Jee Main 2024 01 02 2024 Shift 1 - Question8
Question 8
$5 f(x)+4 f\left(\frac{1}{x}\right)=x^{2}-4$ and $y=9 f(x) \cdot x^{2}$. If $y$ is strictly increasing function, find interval of $x$.
(1) $\left(-\infty, \frac{-1}{\sqrt{5}}\right] \cup\left(\frac{-1}{\sqrt{5}}, 0\right)$
(2) $\left(\frac{-1}{\sqrt{5}}, 0\right) \cup\left(0, \frac{1}{\sqrt{5}}\right)$
(3) $\left(0, \frac{1}{\sqrt{5}}\right) \cup\left(\frac{1}{\sqrt{5}}, \infty\right)$
(4) $\left(-\sqrt{\frac{2}{5}}, 0\right) \cup\left(\sqrt{\frac{2}{5}}, \infty\right)$
Show Answer
Answer: (4)
Solution:
$5 f(x)+4 f\left(\frac{1}{x}\right)=x^{2}-4$
Replace $x$ by $\frac{1}{x}$
$5 f\left(\frac{1}{x}\right)+4 f(x)=\frac{1}{x^{2}}-4$
$5 \times$ equation (1) $-4 \times$ equation (2) $9 f(x)=5 x^{2}-\frac{4}{x^{2}}-4$
$y=9 f(x) \cdot x^{2}=\frac{5 x^{4}-4-4 x^{2}}{x^{2}} x^{2}$
$y=5 x^{4}-4-4 x^{2}$
$y^{\prime}=20 x^{3}-8 x>0$
$4 x\left(5 x^{2}-2\right)>0$
$x \in\left(-\sqrt{\frac{2}{5}}, 0\right) \cup\left(\sqrt{\frac{2}{5}}, \infty\right)$
BETA
