Aspire Faculty ID #18086 · Topic: JEE Main 2026 (24 January Evening Shift) · Just now
JEE Main 2026 (24 January Evening Shift)

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

Solution

$ g(2) = 13 $ $ f(g(2)) = f(13) $ Now $ 4g(f(x)) = 3x^2 - 32x + 72 $ $ 4[3f(x)^2 + 2f(x) - 3] = 3x^2 - 32x + 72 $ Let $ f(x) = t $ $ 12t^2 + 8t - (3x^2 - 32x + 84) = 0 $ $ f(x) = \frac{-8 \pm \sqrt{64 + 48(3x^2 - 32x + 84)}}{24} $ $ f(x) = \frac{-8 \pm 4(3x - 16)}{24} $ $ f(x) = \frac{-8 + 4(3x - 16)}{24} $ $ \because f(0) = -3 \Rightarrow \text{we take +ve sign} $ $ f(x) = \frac{-8 + 4(3x - 16)}{24} $ $ f(13) = \frac{-8 + 4 \cdot 23}{24} = \frac{84}{24} = \frac{7}{2} $
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