Question

The number of distinct real roots of $\begin{vmatrix}\sin x&\cos x&\cos x\\\ \cos x&\sin x&\cos x\\\ \cos x &\cos x&\sin x\end{vmatrix}$ in the interval $\left[\frac{-\pi}{4}, \frac{\pi}{4}\right]$ is

• A. 0
• B. 1
• C. 2
• D. 5

Answer 1

Answer: (B)

1

Answer 2

We have, $\begin{vmatrix}\sin x&\cos x&\cos x\\\ \cos x&\sin x&\cos x\\\ \cos x &\cos x&\sin x\end{vmatrix}$
Applying $C_1 \to C_1 + C_2 + C_3$, we get
$\begin{vmatrix}\sin x + 2\cos x&\cos x&\cos x\\\ \sin x + 2\cos x&\sin x&\cos x\\\ \sin x + 2\cos x &\cos x&\sin x\end{vmatrix}$
$\Rightarrow \ (\sin x + 2 \cos x) \begin{vmatrix} 1 &\cos x&\cos x\\\ 1 &\sin x&\cos x\\\ 1 &\cos x&\sin x\end{vmatrix}$
Applying $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$, we get
$\Rightarrow \ (\sin x + 2 \cos x) \begin{vmatrix} 1 &\cos x&\cos x\\\ 0 &\sin x - \cos x &0 \\\0 &0 &\sin x - \cos x\end{vmatrix}$
Now expanding along $C_1$, we get
$\Rightarrow (\sin x + 2 \cos x) (\sin x - \cos x)2 = 0$
Either $\sin x + 2 \cos x = 0$ or $\sin x - \cos x = 0$
$\Rightarrow \ \frac{\sin x}{\cos x} = -2$ or $\frac{\sin x}{\cos x} =1$
$\Rightarrow\ \tan x=-2$ or $\tan x=1$
This is not possible case in $\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$
So, $\tan x = 1 \Rightarrow x = \frac{\pi}{4}$ is the solution
Number of roots lying in $\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$ is 1