MCQ
If $f (x)=x+\frac{1}{x}$, such that $[ f (x)]^3= f \left(x^3\right)+\lambda f \left(\frac{1}{x}\right)$, then $\lambda=$
  • A
    1
  • 3
  • C
    -3
  • D
    -1

Answer

Correct option: B.
3
(B)
$f (x)=x+\frac{1}{x} \Rightarrow f \left(x^3\right)=x^3+\frac{1}{x^3}$
$\therefore[ f (x)]^3=\left(x+\frac{1}{x}\right)^3=\left(x^3+\frac{1}{x^3}\right)+3\left(x+\frac{1}{x}\right)$
$\therefore[ f (x)]^3= f \left(x^3\right)+3 f (x)$
$\therefore[ f (x)]^3= f \left(x^5\right)+3 f \left(\frac{1}{x}\right) \Rightarrow \lambda=3$

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