MCQ
$\int {\frac{{{e^{\sqrt x }}}}{{\sqrt x }}dx} = $
- A${e^{\sqrt x }}$
- B$\frac{{{e^{\sqrt x }}}}{2}$
- ✓$2\,{e^{\sqrt x }}$
- D$\sqrt x \,.\,{e^{\sqrt x }}$
Put $\sqrt x = t$, $\therefore \frac{1}{{2\sqrt x }}dx = dt$
$I = 2\int {{e^t}dt = 2{e^t} + C = 2{e^{\sqrt x }} + C} $.
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$f(x)=\left\{\begin{array}{rc}x^5+5 x^4+10 x^3+10 x^2+3 x+1, & x<0 \\ x^2-x+1, & 0 \leq x<1 \\ \frac{2}{3} x^3-4 x^2+7 x-\frac{8}{3}, & 1 \leq x<3 \\ (x-2) \log _e(x-2)-x+\frac{10}{3}, & x \geq 3\end{array}\right.$
Then which of the following options is/are correct?
$(1)$ $f^{\prime}$ has a local maximum at $x =1$ $(2)$ $f$ is onto
$(3)$ $f$ is increasing on $(-\infty, 0)$ $(4)$ $f^{\prime}$ is $NOT$ differentiable at $x =1$