MCQ
$\mathop \smallint \limits_3^6 \frac{{\sqrt x }}{{\sqrt {9 - x} + \sqrt x }}\;dx = $
- A$\frac{1}{2}$
- ✓$\frac{3}{2}$
- C$2$
- D$1$
$=\int_{3}^{6} \frac{\sqrt{9-x}}{\sqrt{9-9+x}+\sqrt{9-x}} d x$
$\Rightarrow I=\int_{3}^{6} \frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}} d x \ldots(i i)$
On adding Eqs. (i) and (ii), we get
$2 I=\int_{3}^{6} \frac{\sqrt{x}+\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}} d x$
$=\int_{3}^{6} 1 d x=[x]_{3}^{6}$
$=6-3=3$
$\Rightarrow I=\frac{3}{2}$
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$f(x)=\max \{\sin t: 0 \leq t \leq x\}, \quad 0 \leq x \leq \pi$
$\quad \quad \quad \quad \quad \quad 2+\cos x,\quad \quad \quad \quad x>\pi$
Then which of the following is true?
| $X$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ |
| $P(X)$ | $0.15$ | $0.23$ | $0.12$ | $0.10$ | $0.20$ | $0.08$ | $0.07$ | $0.05$ |