Question

$\text{Range of the function}$ $f\left(x\right)=x^2+\frac{1}{x^2+1},is$

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

$f\left(x\right)=x^2+1+\frac{1}{x^2+1}-1x^2+1+\frac{1}{x^2+1}\ge 2[\because AM\ge GM]x^2+\frac{1}{x^2+1}\ge 1\therefore f\left(x\right)ϵ[1,\infty )$