CUET Preparation Today
Find the range of $\tan^{-1}(\frac{2x}{1+x^2})$ |
$[-\frac{π}{4},\frac{π}{4}]$ $[-\frac{π}{4},\frac{3π}{4}]$ $[\frac{π}{4},\frac{π}{6}]$ $[-\frac{π}{4},-\frac{π}{6}]$ |
$[-\frac{π}{4},\frac{π}{4}]$ |
First, we must get the range of $\frac{2x}{1+x^2}=y$ We have $yx^2 - 2x +y = 0$ Since x is real, D ≥ 0, i.e., $4 - 4y^2 ≥0$ or $-1≤ y ≤1$. So, $\tan^{-1}(y)∈[-\frac{π}{4},\frac{π}{4}]$ (As tan x is an increasing function) |
