Question

Suppose $f$ and $g$ are differentiable functions on $(0,\infty )$ such that $f\text{'}\left(x\right)=-\frac{g\left(x\right)}{x}$ and $g\text{'}\left(x\right)=-\frac{f\left(x\right)}{x}$, for all $x>0$. Further, $f\left(1\right)=3$ and $g\left(1\right)=-1$.

$f\left(10\right)$ is equal to

• A. $\frac{101}{20}$
• B. $10$
• C. $20$
• D. $\frac{201}{20}$

Answer 1

The correct option is D $\frac{201}{20}$

We know that,
$f\left(x\right)+g\left(x\right)=\frac{A}{x}\cdots \left(1\right)f\left(x\right)-g\left(x\right)=Bx\cdots \left(2\right)$
Adding $\left(1\right)$ and $\left(2\right)$
$\Rightarrow 2f\left(x\right)=\frac{A}{x}+Bx\Rightarrow f\left(x\right)=\frac{p}{x}+qx(\text{Here}p=\frac{A}{2},q=\frac{B}{2})\Rightarrow f\left(1\right)=p+q=3\cdots \left(3\right)$

Subtracting $\left(2\right)$ from $\left(1\right)$,
$2g\left(x\right)=\frac{A}{x}-Bx\Rightarrow g\left(x\right)=\frac{p}{x}-qx\Rightarrow g\left(1\right)=p-q=-1\cdots \left(4\right)$

Using $\left(3\right)$ and $\left(4\right)$,
$2p=2\Rightarrow p=1\Rightarrow q=2\Rightarrow f\left(x\right)=\frac{1}{x}+2x;g\left(x\right)=\frac{1}{x}-2x\Rightarrow f\left(10\right)=\frac{1}{10}+20=\frac{201}{20}$