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JEE Main Mathematics Functions 2026 JEE Main 2026 (28 January Shift 1)

JEE Main Mathematics Question (2026) — Solution

Question

If g(x)=3 x^ 2 +2 x-3, f(0)=-3 and 4 g(f(x))=3 x^ 2 -32 x+72, then f(g(2)) is equal to:

Options

  1. A. - 7 2
  2. B. - 25 6
  3. C. 7 2
  4. D. 25 6

Answer

C. 7 2

Step-by-step solution

Let f(x) = ax + b. Then 4g(f(x)) = 12(ax+b)^2 + 8(ax+b) - 12. Comparing with 3x^2 - 32x + 72: x^2: 12a^2 = 3 a = 1 2 Constant: 12b^2 + 8b - 12 = 72 3b^2 + 2b - 21 = 0 b = 7 3 or b = -3. Since f(0) = -3, b = -3. x: 8a(3(-3)+1) = -32 -8a = -4 a = 1 2 . So f(x) = x 2 - 3. g(2) = 3(4) + 4 - 3 = 13. f(g(2)) = f(13) = 13 2 - 3 = 7 2 .

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