If 2f(x) + 3fleft(frac{1}{x}right) = x^2 + 1, | vedclass

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(A) Given $2f(x) + 3f\left(\frac{1}{x}\right) = x^2 + 1$ (Equation $1$).Replace $x$ with $\frac{1}{x}$: $2f\left(\frac{1}{x}\right) + 3f(x) = \frac{1}

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$\left(0, \frac{1}{2}\right)$

Vedclass · Answer

(A) Given $2f(x) + 3f\left(\frac{1}{x}\right) = x^2 + 1$ (Equation $1$).Replace $x$ with $\frac{1}{x}$: $2f\left(\frac{1}{x}\right) + 3f(x) = \frac{1}{x^2} + 1$ (Equation $2$).Multiply Equation $1$ by $2$ and Equation $2$ by $3$: $4f(x) + 6f\left(\frac{1}{x}\right) = 2x^2 + 2$$9f(x) + 6f\left(\frac{1}{x}\right) = \frac{3}{x^2} + 3$Subtracting the first from the second: $5f(x) = \frac{3}{x^2} + 3 - 2x^2 - 2 = \frac{3}{x^2} - 2x^2 + 1$.Then $y = 5x^2 f(x) = x^2 \left(\frac{3}{x^2} - 2x^2 + 1\right) = 3 - 2x^4 + x^2$.To find where $y$ is strictly increasing,calculate $\frac{dy}{dx} = -8x^3 + 2x = 2x(1 - 4x^2) = 2x(1 - 2x)(1 + 2x)$.Set $\frac{dy}{dx} > 0$: $2x(1 - 2x)(1 + 2x) > 0$.Critical points are $x = 0, \frac{1}{2}, -\frac{1}{2}$.Testing intervals: $(-\infty, -1/2) \implies (-)(-)(-)  0$; $(0, 1/2) \implies (+)(+)(+) > 0$; $(1/2, \infty) \implies (+)(-)(+) Thus,$y$ is strictly increasing in $(-\infty, -1/2) \cup (0, 1/2)$.Comparing with options,the interval $(0, 1/2)$ is correct.

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

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