Question

$If3f\left(x\right)+5f\left(\frac{1}{x}\right)=\frac{1}{x}-3,$
$\forall x\ne 0ϵR,thenf\left(x\right)=$

• A. $\frac{1}{16}(\frac{3}{x}+5x-6)$
• B. $\frac{1}{16}(-\frac{3}{x}+5x-6)$
• C. $\frac{1}{14}(-\frac{3}{x}+5x-6)$
• D. $Noneofthese$

Answer 1

The correct option is B $\frac{1}{16}(-\frac{3}{x}+5x-6)$

Best way to solve this kind of problem is by checking options by putting different values of x.
For x = 1, given equation
$3f\left(x\right)+5f\left(\frac{1}{x}\right)=\frac{1}{x}-3,givesf\left(1\right)=\frac{-1}{4}$
Put x = 1 in different options and see which option gives $f\left(1\right)=\frac{-1}{4}$
Option (b) is equal to $\frac{-1}{4}$ for x = 1, hence is the correct answer.

Conventional Approach:
We have, $3f\left(x\right)+5f\left(\frac{1}{x}\right)=\frac{1}{x}-3,\forall x(\ne 0)ϵR$ ....(1)
[Replacing x by $\frac{1}{x}$]
$\Rightarrow 3f\left(\frac{1}{x}\right)+5f\left(x\right)=x-3$ ...(2)
Multiplying (1) by 3 and (2) by 5 subtracting, we get
$\Rightarrow f\left(x\right)=\frac{1}{16}(-\frac{3}{x}+5x-6),\forall x(\ne 0ϵR.)$