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QJEEM24Ma2485: JEE Main 2024 Mathematics - Class 12 JEEM24Ma2485 Solution: If and , then is strictly increasing in :

If 5f(x)+4f(1x)=x2−2,∀x≠05 f(x)+4 f\left(\frac{1}{x}\right)=x^2-2, \forall x \neq 0 and y=9x2f(x)y=9 x^2 f(x) , then yy is strictly increasing in :

Multiple Choice Options

  • A (0,15)∪(15,∞)\left(0, \frac{1}{\sqrt{5}}\right) \cup\left(\frac{1}{\sqrt{5}}, \infty\right)
  • B (−15,0)∪(15,∞)\left(-\frac{1}{\sqrt{5}}, 0\right) \cup\left(\frac{1}{\sqrt{5}}, \infty\right)
  • C (−15,0)∪(0,15)\left(-\frac{1}{\sqrt{5}}, 0\right) \cup\left(0, \frac{1}{\sqrt{5}}\right)
  • D (−∞,15)∪(0,15)\left(-\infty, \frac{1}{\sqrt{5}}\right) \cup\left(0, \frac{1}{\sqrt{5}}\right)

Step-by-Step Verified Solution

5f(x)+4f(1/x)=x2−25 f(x)+4 f(1 / x)=x^2-2 ........(1)

Replace xx by 1/x1 / x

5f(1/x)+4f(x)=1x2−25 f(1 / x)+4 f(x)=\frac{1}{x^2}-2 ..........(2)

Multiply equation (1) by 5 and multiply equation (2) by 4 and then subtract equation (2) from (1)

amp;25f(x)−16f(x)=5x2−10−4x2+8amp;9f(x)=5x2−4x2−2amp;9f(x)=5x4−4−2x2x2\begin{aligned} & 25 f(x)-16 f(x)=5 x^2-10-\frac{4}{x^2}+8 \\\\ & 9 f(x)=5 x^2-\frac{4}{x^2}-2 \\\\ & 9 f(x)=\frac{5 x^4-4-2 x^2}{x^2}\end{aligned}

amp;y=9x2f(x)amp;y=5x4−2x2−4amp;y′=20x3−4xamp; Put y′gt;0amp;20x3−4xgt;0amp;5x3−xgt;0amp;x(5x2−1)gt;0\begin{aligned} & y=9 x^2 f(x) \\\\ & y=5 x^4-2 x^2-4 \\\\ & y^{\prime}=20 x^3-4 x \\\\ & \text { Put } y^{\prime}>0 \\\\ & 20 x^3-4 x>0 \\\\ & 5 x^3-x>0 \\\\ & x\left(5 x^2-1\right)>0\end{aligned}

JEE Main 2024 (Online) 1st February Morning Shift Mathematics - Application of Derivatives Question 32 English Explanation

x∈(−15,0)∪(15,∞)x \in\left(-\frac{1}{\sqrt{5}}, 0\right) \cup\left(\frac{1}{\sqrt{5}}, \infty\right)

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