JEE MAIN - Mathematics Hindi (2024 - 1st February Morning Shift - No. 19)
यदि $5 f(x)+4 f\left(\frac{1}{x}\right)=x^2-2, \forall x \neq 0$ और $y=9 x^2 f(x)$, तब $y$ कठोरता से बढ़ता है :
$\left(0, \frac{1}{\sqrt{5}}\right) \cup\left(\frac{1}{\sqrt{5}}, \infty\right)$
$\left(-\frac{1}{\sqrt{5}}, 0\right) \cup\left(\frac{1}{\sqrt{5}}, \infty\right)$
$\left(-\frac{1}{\sqrt{5}}, 0\right) \cup\left(0, \frac{1}{\sqrt{5}}\right)$
$\left(-\infty, \frac{1}{\sqrt{5}}\right) \cup\left(0, \frac{1}{\sqrt{5}}\right)$
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