- A${\left( {{y \over x}} \right)^{1/3}}$
- ✓$ - {\left( {{y \over x}} \right)^{1/3}}$
- C${\left( {{x \over y}} \right)^{1/3}}$
- D$ - {\left( {{x \over y}} \right)^{1/3}}$
==> $\frac{2}{3}{x^{ - 1/3}} + \frac{2}{3}{y^{ - 1/3}}\frac{{dy}}{{dx}} = 0$ or
$\frac{{dy}}{{dx}} = - {\left( {\frac{x}{y}} \right)^{ - 1/3}} = - {\left( {\frac{y}{x}} \right)^{1/3}}$.
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$\left[\frac{x}{\sqrt{x^{2}-y^{2}}}+e^{\frac{y}{x}}\right] x \frac{d y}{d x}=x+\left[\frac{x}{\sqrt{x^{2}-y^{2}}}+e^{\frac{y}{x}}\right] y$
pass through the points $(1,0)$ and $(2 \alpha, \alpha), \alpha>0$.
Then $\alpha$ is equal to
$1.$ 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 $2.$