Question

Range of the function \(y=\sin^{-1}\left(\dfrac{x^2}{1+x^2}\right)\), is

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

Hint:

If \(u=\frac{x^2}{1+x^2}\), then \(u\in[0,1)\). Applying \(\sin^{-1}\) gives range \([0,\pi/2)\).

Answer 1

The Correct Option is B

Step 1: Analyze inner expression.
\[ u=\frac{x^2}{1+x^2} \] Since \(x^2\ge 0\), we have: 
\[ 0\le \frac{x^2}{1+x^2} <1 \] because denominator is always larger than numerator unless \(x\to\infty\). 
So range of \(u\) is: 
\[ u\in[0,1) \] 

Step 2: Apply \(\sin^{-1}\) to this interval. 
\[ y=\sin^{-1}(u) \] Since \(\sin^{-1}\) is increasing on \([0,1]\): 
\[ u\in[0,1)\Rightarrow y\in\left[\sin^{-1}(0),\sin^{-1}(1)\right) \] \[ y\in\left[0,\frac{\pi}{2}\right) \]

 Final Answer: \[ \boxed{\left[0,\frac{\pi}{2}\right)} \]