MCQ
If $\ln \left( {\left( {e - 1} \right){e^{xy}} + {x^2}} \right) = {x^2} + {y^2}$ , then ${\left. {\frac{{dy}}{{dx}}} \right|_{\left( {1,0} \right)}}$ is
- A$0$
- B$1$
- ✓$2$
- D$4$
$(e - 1){e^{xy}}\left( {\frac{{xdy}}{{dx}} + y} \right) + 2x = {e^{{x^2} + {y^2}}}\left( {2x + 2y\frac{{dy}}{{dx}}} \right)$
$\left.(e-1)\left(\frac{d y}{d x}\right)\right|_{(1,0)}+2=e(2)$
$\left.\frac{d y}{d x}\right|_{(1,0)}=2$
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| $\text{X}:$ | $1$ | $2$ | $3$ | $4$ |
| $\text{P}(\text{X}):$ | $\frac{1}{10}$ | $\frac{1}{5}$ | $\frac{3}{10}$ | $\frac{2}{5}$ |
The value of E(X2) is: