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  • The number of distinct real roots of in the interval is A. 0 B. –1 C. 1 D. None of these

The number of distinct real roots of  \left|\begin{array}{ccc} \sin x & \cos & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \end{array}\right|=0 in the interval -\frac{\pi}{4} \leq x \leq \frac{\pi}{4} is

A. 0

B. –1
C. 1
D. None of these

Answers (1)

C)

Given

\left|\begin{array}{ccc} \sin x & \cos & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \end{array}\right|=0

\begin{aligned} &\text { Apply } \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}\\ &\Rightarrow\left|\begin{array}{lll} \sin x+\cos x+\cos x & \cos x & \cos x \\ \cos x+\sin x+\cos x & \sin x & \cos x \\ \cos x+\cos x+\sin x & \cos x & \sin x \end{array}\right|=0\\ &\Rightarrow\left|\begin{array}{ccc} 2 \cos x+\sin x & \cos x & \cos x \\ 2 \cos x+\sin x & \sin x & \cos x \\ 2 \cos x+\sin x & \cos x & \sin x \end{array}\right|=0\\ &\text { Take }(2 \cos x+\sin x) \text { common from Column } 1\\ &\Rightarrow 2 \cos x+\sin x\left|\begin{array}{ccc} 1 & \cos x & \cos x \\ 1 & \sin x & \cos x \\ 1 & \cos x & \sin x \end{array}\right|=0\\ &\text { Apply } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}\\ &\Rightarrow 2 \cos x+\sin x\left|\begin{array}{ccc} 1 & \cos x & \cos x \\ 1-1 & \sin x-\cos x & \cos x-\cos x \\ 1 & \cos x & \sin x \end{array}\right|=0 \end{aligned}

\\\Rightarrow 2 \cos x+\sin x\left|\begin{array}{ccc} 1 & \cos x & \cos x \\ 0 & \sin x-\cos x & 0 \\ 1 & \cos x & \sin x \end{array}\right|=0 \\ \text { Apply } R_{3} \rightarrow R_{3}-R_{1} \\ \Rightarrow 2 \cos x+\sin x\left|\begin{array}{ccc} 1 & \cos x & \cos x \\ 0 & \sin x-\cos x & 0 \\ 1-1 & \cos x-\cos x & \sin x-\cos x \end{array}\right|=0 \\

 

\Rightarrow 2 \cos x+\sin x\left|\begin{array}{ccc} 1 & \cos x & \cos x \\ 0 & \sin x-\cos x & 0 \\ 0 & 0 & \sin x-\cos x \end{array}\right|=0

Expand along Column 1

\\\vspace{\baselineskip} (2cos X + sin X) [(1)$ \{ $ (sin X -cos X)(sin X -cos X)$ \} $ ]\\ \\ \vspace{\baselineskip}$ \Rightarrow $ (2cos X + sin X)(sin X -cos X)\textsuperscript{2} = 0\\ \\ \vspace{\baselineskip}$ \Rightarrow $ 2cos X = -sin X or (sin X -cos X)\textsuperscript{2} = 0\\

\\ \Rightarrow 2=-\frac{\sin \mathrm{x}}{\cos \mathrm{x}}$ or $\sin \mathrm{x}=\cos \mathrm{x}$\\ $\Rightarrow \tan x=-2$ or tan $x=1\left[\because \tan x=\frac{\sin x}{\cos x}\right]$ \\but tan $x=-2$ is not possible as for $-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}$ \\So, $\tan x=1$ $\\\therefore x=\frac{\pi}{4}$

Only one real and distinct root occurs.

 

 

Posted by

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