f(x) = 3x^ 1/3 - 12x^ 1/6 + 24 (x^ 1/6 +2) + C. From f(0) = -26 + 24 (2): 24 (2) + C = -26 + 24 (2), so C = -26. At x = 1: f(1) = 3 - 12 + 24 (3) - 26 = -35 + 24 (3). Thus a = -35, b = 24, and a + b = -11.

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