Let f(x)= 7x^10+9x^8(1+x^2+2x^9)^2 dx, x0, and A= bmatrix 0 & 0 & 1 14 & f'(1) & 1 & 4 & 1 bmatrix. If B= adj( adj A), then the value of for which (B)=1 is

Given: - f(x)= 7x^10+9x^8(1+x^2+2x^9)^2 dx with x0 - A= bmatrix0&0&1 14&f'(1)&1 &4&1 bmatrix - B= adj( adj A) - (B)=1 Find: The value of. First, by the Fundamental Theorem of Calculus, f'(x)= 7x^10+9x^8(1+x^2+2x^9)^2 So at x=1, f'(1)= 7+9(1+1+2)^2= 1616=1 Hence, A= bmatrix0&0&1 14&1&1 &4&1 bmatrix For a 3 3 matrix, ( adj( adj A))=( A)^(3-1)^2=( A)^4 Therefore, (B)=( A)^4=1 So, A= 1 Now compute A by expanding along the first row: A=0-0+1 vmatrix 14&1 &4 vmatrix Thus, A= 14 4- =1- Therefore, (1-)^4=1 which gives 1- = 1 So, =0 or 2 Among the given options, only 2 is present. The solution lists option C and boxed 3, but that contradicts the determinant calculation shown. Therefore, the defensible correct option from the working is B.

Let f(x)= 7x^10+9x^8(1+x^2+2x^9)^2 dx, x0, and A= bmatrix 0 & 0 &… — Solution | JEE Mathematics

MCQMediumJEE 2026Properties of Determinants

Let $f(x)=\int \frac{7x^{10}+9x^8}{(1+x^2+2x^9)^2}\,dx,\quad x>0,$ and $A= \begin{bmatrix} 0 & 0 & 1\\ \frac14 & f'(1) & 1\\ \alpha & 4 & 1 \end{bmatrix}.$ If $B=\operatorname{adj}(\operatorname{adj} A)$ , then the value of $\alpha$ for which $\det(B)=1$ is

A
$1$
B
$2$
C
$3$
D
$4$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given:

$f(x)=\int \frac{7x^{10}+9x^8}{(1+x^2+2x^9)^2}\,dx$ with $x>0$
$A=\begin{bmatrix}0&0&1\\ \frac14&f'(1)&1\\ \alpha&4&1\end{bmatrix}$
$B=\operatorname{adj}(\operatorname{adj} A)$
$\det(B)=1$

Find: The value of $\alpha$ .

First, by the Fundamental Theorem of Calculus,

f'(x)=\frac{7x^{10}+9x^8}{(1+x^2+2x^9)^2}

So at $x=1$ ,

f'(1)=\frac{7+9}{(1+1+2)^2}=\frac{16}{16}=1

Hence,

A=\begin{bmatrix}0&0&1\\ \frac14&1&1\\ \alpha&4&1\end{bmatrix}

For a $3\times 3$ matrix,

\det(\operatorname{adj}(\operatorname{adj} A))=(\det A)^{(3-1)^2}=(\det A)^4

Therefore,

\det(B)=(\det A)^4=1

So,

\det A=\pm 1

Now compute $\det A$ by expanding along the first row:

\det A=0-0+1\begin{vmatrix}\frac14&1\\ \alpha&4\end{vmatrix}

Thus,

\det A=\frac14\cdot 4-\alpha=1-\alpha

Therefore,

(1-\alpha)^4=1

which gives

1-\alpha=\pm 1

So,

\alpha=0 \text{ or } 2

Among the given options, only $2$ is present. The solution lists option $C$ and boxed $3$ , but that contradicts the determinant calculation shown. Therefore, the defensible correct option from the working is B.

Common mistakes

Using the antiderivative $f(x)$ instead of the integrand for $f'(x)$ . This is wrong because the question asks for $f'(1)$ , and by differentiation of an indefinite integral, $f'(x)$ equals the integrand. Evaluate the given rational expression at $x=1$ directly.
Applying the adjugate determinant property incorrectly. For an $n\times n$ matrix, $\det(\operatorname{adj} A)=(\det A)^{n-1}$ , so for $\operatorname{adj}(\operatorname{adj} A)$ in the $3\times 3$ case, the exponent becomes $(n-1)^2=4$ . Do not use exponent $2$ or $3$ here.
Making a sign error while expanding $\det A$ along the first row. The only nonzero entry in the first row is the third entry, and its cofactor sign is positive. Hence $\det A=\begin{vmatrix}\frac14&1\\ \alpha&4\end{vmatrix}=1-\alpha$ , not $\alpha-1$ .

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