MCQ
If $y = {(1 + x)^x},$ then ${{dy} \over {dx}} = $
- A${(1 + x)^x}\left[ {{x \over {1 + x}} + \log ex} \right]$
- B${x \over {1 + x}} + \log (1 + x)$
- ✓${(1 + x)^x}\left[ {{x \over {1 + x}} + \log (1 + x)} \right]$
- DNone of these
Taking log on both sides, $\log y = x\log (1 + x)$
Differentiating w.r.t. $x,$ we get
$\frac{1}{y}\frac{{dy}}{{dx}} = \log (1 + x) + x\frac{1}{{(1 + x)}}$
Thus $\frac{{dy}}{{dx}} = {(1 + x)^x}\left[ {\frac{x}{{1 + x}} + \log (1 + x)} \right]$
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Statement $-I$ : Let $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$ and $\vec{b}=2 \hat{i}+\hat{j}-\hat{k}$. Then the vector $\vec{r}$ satisfying $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{r}}=0$ is of magnitude $\sqrt{10}$.
Statement $-II$ : In a triangle $A B C, \cos 2 A+\cos 2 B$ $+\cos 2 \mathrm{C} \geq-\frac{3}{2}$