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If 5f(x)+4f( 1 x)=x^2-2,forall x≠ 0 and y=9x^2f(x), then y is strictly increasing in:

JEE Main 2024 — Mathematics Calculus

2024mcqeasy

If 5f(x)+4f(1x)=x2−2,∀x≠05f(x)+4f(\frac{1}{x})={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:

Official previous-year question

Held on 1 Feb 2024 · Verified 6 Jul 2026.

Options

  1. A

    (0,15)∪(15,∞)(0,\frac{1}{\sqrt{5}})\cup (\frac{1}{\sqrt{5}},\infty )(0,5​1​)∪(5​1​,∞)

  2. B

    (−15,0)∪(15,∞)(-\frac{1}{\sqrt{5}},0)\cup (\frac{1}{\sqrt{5}},\infty )(−5​1​,0)∪(5​1​,∞)

  3. C

    (−15,0)∪(0,15)(-\frac{1}{\sqrt{5}},0)\cup (0,\frac{1}{\sqrt{5}})(−5​1​,0)∪(0,5​1​)

  4. D

    (−∞,15)∪(0,15)(-\infty ,\frac{1}{\sqrt{5}})\cup (0,\frac{1}{\sqrt{5}})(−∞,5​1​)∪(0,5​1​)

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