Sets, Relations & Functions
Functional Equation / Limits
nta_pyq_2025_apr
Grade 11
Question:
Let $f : \mathbb{R} - \{0\} \to \mathbb{R}$ be a function such that $f(x) - 6f\!\left(\frac{1}{x}\right) = \frac{35}{3x} - \frac{5}{2}$. If $\lim_{x \to 0}\left(\frac{1}{\alpha x} + f(x)\right) = \beta$; $\alpha, \beta \in \mathbb{R}$, then $\alpha + 2\beta$ is equal to:
Step-by-Step Solution
Key Concept: Solve the functional equation by substituting $x\to1/x$ to get a second equation, then solve the system for $f(x)$. Analyse the limit to find $\alpha$.
System: $f(x)-6f(1/x)=35/(3x)-5/2$ and $6(f(1/x)-6f(x))=210x/3-30/2$. Solving: $f(x)=-\frac{1}{3x}-2x+\frac{1}{2}$. $\lim_{x\to0}\left(\frac{1}{\alpha x}-\frac{1}{3x}-2x+\frac{1}{2}\right)=\beta$. For finite limit: $\alpha=3$, giving $\beta=1/2$. $\alpha+2\beta=3+1=4$.
Correct Answer: 4