Medium MCQ +4 / -1 PYQ · JEE Mains 2021

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 :

  1. A 4
  2. B 1 Correct answer
  3. C 2
  4. D 3

Solution

$$\left| {\matrix{ {\sin x} &amp; {\cos x} &amp; {\cos x} \cr {\cos x} &amp; {\sin x} &amp; {\cos x} \cr {\cos x} &amp; {\cos x} &amp; {\sin x} \cr } } \right| = 0, - {\pi \over 4} \le x \le {\pi \over 4}$$<br><br>Apply : ${R_1} \to {R_1} - {R_2}$ &amp; ${R_2} \to {R_2} - {R_3}$<br><br>$\Rightarrow$ $$\left| {\matrix{ {\sin x - \cos x} &amp; {\cos x - \sin x} &amp; 0 \cr 0 &amp; {\sin x - \cos x} &amp; {\cos x - \sin x} \cr {\cos x} &amp; {\cos x} &amp; {\sin x} \cr } } \right| = 0$$<br><br>$\Rightarrow$ $${(\sin x - \cos x)^2}\left| {\matrix{ 1 &amp; { - 1} &amp; 0 \cr 0 &amp; 1 &amp; { - 1} \cr {\cos x} &amp; {\cos x} &amp; {\sin x} \cr } } \right| = 0$$<br><br>$\Rightarrow$ ${(\sin x - \cos x)^2}(\sin x + 2\cos x) = 0$<br><br>$\therefore$ $x = {\pi \over 4}$

About this question

Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations

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