Question

Find the value of $f\left(x\right)=x^2+\frac{1}{x^2}$ at $x=1,2$.

• A. $f\left(1\right)=2,f\left(2\right)=\frac{17}{4}$
• B. $f\left(2\right)=2,f\left(1\right)=\frac{17}{4}$
• C. $f\left(1\right)=-2,f\left(2\right)=\frac{17}{4}$
• D. $f\left(2\right)=2,f\left(1\right)=-\frac{17}{4}$

Answer 1

The correct option is A $f\left(1\right)=2,f\left(2\right)=\frac{17}{4}$

Given: $f\left(x\right)=x^2+\frac{1}{x^2}$

At $x=1;$

$f\left(1\right)=1^2+\frac{1}{1^2}$

$\Rightarrow f\left(1\right)=2$

At $x=2;$

$f\left(2\right)=2^2+\frac{1}{2^2}$

$\Rightarrow f\left(2\right)=\frac{17}{4}$