Aspire Faculty ID #18061 · Topic: JEE Main 2026 (24 January Evening Shift) · Just now
JEE Main 2026 (24 January Evening Shift)

Let $ f(x) = \int \frac{7x^{10} + 9x^8}{(1 + x^2 + 2x^9)^2} ,dx $, $ x > 0 $, $ \lim_{x \to 0} f(x) = 0 $ and $ f(1) = \frac{1}{4} $. If $ A = \left[ \matrix{ 0 & 0 & 1 \cr \frac{1}{4} & f'(1) & 1 \cr \alpha^2 & 4 & 1 } \right] $ and $ B = \text{adj(adj } A) $ be such that $ |B| = 81 $, then $ \alpha^2 $ is equal to

Solution

$ f(x) = \int \frac{7x^{10} + 9x^8}{(1 + x^2 + 2x^9)^2} ,dx $

$ = \int \left( \frac{7}{x^8} - \frac{9}{x^{10}} \right) \frac{dx}{\left( \frac{1}{x^9} + x^7 + 2 \right)^2} $

Put $ t = \frac{1}{x^9} + x^7 + 2 $

$ \frac{dt}{dx} = -\frac{9}{x^{10}} + \frac{7}{x^8} $

$ f(x) = \int \frac{-dt}{t^2} $

$ = \frac{1}{t} + C $

$ = \frac{1}{\frac{1}{x^9} + x^7 + 2} + C $

$ = \frac{x^9}{1 + x^2 + 2x^9} + C $

Given $ f(1) = \frac{1}{4} $

$ \frac{1}{4} = \frac{1}{4} + C \Rightarrow C = 0 $

$ f(x) = \frac{x^9}{1 + x^2 + 2x^9} $

Differentiate:

$ f'(x) = \frac{(1 + x^2 + 2x^9)(9x^8) - x^9(2x + 18x^8)}{(1 + x^2 + 2x^9)^2} $

At $ x = 1 $:

$ f'(1) = \frac{36 - 20}{16} = 1 $

Now,

$ A = \left[ \matrix{ 0 & 0 & 1 \cr \frac{1}{4} & 1 & 1 \cr \alpha^2 & 4 & 1 } \right] $

Determinant:

$ |A| = 1 - \alpha^2 $

Given:

$ B = \text{adj(adj } A) $

$ |B| = |A|^2 = 81 $

$ |A| = 3 $

$ 1 - \alpha^2 = -3 \Rightarrow \alpha^2 = 4 $
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