Question

If a function f(x) is such that $f(x+\frac{1}{x})=x^2+\frac{1}{x^2}$; then (fof) $\left(\sqrt{11}\right)$ is equal to

• A. 9
• B. 81
• C. 79
• D. $\sqrt{11}$

Answer 1

The correct option is C 79

$f(x+\frac{1}{x})=x^2+\frac{1}{x^2}={(x+\frac{1}{x})}^2-2$

Now replace $x+\frac{1}{x}$ by $x$

$\Rightarrow f\left(x\right)=x^2-2$

$\Rightarrow fof\left(x\right)=(x^2-2{)}^2-2$

$\therefore fof\left(\sqrt{11}\right)=9^2-2=79$

$fof\left(\sqrt{11}\right)=79$