Limits, Continuity & Differentiability
Functional Equation with Limit — f'(1) Recovery
nta_pyq_2024_apr
Grade 12
Question:
Let $f:(−\infty,\infty)−\{0\}\to\mathbb{R}$ be a differentiable function such that $f'(1)=\displaystyle\lim_{a\to\infty}a^2f\!\left(\frac{1}{a}\right)$. Then $\displaystyle\lim_{a\to\infty}\frac{a(a+1)}{2}\tan^{-1}\!\left(\frac{1}{a}\right)+a^2-2\log_e a$ is equal to
$\dfrac{3}{2}+\dfrac{\pi}{4}$
$\dfrac{3}{4}+\dfrac{\pi}{8}$
$\dfrac{3}{8}+\dfrac{\pi}{4}$
$\dfrac{5}{2}+\dfrac{\pi}{8}$
Step-by-Step Solution
Key Concept: From the limit condition, identify $f(x)$ by writing the limit as $\lim_{a\to\infty}a^2\cdot\frac{1}{2}(1+\frac{1}{a})\tan^{-1}(\frac{1}{a})+1-\frac{2}{a^2}\ln a$, leading to $f(x)=\frac{1}{2}(1+x)\tan^{-1}(x)+1-2x^2\ln x$.
$f'(1)=\frac{5}{2}+\frac{\pi}{8}$.
Correct Answer: 4