Let f:R-0R be a function such that f(x)-6f(1x)=353x-52.

Question

Let $f:\mathbb{R}-\{0\}	o\mathbb{R}$ be a function such that $f(x)-6f\left(\dfrac1xight)=\dfrac{35}{3x}-\dfrac52$. If $\displaystyle\lim_{x	o0}\left(\frac1{\alpha x}+f(x)ight)=\beta$, $\alpha,\beta\in\mathbb{R}$, then $\alpha+2\beta$ is equal to

Accepted Answer

Replace $x\to\dfrac1x$ in the given relation:
$$f\left(\frac1x\right)-6f(x)=\frac{35x}{3}-\frac52 \quad\text{...(2)}$$
together with the original $f(x)-6f\left(\dfrac1x\right)=\dfrac{35}{3x}-\dfrac52$ ...(1).

From (2), $f(1/x)=6f(x)+\dfrac{35x}{3}-\dfrac52$. Substituting into (1):
$$f(x)-6\left(6f(x)+\frac{35x}{3}-\frac52\right)=\frac{35}{3x}-\frac52 \implies -35f(x)=\frac{35}{3x}+70x-\frac{35}{2}$$
$$f(x)=-2x-\frac{1}{3x}+\frac12$$

For $\displaystyle\lim_{x\to0}\left(\frac1{\alpha x}+f(x)\right)$ to exist, the $\dfrac1x$ singularities must cancel: $\dfrac1\alpha=\dfrac13\implies\alpha=3$. Then
$$\beta=\lim_{x\to0}\left(-2x+\frac12\right)=\frac12$$

$$\alpha+2\beta=3+1=4$$

Solutions · 1