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Q)

If $f '(x)=\sqrt{x}$ and $f(1)=2$ then $f(x)$ is

\[\begin{array}{1 1}(1)\frac{-2}{3}(x\sqrt{x}+2)&(2)\frac{3}{2}(x\sqrt{x}+2)\\(3)\frac{2}{3}(x(\sqrt{x}+2)&(4)\frac{2}{3}x(\sqrt{x}+2)\end{array}\]

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A)
$f'(x)= \sqrt x$
$f(x) =\int \sqrt x dx $
$\qquad =\large\frac{x^{1/2+1}}{1/2+1}$$+c$
$f(x) =\large\frac{2}{3} $$x^{3/2} +c$
$f(1)= \large\frac{2}{3} $$ \times 1 +c$
$2= \large\frac{2}{3} $$+c$
$c= 2 - \large\frac{2}{3}$
$\quad= \large\frac{6-2}{3} =\frac{4}{3}$
$f(x) =\large\frac{2}{3} $$x^{2/3} +\large\frac{4}{3} $
$\qquad= \large\frac{2}{3} $$ (x \sqrt x +2)$
Hence 3 is the correct answer.
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