JEE MAIN - Mathematics (2021 - 25th July Evening Shift - No. 10)
The number of distinct real roots
of $$\left| {\matrix{ {\sin x} & {\cos x} & {\cos x} \cr {\cos x} & {\sin x} & {\cos x} \cr {\cos x} & {\cos x} & {\sin x} \cr } } \right| = 0$$ in the interval $$ - {\pi \over 4} \le x \le {\pi \over 4}$$ is :
of $$\left| {\matrix{ {\sin x} & {\cos x} & {\cos x} \cr {\cos x} & {\sin x} & {\cos x} \cr {\cos x} & {\cos x} & {\sin x} \cr } } \right| = 0$$ in the interval $$ - {\pi \over 4} \le x \le {\pi \over 4}$$ is :
4
1
2
3
Explanation
$$\left| {\matrix{
{\sin x} & {\cos x} & {\cos x} \cr
{\cos x} & {\sin x} & {\cos x} \cr
{\cos x} & {\cos x} & {\sin x} \cr
} } \right| = 0, - {\pi \over 4} \le x \le {\pi \over 4}$$
Apply : $${R_1} \to {R_1} - {R_2}$$ & $${R_2} \to {R_2} - {R_3}$$
$$ \Rightarrow $$ $$\left| {\matrix{ {\sin x - \cos x} & {\cos x - \sin x} & 0 \cr 0 & {\sin x - \cos x} & {\cos x - \sin x} \cr {\cos x} & {\cos x} & {\sin x} \cr } } \right| = 0$$
$$ \Rightarrow $$ $${(\sin x - \cos x)^2}\left| {\matrix{ 1 & { - 1} & 0 \cr 0 & 1 & { - 1} \cr {\cos x} & {\cos x} & {\sin x} \cr } } \right| = 0$$
$$ \Rightarrow $$ $${(\sin x - \cos x)^2}(\sin x + 2\cos x) = 0$$
$$\therefore$$ $$x = {\pi \over 4}$$
Apply : $${R_1} \to {R_1} - {R_2}$$ & $${R_2} \to {R_2} - {R_3}$$
$$ \Rightarrow $$ $$\left| {\matrix{ {\sin x - \cos x} & {\cos x - \sin x} & 0 \cr 0 & {\sin x - \cos x} & {\cos x - \sin x} \cr {\cos x} & {\cos x} & {\sin x} \cr } } \right| = 0$$
$$ \Rightarrow $$ $${(\sin x - \cos x)^2}\left| {\matrix{ 1 & { - 1} & 0 \cr 0 & 1 & { - 1} \cr {\cos x} & {\cos x} & {\sin x} \cr } } \right| = 0$$
$$ \Rightarrow $$ $${(\sin x - \cos x)^2}(\sin x + 2\cos x) = 0$$
$$\therefore$$ $$x = {\pi \over 4}$$
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