Question

If \(2 \mathrm{f}(x)+3 \mathrm{f}\left(\frac{1}{x}\right)=x^2+1, x \neq 0\) and \(y=5 x^2 \mathrm{f}(x)\), then \(y\) is strictly increasing in

• A. \((0,\frac{1}{2})\)
• B. \((\frac{-2}{5},0)\)
• C. \((\frac{1}{2},\frac{\sqrt{5}}{2})\)
• D. \((\frac{-1}{2},0)\)

Hint:

Use the given relation with x and 1/x to find f(x), then find where y' is positive.

Answer 1

The Correct Option is A

Step 1: Formula-first
Solving the pair of equations gives $f(x)=\frac{1}{5}\left(\frac{3}{x^2}+1-2x^2\right)$.

Step 2: Form y
$y=5x^2f(x)=3+x^2-2x^4$, so $\frac{dy}{dx}=2x(1-4x^2)$.

Step 3: Test the intervals
For $x\in(0,\frac12)$: $x>0$ and $1-4x^2>0$, so $y'>0$.
For $x\in(-\frac12,0)$: $x<0$ and $1-4x^2>0$, so $y'<0$.

Step 4: Conclusion
Hence the increasing interval from the options is $\left(0,\frac12\right)$.

Final Answer:
y increases strictly in $\left(0,\frac12\right)$. \[ \boxed{\text{(A) }\left(0,\dfrac{1}{2}\right)} \]