Let f(x)=∫ 7 x^ 10 +9 x^ 8 (1+x^ 2 +2 x^ 9 )^ 2 d x, x>0, _ x arrow 0 f(x)=0 and f(1)=(1)/(4) . If A=[ array ccc 0 & 0 & 1 (1)/(4) & f^ (1) & 1 α^ 2 & 4 & 1 array ] and B=adj(adj A) be such that |B|=81 , then α^ 2 is

Correct answer: D. 4 Divide numerator and denominator by x^ 18 , let t = x^ -9 + x^ -7 + 2 , then dt = -(9x^ -10 +7x^ -8 )dx . f(x) = (1)/(t) + C = (x^9)/(1+x^2+2x^9) + C . _ x → 0 f(x) = 0 C = 0 . f(1) = (1)/(4) (verified). f'(x) = (9x^8(1+x^2+2x^9) - x^9(2x+18x^8))/((1+x^2+2x^9)^2) . f'(1) = (36-20)/(16) = 1 . |A| = vmatrix 0 & 0 & 1 1/4 & 1 & 1 α^2 & 4 & 1 vmatrix = 1 - α^2 . |B| = |adj(adj A)| = |A|^ (3-1)^2 = (1-α^2)^4 = 81 = 3^4 . |1-α^2| = 3 α^2 = 4 .

JEE Main 202624 JanuaryEvening Shift

Exam: JEE Main
Year: 2026
Date: 24 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Matrices

JEE Main 2026 · 24 January · Evening Shift · Mathematics · Matrices

Let f(x)=∫ 7 x^ 10 +9 x^ 8 (1+x^ 2 +2 x^ 9 )^ 2 d x, x>0, _ x arrow...

Let

f(x)=\int \frac{7 x^{10}+9 x^{8}}{\left(1+x^{2}+2 x^{9}\right)^{2}} d x, x>0, \lim _{x \rightarrow 0} f(x)=0

and

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

. If

\mathrm{A}=\left[

\begin{array}{ccc}0 & 0 & 1 \\ \frac{1}{4} & f^{\prime}(1) & 1 \\ \alpha^{2} & 4 & 1\end{array}

\right]

and

\mathrm{B}=\operatorname{adj}(\operatorname{adj} \mathrm{A})

be such that

|\mathrm{B}|=81

, then

\alpha^{2}

is equal to

Options

Correct answer

D. 4

Step-by-step solution

Divide numerator and denominator by

x^{18}

, let

t = x^{-9} + x^{-7} + 2

, then

dt = -(9x^{-10}+7x^{-8})dx

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

\lim_{x \to 0} f(x) = 0 \Rightarrow C = 0

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

(verified).

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

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

|A| = \begin{vmatrix} 0 & 0 & 1 \\ 1/4 & 1 & 1 \\ \alpha^2 & 4 & 1 \end{vmatrix} = 1 - \alpha^2

|B| = |\text{adj}(\text{adj } A)| = |A|^{(3-1)^2} = (1-\alpha^2)^4 = 81 = 3^4

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

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