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

Matrices and Determinants PYQ — JEE Main 2016 · 9 Apr · Shift 1 · Q25 with Solution

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Correct answer: C

\cos x & \sin x & \sin x\\ \sin x & \cos x & \sin x\\ \sin x & \sin x & \cos x \end{vmatrix}.$$ We need to solve $$D=0$$ in the interval $$\left[-\frac{\pi}{4},\frac{\pi}{4}\right].$$ 2. Observe that this is a $3\times 3$ matrix with diagonal entries $a=\cos x$ and off-diagonal entries $b=\sin x$. For a matrix of the form $$\begin{pmatrix} a & b & b\\ b & a & b\\ b & b & a \end{pmatrix},$$ its determinant is $$ (a-b)^2(a+2b). $$ So here, $$D=(\cos x-\sin x)^2(\cos x+2\sin x).$$ 3. Hence the equation becomes $$(\cos x-\sin x)^2(\cos x+2\sin x)=0.$$ Therefore, - either $\cos x-\sin x=0$, or - $\cos x+2\sin x=0$. 4. Solve each factor. ### Case 1: $$\cos x-\sin x=0$$ $$\cos x=\sin x$$ Assuming $\cos x\neq 0$ in the given interval (true since $x\in[-\pi/4,\pi/4]$), divide by $\cos x$: $$1=\tan x$$ $$x=\frac{\pi}{4}+n\pi.$$ In the interval $\left[-\frac{\pi}{4},\frac{\pi}{4}\right]$, this gives only $$x=\frac{\pi}{4}.$$ 5. ### Case 2: $$\cos x+2\sin x=0$$ $$\cos x=-2\sin x$$ Again divide by $\cos x$: $$1=-2\tan x$$ $$\tan x=-\frac12.$$ Thus $$x=\tan^{-1}\left(-\frac12\right)+n\pi.$$ Since $$\tan^{-1}\left(-\frac12\right)\in\left[-\frac{\pi}{4},\frac{\pi}{4}\right],$$ there is exactly one solution in the interval: $$x=\tan^{-1}\left(-\frac12\right).$$ 6. Total distinct real roots in the interval: - $x=\frac{\pi}{4}$ - $x=\tan^{-1}(-1/2)$ So the number of distinct real roots is $$2.$$ 7. Therefore, the correct option is $$\boxed{\text{C}}.$$

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Correct answer: C

\cos x & \sin x & \sin x\\ \sin x & \cos x & \sin x\\ \sin x & \sin x & \cos x \end{vmatrix}.$$ We need to solve $$D=0$$ in the interval $$\left[-\frac{\pi}{4},\frac{\pi}{4}\right].$$ 2. Observe that this is a $3\times 3$ matrix with diagonal entries $a=\cos x$ and off-diagonal entries $b=\sin x$. For a matrix of the form $$\begin{pmatrix} a & b & b\\ b & a & b\\ b & b & a \end{pmatrix},$$ its determinant is $$ (a-b)^2(a+2b). $$ So here, $$D=(\cos x-\sin x)^2(\cos x+2\sin x).$$ 3. Hence the equation becomes $$(\cos x-\sin x)^2(\cos x+2\sin x)=0.$$ Therefore, - either $\cos x-\sin x=0$, or - $\cos x+2\sin x=0$. 4. Solve each factor. ### Case 1: $$\cos x-\sin x=0$$ $$\cos x=\sin x$$ Assuming $\cos x\neq 0$ in the given interval (true since $x\in[-\pi/4,\pi/4]$), divide by $\cos x$: $$1=\tan x$$ $$x=\frac{\pi}{4}+n\pi.$$ In the interval $\left[-\frac{\pi}{4},\frac{\pi}{4}\right]$, this gives only $$x=\frac{\pi}{4}.$$ 5. ### Case 2: $$\cos x+2\sin x=0$$ $$\cos x=-2\sin x$$ Again divide by $\cos x$: $$1=-2\tan x$$ $$\tan x=-\frac12.$$ Thus $$x=\tan^{-1}\left(-\frac12\right)+n\pi.$$ Since $$\tan^{-1}\left(-\frac12\right)\in\left[-\frac{\pi}{4},\frac{\pi}{4}\right],$$ there is exactly one solution in the interval: $$x=\tan^{-1}\left(-\frac12\right).$$ 6. Total distinct real roots in the interval: - $x=\frac{\pi}{4}$ - $x=\tan^{-1}(-1/2)$ So the number of distinct real roots is $$2.$$ 7. Therefore, the correct option is $$\boxed{\text{C}}.$$