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3 and 4 . determinant and metrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths3 and 4 . determinant and metrices1 Mark
MCQ
The solutions of the equation $\left|\begin{array}{ccc}1+\sin ^{2} x & \sin ^{2} x & \sin ^{2} x \\ \cos ^{2} x & 1+\cos ^{2} x & \cos ^{2} x \\ 4 \sin 2 x & 4 \sin 2 x & 1+4 \sin 2 x\end{array}\right|=0,(0< x< \pi), \operatorname{are}$
  • A
    $\frac{\pi}{12}, \frac{\pi}{6}$
  • B
    $\frac{\pi}{6}, \frac{5 \pi}{6}$
  • C
    $\frac{5 \pi}{12}, \frac{7 \pi}{12}$
  • ✓
    $\frac{7 \pi}{12}, \frac{11 \pi}{12}$
✓

Answer

Correct option: D.
$\frac{7 \pi}{12}, \frac{11 \pi}{12}$
d
$\left|\begin{array}{ccc}1+\sin ^{2} x & \sin ^{2} x & \sin ^{2} x \\ \cos ^{2} x & 1+\cos ^{2} x & \cos ^{2} x \\ 4 \sin 2 x & 4 \sin 2 x & 1+4 \sin 2 x\end{array}\right|=0$

use $R _{1} \rightarrow R _{1}+ R _{2}+ R _{3}$

$\Rightarrow(2+4 \sin 2 x )\left|\begin{array}{ccc}1 & 1 & 1 \\ \cos ^{2} x & 1+\cos ^{2} x & \cos ^{2} x \\ 4 \sin 2 x & 4 \sin 2 x & 1+4 \sin 2 x \end{array}\right|=0$

$\Rightarrow \sin 2 x =-\frac{1}{2}$

$\Rightarrow 2 x =\pi+\frac{\pi}{6}, 2 \pi-\frac{\pi}{6}$

$x =\frac{\pi}{2}+\frac{\pi}{12}, \pi-\frac{\pi}{12}$

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