Let $f:R-\{0\} o R$ be a function such that $f(x)-6f\left(\frac1x ight)=\frac{35}{3x}-\frac52$. If the $\lim_{x o0}\left(\frac1{\alpha x}+f(x) ight)=\beta;\ \alpha,\beta\in R$, then $\alpha+2\beta$ is equal to

Question

Let $f:R-\{0\}	o R$ be a function such that $f(x)-6f\left(\frac1xight)=\frac{35}{3x}-\frac52$. If the $\lim_{x	o0}\left(\frac1{\alpha x}+f(x)ight)=\beta;\ \alpha,\beta\in R$, then $\alpha+2\beta$ is equal to

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] Replacing $x	o\frac1x$: $f\left(\frac1xight)-6f(x)=\frac{35x}3-\frac52$. Eliminating $f(1/x)$ gives $-35f(x)=70x+\frac{35}{3x}-\frac{35}2$, so $f(x)=-2x-\frac1{3x}+\frac12$. For the limit to exist, $\frac1\alpha=\frac13$, so $\alpha=3$ and $\beta=\frac12$. $\alpha+2\beta=4$. Answer (3).

Q. Let $f:R-\{0\}\to R$ be a function such that $f(x)-6f\left(\frac1x\right)=\frac{35}{3x}-\frac52$. If the $\lim_{x\to0}\left(\frac1{\alpha x}+f(x)\right)=\beta;\ \alpha,\beta\in R$, then $\alpha+2\beta$ is equal to

Explanation

Reviewed by Dr. Abubacker

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