- 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
$2 I=\int_{3}^{6} \frac{\sqrt{x}+\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}} d x$
$=\int_{3}^{6} 1 d x=[x]_{3}^{6}$
$=6-3=3$
$\Rightarrow I=\frac{3}{2}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$g(x)=\left\{\begin{array}{ccc}0 & \text { if } & x < a, \\ \int_a^x f(t) d t & \text { if } & a \leq x \leq b, \\ \int_a^b f(t) d t & \text { if } & x > b .\end{array}\right.$, Then
$(A)$ $g(x)$ is continuous but not differentiable at a
$(B)$ $g(x)$ is differentiable on $R$
$(C)$ $g(x)$ is continuous but not differentiable at $b$
$(D)$ $g(x)$ is continuous and differentiable at either a or $b$ but not both
$1.$ If $1$ ball is drawn from each of the boxes $B_1, B_2$ and $B_3$, the probability that all $3$ drawn balls are of the same colour is
$(A)$ $\frac{82}{648}$ $(B)$ $\frac{90}{648}$ $(C)$ $\frac{558}{648}$ $(D)$ $\frac{566}{648}$
$2.$ If $2$ balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these $2$ balls are drawn from bo $B _2$ is
$(A)$ $\frac{116}{181}$ $(B)$ $\frac{126}{181}$ $(C)$ $\frac{65}{181}$ $(D)$ $\frac{55}{181}$
Give the answer question $1$ and $2.$