CUET Preparation Today
Range of $f(x)=\sin^{-1}(\sqrt{x^2+x+1})$ is |
$\left(0,\frac{π}{2}\right]$ $\left(0,\frac{π}{3}\right]$ $\left[\frac{π}{3},\frac{π}{2}\right]$ $\left[\frac{π}{3},\frac{π}{3}\right]$ |
$\left[\frac{π}{3},\frac{π}{2}\right]$ |
$0 ≤ x^2 + 2×\frac{1}{2}x + 1 ≤ 1$ $⇒\frac{3}{4}≤(x+\frac{1}{2})^2+\frac{3}{4}≤ 1$ so $\frac{\sqrt 3}{2}≤\sqrt{x^2+x+1}≤ 1$ so $\frac{π}{3}≤\sin^{-1}\sqrt{x^2+x+1}≤\frac{π}{2}$ |
