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KEAM 2013 Math Paper

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Question : 30 of 120

Marks: +1, -0

The value of

x

satisfying the equation

\begin{vmatrix} \cos 2x & \sin 2x & \sin 2x \\ \sin 2x & \cos 2x & \sin 2x \\ \sin 2x & \sin 2x & \cos 2x \end{vmatrix} = 0

x \in \left[0, \frac{\pi}{4}\right]

is

$\frac{\pi}{4}$
$\frac{\pi}{2}$
$\frac{\pi}{16}$
$\frac{\pi}{3}$
$\frac{\pi}{8}$

Solution:

Given,

\begin{vmatrix} \cos 2x & \sin 2x & \sin 2x \\ \sin 2x & \cos 2x & \sin 2x \\ \sin 2x & \sin 2x & \cos 2x \end{vmatrix} = 0 ; x \in \left[0, \frac{\pi}{4}\right]

Applying operation:

R_1 \rightarrow R_1 + R_2 + R_3,

we get

\begin{vmatrix} \cos 2x + 2\sin 2x & 2\sin 2x + \cos 2x & 2\sin 2x + \cos 2x \\ \sin 2x & \cos 2x & \sin 2x \\ \sin 2x & \sin 2x & \cos 2x \end{vmatrix}

\Rightarrow (2 \sin 2x + \cos 2x) \begin{vmatrix} 1 & 1 & 1 \\ \sin 2x & \cos 2x & \sin 2x \\ \sin 2x & \sin 2x & \cos 2x \end{vmatrix} = 0

Now, apply operation

C_2 \rightarrow C_2 - C_1

and

C_3 \rightarrow C_3 - C_1,

we get

(2 \sin 2x + \cos 2x)

\begin{vmatrix} 1 & 0 & 0 \\ \sin 2x & \cos 2x - \sin 2x & 0 \\ \sin 2x & 0 & \cos 2x - \sin 2x \end{vmatrix} = 0

Expanding along

R_1

, we get

(2 \sin 2x + \cos 2x)(\cos 2x - \sin 2x)^2 = 0

\Rightarrow \;\; 2 \sin 2x + \cos 2x = 0

or

\cos 2x - \sin 2x = 0

\Rightarrow \;\; \tan 2x = -\frac{1}{2}

or

\;\; \tan 2x = 1 = \tan \frac{\pi}{4}

\because \;\; x \neq \frac{1}{2} \tan^{-1}\left(\frac{-1}{2}\right)

\therefore \;\; x = \frac{\pi}{8} \in \left[0, \frac{\pi}{4}\right]

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