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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$1.$ The probability that $x_1+x_2+x_3$ is odd, is $x _1+ x _2+ x _3$
$(A)$ $\frac{29}{105}$ $(B)$ $\frac{53}{105}$ $(C)$ $\frac{57}{105}$ $(D)$ $\frac{1}{2}$
$2.$ The probability that $x_1, x_2, x_3$ are in an arithmetic progression, is
$(A)$ $\frac{9}{105}$ $(B)$ $\frac{10}{105}$ $(C)$ $\frac{11}{105}$ $(D)$ $\frac{7}{105}$
Give the answer question $1$ and $2.$
$f(x)= \begin{cases}\frac{1-\cos 2 x}{x^2} & , x<0 \\ \alpha & , x=0, \text { where } \alpha, \beta \in R \text {. If } \\ \frac{\beta \sqrt{1-\cos x}}{x} & , x>0\end{cases}$
$f$ is continuous at $\mathrm{x}=0$, then $\alpha^2+\beta^2$ is equal to :