If and then is strictly increasing in, Differential Calculus jee mains questions | JEE Main Maths Quadratic Equations Previous Year Question-1-2024

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Solution

Q.
If $5 f (x) + 4 f (\frac{1}{x}) = x^{2} - 2, \forall x \neq 0$ and $y = 9 x^{2} f (x),$ then $y$ is strictly increasing in [2024]

1 $(- \infty, \frac{1}{\sqrt{5}}) \cup (0, \frac{1}{\sqrt{5}})$

2 $(- \frac{1}{\sqrt{5}}, 0) \cup (0, \frac{1}{\sqrt{5}})$

3 $(0, \frac{1}{\sqrt{5}}) \cup (\frac{1}{\sqrt{5}}, \infty)$

4 $(- \frac{1}{\sqrt{5}}, 0) \cup (\frac{1}{\sqrt{5}}, \infty)$

Ans.
(4)

$5 f (x) + 4 f (\frac{1}{x}) = x^{2} - 2$ ...(i)

Replace x by $\frac{1}{x}$ , we get:

$5 f (\frac{1}{x}) + 4 f (x) = \frac{1}{x^{2}} - 2$ ...(ii)

$5 \times (i)$ and $4 \times (i i)$ , we get

$\Rightarrow$ $f (x) = \frac{5}{9} x^{2} - \frac{4}{9 x^{2}} - \frac{2}{9}$ $∴$ $y = 9 x^{2} (\frac{5}{9} x^{2} - \frac{4}{9 x^{2}} - \frac{2}{9})$

$y = 5 x^{4} - 4 - 2 x^{2}$

As $y$ is strictly increasing when $y^{'} > 0,$ so

$y' = 20 x^{3} - 4 x > 0$

$\Rightarrow$ $4 x (5 x^{2} - 1) > 0$

$x \in (\frac{- 1}{\sqrt{5}}, 0) \cup (\frac{1}{\sqrt{5}}, \infty)$

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