Matrices & Determinants
Matrices and Determinants
star_batch_jee_advanced_2025
Grade 12
Question:
Atleast one root of the equation $\begin{vmatrix} x^2 + \sin x \cos x & x(1+\sin x) \\ x + \cos x & x+1 \end{vmatrix} = 0$ lies in:
$\left(0, \frac{\pi}{6}\right)$
$\left(\frac{\pi}{6}, \frac{\pi}{3}\right)$
$\left(-\frac{\pi}{3}, -\frac{\pi}{4}\right)$
$\left(-\frac{\pi}{4}, \frac{\pi}{6}\right)$
Step-by-Step Solution
Key Concept: Use the Intermediate Value Theorem by computing $f(x)$ at interval endpoints to detect sign changes indicating roots.
Expanding the determinant: $\begin{vmatrix} x^2 + \sin x \cos x & x(1+\sin x) \\ x + \cos x & x+1 \end{vmatrix} = (x^2 + \sin x \cos x)(x+1) - x(1+\sin x)(x + \cos x) = 0$. Simplifying: $x^3 + x^2 + x\sin x \cos x + \sin x \cos x - x^2 - x\cos x - x^2\sin x - x\sin x \cos x = 0$, which reduces to $x^3 - x^2\sin x - x\cos x + \sin x \cos x = 0$ or $x^3 + \sin x \cos x - x(x\sin x + \cos x) = 0$. Let $f(x) = x^3 + \sin x \cos x - x^2\sin x - x\cos x$. By checking sign changes in the given intervals using the Intermediate Value Theorem, roots exist in intervals 2, 3, and 4.
Correct Answer: 2,3,4