JEE Main 20241 Feb, Morning shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002If 5f(x)+4f(1x)=x2−2,∀x≠05 f(x)+4 f\left(\frac{1}{x}\right)=x^2-2, \forall x \neq 05f(x)+4f(x1)=x2−2,∀x=0 and y=9x2f(x)y=9 x^2 f(x)y=9x2f(x), then yyy is strictly increasing in :AOption A. (0,15)∪(15,∞)\left(0, \frac{1}{\sqrt{5}}\right) \cup\left(\frac{1}{\sqrt{5}}, \infty\right)(0,51)∪(51,∞)BOption B. (−15,0)∪(15,∞)\left(-\frac{1}{\sqrt{5}}, 0\right) \cup\left(\frac{1}{\sqrt{5}}, \infty\right)(−51,0)∪(51,∞)COption C. (−15,0)∪(0,15)\left(-\frac{1}{\sqrt{5}}, 0\right) \cup\left(0, \frac{1}{\sqrt{5}}\right)(−51,0)∪(0,51)DOption D. (−∞,15)∪(0,15)\left(-\infty, \frac{1}{\sqrt{5}}\right) \cup\left(0, \frac{1}{\sqrt{5}}\right)(−∞,51)∪(0,51)CheckSkip · show solution
