MCQ
if $f \left(x+\frac{1}{x}\right)=x^2+\frac{1}{x^2}$ then $f ( x )=$ ?
  • $\left(x^2-2\right)$
  • B
    $\left(x^2+1\right)$
  • C
    $\left(x^2-1\right)$
  • D
    $x^2$

Answer

Correct option: A.
$\left(x^2-2\right)$
$f\left(x+\frac{1}{x}\right)=x^2+\frac{1}{x^2}$
$=\left(x+\frac{1}{x}\right)^2-2$
$\text { Put, }\left(x+\frac{1}{x}\right)=t$
$\Rightarrow f(t)=t^2-2$
$\therefore f(x)=x^2-2$

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