Application Of Derivatives Question 1
Question 1 - 2024 (01 Feb Shift 1)
If $5 f(x)+4 f\left(\frac{1}{x}\right)=x^{2}-2, \forall x \neq 0$ and $y=9 x^{2} f(x)$, then $y$ is strictly increasing in :
(1) $\left(0, \frac{1}{\sqrt{5}}\right) \cup\left(\frac{1}{\sqrt{5}}, \infty\right)$
(2) $\left(-\frac{1}{\sqrt{5}}, 0\right) \cup\left(\frac{1}{\sqrt{5}}, \infty\right)$
(3) $\left(-\frac{1}{\sqrt{5}}, 0\right) \cup\left(0, \frac{1}{\sqrt{5}}\right)$
(4) $\left(-\infty, \frac{1}{\sqrt{5}}\right) \cup\left(0, \frac{1}{\sqrt{5}}\right)$
Show Answer
Answer (2)
Solution
$5f(x)+4 f\left(\frac{1}{x}\right)=x^{2}-2, \forall x \neq 0$
Substitute $x \rightarrow \frac{1}{x}$
$5 f\left(\frac{1}{x}\right)+4 f(x)=\frac{1}{x^{2}}-2$
On solving (1) and (2)
$f(x)=\frac{5 x^{4}-2 x^{2}-4}{9 x^{2}}$
$y=9 x^{2} f(x)$
$y=5 x^{4}-2 x^{2}-4$.
$\frac{d y}{d x}=20 x^{3}-4 x$
for strictly increasing
$\frac{dy}{dx}>0$
$4 x\left(5 x^{2}-1\right)>0$
$x \in\left(-\frac{1}{\sqrt{5}}, 0\right) \cup\left(\frac{1}{\sqrt{5}}, \infty\right)$
