Question

Let F be a real valued function of real and positive argument such that $F\left(x\right)+3xF\left(\frac{1}{x}\right)=2(x+1)\forall x>0$, then the value of F(10099) is

• A. 5010
• B. 5050
• C. $\frac{10099}{2}$
• D. 5000

Answer 1

The correct option is B

5050

$F\left(x\right)+3xF\left(\frac{1}{x}\right)=2(x+1)\dots \dots \left(1\right)$
Replacing x by $\frac{1}{x}$
$F\left(\frac{1}{x}\right)+\frac{3}{x}F\left(x\right)=2(\frac{1}{x}+1)xF\left(\frac{1}{x}\right)+3F\left(x\right)=1(1+x)\dots \dots \left(2\right)$
On solving (1) and (2)
$8F\left(x\right)=4(1+x)\Rightarrow F\left(x\right)=\frac{x+1}{2}F\left(10099\right)=\frac{10100}{2}=5050$