Question

The range of the real valued function \( f(x)=\sin ^{-1}\left(\sqrt{x^{2}+x+1}\right) \) is

• A. \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)
• B. \( \left[0, \frac{\pi}{2}\right] \)
• C. \( \left[\frac{\pi}{6}, \frac{\pi}{2}\right] \)
• D. \( \left[\frac{\pi}{3}, \frac{\pi}{2}\right] \)

Hint:

Before finding the range of functions involving inverse trigonometric expressions, always verify that the argument satisfies the domain condition \( -1 \le \text{argument} \le 1 \). This restriction often determines the possible values of the function.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:

The given function involves an inverse sine function whose argument contains a square root of a quadratic expression. To determine the range, we must ensure that the argument of \( \sin^{-1} \) lies within its valid domain \( [-1,1] \), and then evaluate the possible outputs of the inverse sine function.
Step 2: Key Formula or Approach:

1. The domain of \( \sin^{-1}(u) \) requires \( u \in [-1,1] \). 
2. A quadratic expression \( ax^2+bx+c \) (with \( a>0 \)) attains its minimum at \( x=-\frac{b}{2a} \). 
3. The function \( \sin^{-1}u \) is strictly increasing with respect to \( u \). 
Step 3: Detailed Explanation:

Let \( u=\sqrt{x^2+x+1} \). 
For the function to exist, the argument of the inverse sine must satisfy \[ -1 \le u \le 1 \] Since \( u \) is a square root quantity, it is always non-negative. Therefore, \[ 0 \le \sqrt{x^2+x+1} \le 1 \] Squaring the inequality gives \[ 0 \le x^2+x+1 \le 1 \] Now consider the quadratic expression inside the root: \[ g(x)=x^2+x+1 \] Completing the square: \[ x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4} \] Hence, the minimum value of \( g(x) \) is \( \frac{3}{4} \), which occurs at \( x=-\frac{1}{2} \). 
Thus, the smallest value of \( u=\sqrt{x^2+x+1} \) is \[ \sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2} \] Combining this with the restriction required for \( \sin^{-1} \), we obtain \[ \frac{\sqrt{3}}{2} \le \sqrt{x^2+x+1} \le 1 \] Therefore, the argument of the inverse sine varies over the interval \( \left[\frac{\sqrt{3}}{2},1\right] \). 
Applying the inverse sine function to this interval, and using the fact that \( \sin^{-1} \) is increasing, we get \[ \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) \le f(x) \le \sin^{-1}(1) \] Using standard trigonometric values: \[ \frac{\pi}{3} \le f(x) \le \frac{\pi}{2} \] 
Step 4: Final Answer:

The range of the function is \( \left[\frac{\pi}{3},\frac{\pi}{2}\right] \).