The number of distinct real roots of | \beginmatrix \sin x & \cos x & \cos x \cos x & \sin x & \cos x \cos x & \cos x & \sin x \endmatrix | = 0 in the interval - \pi \over 4 \le x \le \pi \over 4 is :

Matrices and Determinants PYQ — JEE Main 2021 · 25 Jul · Shift 2 · Q33 with Solution

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Correct answer: B

\sin x & \cos x & \cos x\\ \cos x & \sin x & \cos x\\ \cos x & \cos x & \sin x \end{vmatrix}.$$ We need to solve $D=0$ for $x\in\left[-\frac\pi4,\frac\pi4\right]$. 2. Observe that this is a $3\times 3$ matrix with diagonal entries $a=\sin x$ and off-diagonal entries $b=\cos x$. For such a matrix, $$\begin{vmatrix} a & b & b\\ b & a & b\\ b & b & a \end{vmatrix}=(a-b)^2(a+2b).$$ So here, $$D=(\sin x-\cos x)^2(\sin x+2\cos x).$$ 3. Therefore, $$D=0 \iff (\sin x-\cos x)^2(\sin x+2\cos x)=0.$$ Hence either - $\sin x-\cos x=0$, or - $\sin x+2\cos x=0$. 4. Solve the first equation: $$\sin x=\cos x \implies \tan x=1.$$ So, $$x=\frac\pi4+n\pi.$$ In the interval $\left[-\frac\pi4,\frac\pi4\right]$, this gives only $$x=\frac\pi4.$$ 5. Solve the second equation: $$\sin x+2\cos x=0 \implies \tan x=-2$$ (provided $\cos x\neq 0$, which is true in this interval). Thus, $$x=\tan^{-1}(-2)+n\pi.$$ Now, $$\tan^{-1}(-2)\approx -1.107$$ which lies outside the interval $$\left[-\frac\pi4,\frac\pi4\right]=[-0.785...,0.785...].$$ Adding or subtracting $\pi$ moves it even farther away, so there is no solution from this branch in the given interval. 6. Hence the only root in the interval is $$x=\frac\pi4.$$ Therefore, the number of distinct real roots is $$1.$$ 7. Checking options: - A: 4 ❌ - B: 1 ✅ - C: 2 ❌ - D: 3 ❌ So the correct option is **B**.

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Correct answer: B

\sin x & \cos x & \cos x\\ \cos x & \sin x & \cos x\\ \cos x & \cos x & \sin x \end{vmatrix}.$$ We need to solve $D=0$ for $x\in\left[-\frac\pi4,\frac\pi4\right]$. 2. Observe that this is a $3\times 3$ matrix with diagonal entries $a=\sin x$ and off-diagonal entries $b=\cos x$. For such a matrix, $$\begin{vmatrix} a & b & b\\ b & a & b\\ b & b & a \end{vmatrix}=(a-b)^2(a+2b).$$ So here, $$D=(\sin x-\cos x)^2(\sin x+2\cos x).$$ 3. Therefore, $$D=0 \iff (\sin x-\cos x)^2(\sin x+2\cos x)=0.$$ Hence either - $\sin x-\cos x=0$, or - $\sin x+2\cos x=0$. 4. Solve the first equation: $$\sin x=\cos x \implies \tan x=1.$$ So, $$x=\frac\pi4+n\pi.$$ In the interval $\left[-\frac\pi4,\frac\pi4\right]$, this gives only $$x=\frac\pi4.$$ 5. Solve the second equation: $$\sin x+2\cos x=0 \implies \tan x=-2$$ (provided $\cos x\neq 0$, which is true in this interval). Thus, $$x=\tan^{-1}(-2)+n\pi.$$ Now, $$\tan^{-1}(-2)\approx -1.107$$ which lies outside the interval $$\left[-\frac\pi4,\frac\pi4\right]=[-0.785...,0.785...].$$ Adding or subtracting $\pi$ moves it even farther away, so there is no solution from this branch in the given interval. 6. Hence the only root in the interval is $$x=\frac\pi4.$$ Therefore, the number of distinct real roots is $$1.$$ 7. Checking options: - A: 4 ❌ - B: 1 ✅ - C: 2 ❌ - D: 3 ❌ So the correct option is **B**.