Sets, Relations & Functions
Functional Equation — Elimination Method
nta_pyq_2023_apr
Grade 11
Question:
Let $5f(x)+4f\!\left(\dfrac{1}{x}\right)=\dfrac{1}{x}+3$, $x>0$. Then $18\displaystyle\int_1^2 f(x)\,dx$ is equal to
$5\ln 2+3$
$10\ln 2+6$
$10\ln 2-6$
$5\ln 2-3$
Step-by-Step Solution
Key Concept: Replace $x\to\frac{1}{x}$ to get a second equation. Eliminate $f(\frac{1}{x})$ to find $9f(x)=\frac{5}{x}-4x+3$.
$18f(x)=\frac{10}{x}-8x+6$. $18\int_1^2=(10\ln2-16+12)-(0-4+6)=10\ln2-6$.
Correct Answer: 3