Question

Find the domain of \[ p(x)=\sin^{-1}(1-2x^2). \] Hence, find the value(s) of \(x\) for which \[ p(x)=\frac{\pi}{6}. \] Also, write the range of \[ 2p(x)+\frac{\pi}{2}. \]

Hint:

When solving equations involving \(x^2\), never forget to include the negative square root. Dropping the \(\pm\) sign will cause you to miss half of the valid solutions!

Answer 1

Concept:
• Domain of \(\sin^{-1}(\theta)\): The principal domain for the inverse sine function requires its argument to lie within the closed interval \([-1,1]\). Therefore, for \(\sin^{-1}(f(x))\) to be defined, we must satisfy: \[ -1 \le f(x) \le 1 \] 
• Principal Range of \(\sin^{-1}(\theta)\): The output values of the inverse sine function are restricted to the principal interval: \[ -\frac{\pi}{2}\le \sin^{-1}(\theta)\le \frac{\pi}{2} \] Using linear transformations, we can modify this inequality to determine the range of expressions such as \[ 2p(x)+\frac{\pi}{2}. \] 

Step 1: Determine the analytical Domain of \(p(x)\).
The given function is: \[ p(x)=\sin^{-1}(1-2x^2) \] For this function to be defined over the real numbers, its argument must satisfy \[ -1\le 1-2x^2\le 1. \] Let us solve the two inequalities separately. 
• Right inequality: \[ 1-2x^2\le 1 \] \[ -2x^2\le0 \] \[ x^2\ge0. \] Since \(x^2\ge0\) for every real number, \[ x\in\mathbb{R}. \] 
• Left inequality: \[ -1\le1-2x^2 \] \[ 2x^2\le2 \] \[ x^2\le1 \] \[ |x|\le1 \] \[ -1\le x\le1. \] Combining both conditions, \[ \boxed{\text{Domain}=[-1,1]}. \] 

Step 2: Solve for \(x\) when \(p(x)=\dfrac{\pi}{6}\).
Given, \[ \sin^{-1}(1-2x^2)=\frac{\pi}{6}. \] Taking sine on both sides, \[ 1-2x^2=\sin\left(\frac{\pi}{6}\right). \] Since \[ \sin\left(\frac{\pi}{6}\right)=\frac12, \] we obtain \[ 1-2x^2=\frac12. \] Hence, \[ 2x^2=\frac12 \] \[ x^2=\frac14. \] Therefore, \[ x=\pm\frac12. \] Both values lie in the domain \([-1,1]\). 

Step 3: Determine the analytical Range of \(2p(x)+\dfrac{\pi}{2}\).
Since \[ x\in[-1,1], \] we have \[ 0\le x^2\le1. \] Multiplying by \(-2\), \[ -2\le-2x^2\le0. \] Adding \(1\), \[ -1\le1-2x^2\le1. \] Hence, \[ -\frac{\pi}{2}\le p(x)\le\frac{\pi}{2}. \] Multiplying by \(2\), \[ -\pi\le2p(x)\le\pi. \] Adding \(\frac{\pi}{2}\), \[ -\pi+\frac{\pi}{2} \le 2p(x)+\frac{\pi}{2} \le \pi+\frac{\pi}{2}. \] Therefore, \[ -\frac{\pi}{2} \le 2p(x)+\frac{\pi}{2} \le \frac{3\pi}{2}. \] Hence, \[ \boxed{\text{Range}=\left[-\frac{\pi}{2},\,\frac{3\pi}{2}\right].} \]

Note: While the mathematical range is \(\left[-\frac{\pi}{2},\frac{3\pi}{2}\right]\), choose the option that matches the question's intended answer if the options differ.