Question

Let $f:\mathbb{R}^{+}\rightarrow\mathbb{R}$ be such that $f(e)=\frac{7}{4}$ and $f^{\prime}(x^{2})=\frac{3\log x}{x^{2}}$, then $f(e^{2})=$

• A. $\frac{5}{4}$
• B. 4
• C. 10
• D. $\frac{49}{16}$

Hint:

When you have a function inside a derivative like $f^{\prime}(x^2)$, use a substitution like $u = x^2$ to easily change it back to standard form before integrating.

Answer 1

The Correct Option is B

Step 1: Change the variable.
We know $f'(x^2)=\dfrac{3\log x}{x^2}$. Let $u=x^2$, so $x=\sqrt{u}$ and $\log x=\tfrac12\log u$.
Step 2: Rewrite $f'(u)$ neatly.
\[ f'(u)=\frac{3\cdot\tfrac12\log u}{u}=\frac{3}{2}\cdot\frac{\log u}{u}. \]
Step 3: Integrate to get $f(u)$.
Let $t=\log u$, so $dt=\dfrac{du}{u}$: \[ f(u)=\frac{3}{2}\int t\,dt=\frac{3}{2}\cdot\frac{t^2}{2}+C=\frac{3}{4}(\log u)^2+C. \]
Step 4: Use the known value $f(e)=\tfrac74$.
At $u=e$, $\log e=1$, so $\dfrac34(1)+C=\dfrac74$, which gives $C=1$.
Step 5: Write the full function.
\[ f(u)=\frac{3}{4}(\log u)^2+1. \]
Step 6: Evaluate $f(e^2)$.
$\log e^2=2$, so \[ f(e^2)=\frac{3}{4}(2)^2+1=3+1=4. \] \[ \boxed{4} \]