If 5 f ( x )+4 f (frac 1 x )= x 2 -2, forall x neq 0 and y =9 x 2 f ( x ) , then y is strictly increasing in :

(-frac 1 √(5) , 0 ) cup (frac 1 √(5) , ∞ )

(0, frac 1 √(5) ) cup (frac 1 √(5) , ∞ )

(-frac 1 √(5) , 0 ) cup (0, frac 1 √(5) )

(-∞, frac 1 √(5) ) cup (0, frac 1 √(5) )

Mathematics · Application of Derivatives

JEE Main 2024 — 1 February, Shift 1 — Question 19

If $5 \mathrm{f}(\mathrm{x})+4 \mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=\mathrm{x}^{2}-2, \forall \mathrm{x} \neq 0$ and $\mathrm{y}=9 \mathrm{x}^{2} \mathrm{f}(\mathrm{x})$ , then y is strictly increasing in :

Option A:
$\left(0, \frac{1}{\sqrt{5}}\right) \cup\left(\frac{1}{\sqrt{5}}, \infty\right)$
Option B:
$\left(-\frac{1}{\sqrt{5}}, 0\right) \cup\left(\frac{1}{\sqrt{5}}, \infty\right)$
Correct
Option C:
$\left(-\frac{1}{\sqrt{5}}, 0\right) \cup\left(0, \frac{1}{\sqrt{5}}\right)$
Option D:
$\left(-\infty, \frac{1}{\sqrt{5}}\right) \cup\left(0, \frac{1}{\sqrt{5}}\right)$

Answer: B

Step-by-step solution

$5\mathrm{f}(\mathrm{x})+4 \mathrm{f}\left(\frac{1}{x}\right)=\mathrm{x}^{2}-2, \forall x \neq 0 \ldots$ (1) 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 $\frac{d y}{d x}>0$

$4 \mathrm{x}\left(5 \mathrm{x}^{2}-1\right)>0$

$\mathrm{x} \in\left(-\frac{1}{\sqrt{5}}, 0\right) \cup\left(\frac{1}{\sqrt{5}}, \infty\right)$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 1 February, Shift 1
Subject: Mathematics
Chapter: Application of Derivatives
Topic: Monotonicity