Applications of Derivatives
Monotonicity — Solving Functional Equation
nta_pyq_2024_jan
Grade 12
Question:
If $5f(x)+4f\left(\dfrac{1}{x}\right)=x^2-2$, $\forall x\neq 0$ and $y=9x^2f(x)$, then $y$ is strictly increasing in:
$\left(0,\frac{1}{\sqrt{5}}\right)\cup\left(\frac{1}{\sqrt{5}},\infty\right)$
$\left(-\frac{1}{\sqrt{5}},0\right)\cup\left(\frac{1}{\sqrt{5}},\infty\right)$
$\left(-\frac{1}{\sqrt{5}},0\right)\cup\left(0,\frac{1}{\sqrt{5}}\right)$
$\left(-\infty,\frac{1}{\sqrt{5}}\right)\cup\left(0,\frac{1}{\sqrt{5}}\right)$
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 $f(x)$. Compute $y=9x^2f(x)$ and find $dy/dx>0$.
Substitute $x\to1/x$: $5f(1/x)+4f(x)=1/x^2-2$. Solving: $f(x)=\frac{5x^4-2x^2-4}{9x^2}$. Then $y=9x^2f(x)=5x^4-2x^2-4$. $dy/dx=20x^3-4x=4x(5x^2-1)>0\Rightarrow x\in\left(-\frac{1}{\sqrt5},0\right)\cup\left(\frac{1}{\sqrt5},\infty\right)$.
Correct Answer: 2