The range of the real value function f(x)= ^{-1} ≤ft( {x²+x+1} ) is

Correct Answer is : [π3,π2]≤ft[π/3, π/2][3π,2π]

Correct Answer is : [π3,π2]≤ft[π/3, π/2][3π,2π]

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TG EAMCET 2 May 2025 Shift 1 Paper

Question : 2 of 160

Marks: +1, -0

f(x)= \sin^{-1} \left( \sqrt{x^2+x+1} \right)

Solution:

Given,

f(x)= \sin^{-1} \left( \sqrt{x^2+x+1} \right)

The domain of

\sin^{-1} \left( \sqrt{x^2+x+1} \right)

[-1,1]

\therefore \quad -1 \leq \sqrt{x^2+x+1} \leq 1

Since

\sqrt{x^2+x+1}

is always,

non-negative, so

0 \leq \sqrt{x^2+x+1} \leq 1

Let

g(x)=x^2+x+1= \left(x+\frac{1}{2}\right)^2+\frac{3}{4}

The minimum value of

g(x)

\frac{3}{4}

, which occurs at

x=-\frac{1}{2}

\therefore \quad x^2+x+1 \geq \frac{3}{4}

\Rightarrow \quad \sqrt{x^2+x+1} \geq \sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}

\Rightarrow \quad \frac{\sqrt{3}}{2} \leq \sqrt{x^2+x+1} \leq 1

\Rightarrow \quad \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) \leq \sin^{-1}\left(\sqrt{x^2+x+1}\right)

\Rightarrow \quad \frac{\pi}{3} \leq f(x) \leq \frac{\pi}{2} \leq \sin^{-1}

Thus, the range of

f(x)= \sin^{-1} \left( \sqrt{x^2+x+1} \right)

\left[\frac{\pi}{3}, \frac{\pi}{2}\right]

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