The value of theta lying between 0 and pi / 2 | vedclass

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Question

(A) Let the given determinant be $\Delta = 0$.Applying row operations $R_1 	o R_1 - R_3$ and $R_2 	o R_2 - R_3$:$\Delta = \left| \begin{array}{ccc}

Vedclass · Accepted Answer

$\frac{7\pi}{24}$ or $\frac{11\pi}{24}$

Vedclass · Answer

(A) Let the given determinant be $\Delta = 0$.Applying row operations $R_1 	o R_1 - R_3$ and $R_2 	o R_2 - R_3$:$\Delta = \left| \begin{array}{ccc} 1 & 0 & -1 \ 0 & 1 & -1 \ \sin^2 	heta & \cos^2 	heta & 1 + 4 \sin 4 	heta \end{array} ight| = 0$Expanding along $R_1$:$1(1 + 4 \sin 4 	heta + \cos^2 	heta) - 0 + (-1)(0 - \sin^2 	heta) = 0$$1 + 4 \sin 4 	heta + \cos^2 	heta + \sin^2 	heta = 0$Since $\sin^2 	heta + \cos^2 	heta = 1$,we get:$1 + 4 \sin 4 	heta + 1 = 0$$2 + 4 \sin 4 	heta = 0$$4 \sin 4 	heta = -2$$\sin 4 	heta = -\frac{1}{2}$Given $0 The values of $4 	heta$ for which $\sin 4 	heta = -\frac{1}{2}$ in the interval $(0, 2 \pi)$ are $4 	heta = \frac{7 \pi}{6}$ and $4 	heta = \frac{11 \pi}{6}$.Therefore,$	heta = \frac{7 \pi}{24}$ and $	heta = \frac{11 \pi}{24}$.

Column $I$	Column $II$
$(A)$ The set $\{\operatorname{Re}(\frac{2 i z}{1-z^2}): \|z\|=1, z \neq \pm 1\}$ is	$(p)$ $(-\infty,-1) \cup(1, \infty)$
$(B)$ The domain of $f(x)=\sin ^{-1}(\frac{8(3)^{x-2}}{1-3^{2(x-1)}})$ is	$(q)$ $(-\infty, 0) \cup(0, \infty)$
$(C)$ If $f(\theta)=\left\|\begin{array}{ccc}1 & \tan \theta & 1 \\ -\tan \theta & 1 & \tan \theta \\ -1 & -\tan \theta & 1\end{array}\right\|$,then the set $\{f(\theta): 0 \leq \theta < \frac{\pi}{2}\}$ is	$(r)$ $[2, \infty)$
$(D)$ If $f(x)=x^{3 / 2}(3 x-10), x \geq 0$,then $f(x)$ is increasing in	$(s)$ $(-\infty,-1] \cup[1, \infty)$
	$(t)$ $(-\infty, 0] \cup[2, \infty)$

The value of $\theta$ lying between $0$ and $\pi / 2$ and satisfying the equation $\left| \begin{array}{ccc} 1 + \sin^2 \theta & \cos^2 \theta & 4 \sin 4 \theta \\ \sin^2 \theta & 1 + \cos^2 \theta & 4 \sin 4 \theta \\ \sin^2 \theta & \cos^2 \theta & 1 + 4 \sin 4 \theta \end{array} \right| = 0$ is:

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