Question

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

Hint:

For inverse trigonometric functions, always restrict the argument to the interval \([-1,1]\). Then use algebraic inequalities to determine the domain.

Answer 1

Step 1: Find the domain of \(p(x) = \sin^{-1}(1 - 2x^2)\). 
For the function \( p(x) = \sin^{-1}(1 - 2x^2) \), the argument of the inverse sine function must lie within the domain of \([-1, 1]\). Therefore, we require: \[ -1 \leq 1 - 2x^2 \leq 1 \] 

Now, solve the inequalities: 1. For the inequality \(1 - 2x^2 \geq -1\): \[ 1 - 2x^2 \geq -1 \implies -2x^2 \geq -2 \implies x^2 \leq 1 \] Therefore, \( -1 \leq x \leq 1 \). 2. For the inequality \(1 - 2x^2 \leq 1\): \[ 1 - 2x^2 \leq 1 \implies -2x^2 \leq 0 \implies x^2 \geq 0 \] This is always true for all real values of \(x\). Therefore, the domain of \(p(x)\) is: \[ -1 \leq x \leq 1 \] 

Step 2: Find the value of \(x\) for which \(p(x) = \frac{\pi}{6}\). \br Given that \( p(x) = \sin^{-1}(1 - 2x^2) \), we are asked to find \(x\) when: \[ \sin^{-1}(1 - 2x^2) = \frac{\pi}{6} \] Taking the sine of both sides: \[ 1 - 2x^2 = \sin\left(\frac{\pi}{6}\right) \] \[ 1 - 2x^2 = \frac{1}{2} \] Solving for \(x\): \[ 2x^2 = 1 - \frac{1}{2} = \frac{1}{2} \] \[ x^2 = \frac{1}{4} \] \[ x = \pm \frac{1}{2} \] Since \(x\) lies in the domain \([-1, 1]\), both \(x = \frac{1}{2}\) and \(x = -\frac{1}{2}\) are valid solutions. 

Step 3: Find the range of \(2p(x) + \frac{\pi}{2}\). 
Since \( p(x) = \sin^{-1}(1 - 2x^2) \), the range of \(p(x)\) is determined by the range of the inverse sine function. The inverse sine function, \(\sin^{-1}(y)\), has a range of \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\). The argument of \(\sin^{-1}\), which is \(1 - 2x^2\), varies between \([-1, 1]\) for \(x \in [-1, 1]\). This ensures that the output of \(p(x)\) is also within the range \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\). Now, we are asked to find the range of \(2p(x) + \frac{\pi}{2}\). Since \(p(x) \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), we multiply by 2: \[ 2p(x) \in \left[-\pi, \pi\right] \] Adding \(\frac{\pi}{2}\) to both ends: \[ 2p(x) + \frac{\pi}{2} \in \left[-\pi + \frac{\pi}{2}, \pi + \frac{\pi}{2}\right] \] \[ 2p(x) + \frac{\pi}{2} \in \left[-\frac{\pi}{2}, \frac{3\pi}{2}\right] \] Therefore, the range of \(2p(x) + \frac{\pi}{2}\) is: \[ \left[-\frac{\pi}{2}, \frac{3\pi}{2}\right] \] 

Final Answer: 
The domain of \(p(x)\) is \([-1, 1]\), the values of \(x\) for which \(p(x) = \frac{\pi}{6}\) are \(x = \pm \frac{1}{2}\), and the range of \(2p(x) + \frac{\pi}{2}\) is \(\left[-\frac{\pi}{2}, \frac{3\pi}{2}\right]\).