JEE Main 2024 — Mathematics Calculus

Given: $5f(x)+4f(\frac{1}{x})={x}^{2}-2...(i)$5f(x)+4f(x1​)=x2−2...(i)

Replacing $x$x by $\frac{1}{x}$x1​.

$$\Rightarrow 5f(\frac{1}{x})+4f(x)={(\frac{1}{x})}^{2}-2...(ii)$$⇒5f(x1​)+4f(x)=(x1​)2−2...(ii)

Using, $(i)\times 5-(ii)\times 4$(i)×5−(ii)×4

$$\Rightarrow 25f(x)+20f(\frac{1}{x})-16f(x)-20f(\frac{1}{x})=5{x}^{2}-10-\frac{4}{{x}^{2}}+8$$⇒25f(x)+20f(x1​)−16f(x)−20f(x1​)=5x2−10−x24​+8

$$\Rightarrow 9f(x)=5{x}^{2}-\frac{4}{{x}^{2}}-2$$⇒9f(x)=5x2−x24​−2

Now, $y=9f(x)\times {x}^{2}$y=9f(x)×x2

$$\Rightarrow y=(5{x}^{2}-2-\frac{4}{{x}^{2}})\times {x}^{2}$$⇒y=(5x2−2−x24​)×x2

$$\Rightarrow y=5{x}^{4}-2{x}^{2}-4$$⇒y=5x4−2x2−4

$$\Rightarrow {y}^{'}=20{x}^{3}-4x$$⇒y′=20x3−4x

$$\Rightarrow {y}^{'}=4x(5{x}^{2}-1)$$⇒y′=4x(5x2−1)

For function to be increasing ${y}^{'}>0$y′>0

$$\Rightarrow x\in (-\sqrt{\frac{1}{5}},0)\cup (\sqrt{\frac{1}{5}},\infty )$$⇒x∈(−51​​,0)∪(51​​,∞)