MCQ
The function $f(x) = e^x + x$, being differentiable and one to one, has a differentiable inverse $f^{-1} (x)$. The value of $(f^{-1})$ at the point $f(l n2)$ is
- A$\frac{1}{{\ell n2}}$
- ✓$\frac{1}{3}$
- C$\frac{1}{4}$
- Dnone
$1 = (e^x + 1) \frac{{dx}}{{dy}};\frac{{dx}}{{dy}} =\frac{1}{{{e^x} + 1}} =\frac{1}{{{e^x} + 1}}$
$=>{\left. {\frac{{dx}}{{dy}}} \right]_{x = \ell n2}} =\frac{1}{{{e^{\ell n2}} + 1}}$
Alternate : $\frac{{dy}}{{dx}}\,\, = \,{e^x} + 1\,\,;\,\,{\left. {\frac{{dy}}{{dx}}\,} \right|_{x = \ell n\,2}}\,\, = \,\,3\,\,\, \Rightarrow \,\,\frac{{dx}}{{dy}}\,\,\,\,\, = \,\,\frac{1}{3}$
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$f(x) \rightarrow \frac{\lambda\left|x^{2}-5 x+6\right|}{\mu\left(5 x-x^{2}-6\right)}, x<2$
$\quad\quad\quad\quad e^{\frac{\tan (x-2)}{x-[x]}}, \quad x>2$
$\quad\quad\quad\quad \mu \quad\quad\quad\quad x=2$
Where $[x]$ is the greatest integer less than or equal to $x$. If $f$ is continuous at $x=2$, then $\lambda+\mu$ is equal to: