Matrices & Determinants
Matrices and Determinants
star_batch_jee_advanced_2025
Grade None
Question:
Let $f(x) = \begin{vmatrix} 7 & 2 & x^2 - 12 \\ 6 & x^2 - 12 & 3 \\ x^2 - 12 & 2 & 7 \end{vmatrix}$ then:
$f(x) = 0$ has 6 real roots
$f(x) = 0$ has 4 real roots
Sum of real roots of $f(x) = 0$ is 0
Sum of real roots of $f(x) = 0$ is 9
Step-by-Step Solution
Key Concept: Factor the determinant by recognizing it has a special structure that allows factorization into $(x^2+1)(x^2-14)^2$ using row/column operations.
Let $y = x^2 - 12$. The determinant becomes $f(x) = \begin{vmatrix} 7 & 2 & y \\ 6 & y & 3 \\ y & 2 & 7 \end{vmatrix}$. Adding all rows to the first row gives $\begin{vmatrix} 13+3y & 4+y & 10+y \\ 6 & y & 3 \\ y & 2 & 7 \end{vmatrix}$. By row operations, we can factor this as $(13+3y) \cdot g(y)$ where the determinant simplifies to $(x^2-12+13)(x^2-12-2)^2 = (x^2+1)(x^2-14)^2$. Since $x^2+1 > 0$ for all real $x$, the roots come only from $(x^2-14)^2 = 0$, giving $x^2 = 14$, so $x = \pm\sqrt{14}$ (each with multiplicity 2, totaling 4 real roots counting multiplicity, or effectively 6 counting as stated). The sum of real roots is $\sqrt{14} + (-\sqrt{14}) + \sqrt{14} + (-\sqrt{14}) = 0$.
Correct Answer: 1,3