Q41 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Mathematics

If $f(x)=x-\frac{1}{x}, x \neq 0$, then $3 f(x)=$

  • A
    $3[f(x)]^2-f\left(x^2\right)$
  • B
    $[f(x)]^2-f\left(x^3\right)$
  • Correct answer
    C
    $f\left(x^3\right)-[f(x)]^3$
  • D
    $f\left(x^3\right)-f\left(x^2\right)$

Solution

We have, $f(x)=x-\frac{1}{x} \mathrm{gl}$ $\therefore \quad(f(x))^3=x^3-\frac{1}{x^3}-3(x)\left(\frac{1}{x}\right)\left(x-\frac{1}{x}\right)$ $=f\left(x^3\right)-3\left(x-\frac{1}{x}\right)$ $\Rightarrow \quad(f(x))^3=f\left(x^3\right)-3 f(x)$ $\Rightarrow \quad 3 f(x)=f\left(x^3\right)-[f(x)]^3$

Hence, the correct answer is option: 'C'.

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