ISI2016-MMA-8

Question

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

• A. $64/5$
• B. $32/5$
• C. $37/5$
• D. $67/5$

Comments

Okay.

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Make this a comment.

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Yes, that was a mistake.

Answer 1

$\underline{\mathbf{Answer:D}}$

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

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

$\therefore \mathbf D$ is the correct option.

Comments

Nice solution.

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Thanks.

Answer 2

g'(x²) = x³

let x² = y so given equation becomes g'(y) = y^1.5

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  $\frac{x^{5}}{2.5}$ = $\frac{32}{2.5}$  =  $\frac{64}{5}$

Comments

@Helsenberg: What about the Integration constant? I think the correct answer is 67/5.

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from $g(1) = 1$ and $ g(x) = x^{5/2} + c$ giving

$c=3/5$ thus $g(4) = 67/5$