Definite Integration
Integral Calculus-2
star_batch_jee_advanced_2025
Grade 12
Question:
Let $g(x) = x^c e^{2x}$ & let $f(x) = \int_0^x e^{2t}(3t^2+1)^{1/2} dt$. For a certain value of 'c', the limit of $\frac{f'(x)}{g'(x)}$ as $x \to \infty$ is finite and non-zero, then:
c = 1
\lim_{x \to \infty} \frac{f'(x)}{g'(x)} = \frac{\sqrt{3}}{2}
\lim_{x \to \infty} \frac{f'(x)}{g'(x)} = \frac{2}{\sqrt{3}}
None of these
Step-by-Step Solution
Key Concept: L'Hôpital's rule combined with asymptotic analysis determines the parameter $c$ by matching the limiting ratio.
Given $g'(x) = cx^{c-1}e^{2x} + x^c e^{2x} \cdot 2$ and $f'(x) = e^{2x}(3x^2+1)^{1/2}$. Computing $\frac{f'(x)}{g'(x)}$ and applying L'Hôpital's rule as $x \to \infty$: the limit equals $\frac{x(3+\frac{1}{x^2})^{1/2}}{x^c(\frac{c}{x}+2)}$. For this to be finite and equal $\frac{\sqrt{3}}{2}$, we need $c=1$.
Correct Answer: 1,2