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Let $g: \mathbb{R} \rightarrow \mathbb{R}$ be differentiable with $g'(x^2)=x^3$ for all $x>0$ and $g(1) =1$. Then $g(4)$ equals

  1. $64/5$
  2. $32/5$
  3. $37/5$
  4. $67/5$
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$\underline{\mathbf{Answer:D}}$

$\underline{\mathbf{Solution:}}$

 $\begin{align}\textbf{Given:} \;\;\;\mathrm{x^2 = t > 0 }\\ \text{On integrating above equation, we get:} \\ \ \mathrm{g’(t) = t^{\frac{3}{2}}}\\ \mathrm{g(t) = \frac{2}{5}t^{\frac{5}{2}}} + \mathrm C \\ \textbf{Given} \; \mathrm g(1) = 1\\  \Rightarrow \mathrm C = \dfrac{3}{5} \\ \Rightarrow \mathrm{g(x^2) = \dfrac{2}{5}x^5 + \dfrac{3}{5} \\ \text{and,}\; g(2^2) = \dfrac{2}{5}2^5 + \dfrac{3}{5} \\= \dfrac{67}{5}} \end {align}$

$\therefore \mathbf D$ is the correct option.
edited by
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g'(x2) = x3

let x2 = y so given equation becomes g'(y) = y1.5

Inegrating it wrt y,   g(y) = $\frac{y^{2.5}}{2.5}$  =>   g(x2) = $\frac{x^{5}}{2.5}$

g(4) implies x=2, Therefore  $\frac{x^{5}}{2.5}$ = $\frac{32}{2.5}$  =  $\frac{64}{5}$

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