JEE MAIN - Mathematics (2024 - 1st February Morning Shift - No. 19)
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 :
$\left(0, \frac{1}{\sqrt{5}}\right) \cup\left(\frac{1}{\sqrt{5}}, \infty\right)$
$\left(-\frac{1}{\sqrt{5}}, 0\right) \cup\left(\frac{1}{\sqrt{5}}, \infty\right)$
$\left(-\frac{1}{\sqrt{5}}, 0\right) \cup\left(0, \frac{1}{\sqrt{5}}\right)$
$\left(-\infty, \frac{1}{\sqrt{5}}\right) \cup\left(0, \frac{1}{\sqrt{5}}\right)$
Explanation
$$
5 f(x)+4 f(1 / x)=x^2-2
$$ ........(1)
Replace $x$ by $1 / x$
$$ 5 f(1 / x)+4 f(x)=\frac{1}{x^2}-2 $$ ..........(2)
Multiply equation (1) by 5 and multiply equation (2) by 4 and then subtract equation (2) from (1)
$\begin{aligned} & 25 f(x)-16 f(x)=5 x^2-10-\frac{4}{x^2}+8 \\\\ & 9 f(x)=5 x^2-\frac{4}{x^2}-2 \\\\ & 9 f(x)=\frac{5 x^4-4-2 x^2}{x^2}\end{aligned}$
$\begin{aligned} & y=9 x^2 f(x) \\\\ & y=5 x^4-2 x^2-4 \\\\ & y^{\prime}=20 x^3-4 x \\\\ & \text { Put } y^{\prime}>0 \\\\ & 20 x^3-4 x>0 \\\\ & 5 x^3-x>0 \\\\ & x\left(5 x^2-1\right)>0\end{aligned}$
_1st_February_Morning_Shift_en_19_1.png)
$x \in\left(-\frac{1}{\sqrt{5}}, 0\right) \cup\left(\frac{1}{\sqrt{5}}, \infty\right)$
Replace $x$ by $1 / x$
$$ 5 f(1 / x)+4 f(x)=\frac{1}{x^2}-2 $$ ..........(2)
Multiply equation (1) by 5 and multiply equation (2) by 4 and then subtract equation (2) from (1)
$\begin{aligned} & 25 f(x)-16 f(x)=5 x^2-10-\frac{4}{x^2}+8 \\\\ & 9 f(x)=5 x^2-\frac{4}{x^2}-2 \\\\ & 9 f(x)=\frac{5 x^4-4-2 x^2}{x^2}\end{aligned}$
$\begin{aligned} & y=9 x^2 f(x) \\\\ & y=5 x^4-2 x^2-4 \\\\ & y^{\prime}=20 x^3-4 x \\\\ & \text { Put } y^{\prime}>0 \\\\ & 20 x^3-4 x>0 \\\\ & 5 x^3-x>0 \\\\ & x\left(5 x^2-1\right)>0\end{aligned}$
$x \in\left(-\frac{1}{\sqrt{5}}, 0\right) \cup\left(\frac{1}{\sqrt{5}}, \infty\right)$
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