Range of $f(x)=\sin^{-1}(\sqrt{x^2+x+1})

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Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Relations and Functions

Question:

Range of $f(x)=\sin^{-1}(\sqrt{x^2+x+1})$ is

Options:

$\left(0,\frac{π}{2}\right]$

$\left(0,\frac{π}{3}\right]$

$\left[\frac{π}{3},\frac{π}{2}\right]$

$\left[\frac{π}{3},\frac{π}{3}\right]$

Correct Answer:

$\left[\frac{π}{3},\frac{π}{2}\right]$

Explanation:

$0 ≤ x^2 + 2×\frac{1}{2}x + 1 ≤ 1$

$⇒\frac{3}{4}≤(x+\frac{1}{2})^2+\frac{3}{4}≤ 1$

so $\frac{\sqrt 3}{2}≤\sqrt{x^2+x+1}≤ 1$

so $\frac{π}{3}≤\sin^{-1}\sqrt{x^2+x+1}≤\frac{π}{2}$