MCQ
${d \over {dx}}\sqrt {x\sin x} = $
- ✓${{\sin x + x\cos x} \over {2\sqrt {x\sin x} }}$
- B${{\sin x + x\cos x} \over {\sqrt {x\sin x} }}$
- C${{x\sin x + \cos x} \over {\sqrt {2\sin x} }}$
- D${{\sin x + x\cos x} \over {\sqrt {2x\sin x} }}$
$\therefore \frac{{dy}}{{dx}} = \frac{{[\sin x + x\cos x]}}{{2\sqrt {x\sin x} }}$.
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If $\text{A}=\begin{bmatrix}1&0\\0&1\end{bmatrix},$ then A2 is equal to:
$\begin{bmatrix}0&1\\1&0\end{bmatrix}$
$\begin{bmatrix}1&0\\1&0\end{bmatrix}$
$\begin{bmatrix}0&1\\0&1\end{bmatrix}$
$\begin{bmatrix}1&0\\0&1\end{bmatrix}$