MCQ
${d \over {dx}}\left\{ {\log \left( {{{{e^x}} \over {1 + {e^x}}}} \right)} \right\} = $
- A${1 \over {1 - {e^x}}}$
- B$ - {1 \over {1 + {e^x}}}$
- C$ - {1 \over {1 - {e^x}}}$
- ✓None of these
$ = \frac{{1 + {e^x}}}{{{e^x}}} \times \frac{{{e^x}}}{{{{(1 + {e^x})}^2}}} = \frac{1}{{1 + {e^x}}}$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$ \mathrm{L}_1: \overrightarrow{\mathrm{r}}=(2+\lambda) \hat{\mathrm{i}}+(1-3 \lambda) \hat{\mathrm{j}}+(3+4 \lambda) \hat{\mathrm{k}}, \lambda \in \mathbb{R} $
$ \mathrm{L}_2: \overrightarrow{\mathrm{r}}=2(1+\mu) \hat{\mathrm{i}}+3(1+\mu) \hat{\mathrm{j}}+(5+\mu) \hat{k}, \mu \in \mathbb{R}$
is $\frac{\mathrm{m}}{\sqrt{\mathrm{n}}}$, where $\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$, then the value of $\mathrm{m}+\mathrm{n}$ equals.