MCQ
$\int_0^2 {\sqrt {\frac{{2 + x}}{{2 - x}}} } \,dx = $
- ✓$\pi + 2$
- B$\pi + \frac{3}{2}$
- C$\pi + 1$
- DNone of these
$\Rightarrow dx = - 2\sin \theta \,d\theta ,$ then
$\int_0^2 {\sqrt {\frac{{2 + x}}{{2 - x}}} } dx = - 2\int_{\pi /2}^0 {\sqrt {\frac{{1 + \cos \theta }}{{1 - \cos \theta }}} } \sin \theta \,d\theta $
$ = 4\int_0^{\pi /2} {\frac{{\cos (\theta /2)}}{{\sin (\theta /2)}}\sin \frac{\theta }{2}\cos \frac{\theta }{2}d\theta } $
$ = 2\int_0^{\pi /2} {(1 + \cos \theta )\,d\theta } $
$ = 2[\theta + \sin \theta ]_0^{\pi /2} $
$= 2\left[ {\frac{\pi }{2} + 1} \right] = \pi + 2$.
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$1.$ One of the two boxes, box $I$ and box $II$, was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this red ball was drawn from box $II$ is $\frac{1}{3}$, then the correct option$(s)$ with the possible values of $n_1, n_2, n_3$ and $n_4$ is(are)
$(A)$ $n_1=3, n_2=3, n_3=5, n_4=15$
$(B)$ $n_1=3, n_2=6, n_3=10, n_4=50$
$(C)$ $n_1=8, n_2=6, n_3=5, n_4=20$
$(D)$ $n_1=6, n_2=12, n_3=5, n_4=20$
$2.$ A ball is drawn at random from box $I$ and transferred to box $II$. If the probability of drawing a red ball from box $I$, after this transfer, is $\frac{1}{3}$, then the correct option$(s)$ with the possible values of $n_1$ and $n_2$ is(are)
$(A)$ $n_1=4, n_2=6$ $(B)$ $n_1=2, n_2=3$
$(C)$ $n_1=10, n_2=20$ $(D)$ $n_1=3, n_2=6$
Give the answer question $1$ and $2.$