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: Use the definition of inverse sine.
The function \(\sin^{-1}(t)\) is defined only when \[ -1 \le t \le 1 \] Here \[ t = 1 - 2x^2 \] Thus \[ -1 \le 1 - 2x^2 \le 1 \] Step 2: Solve the inequality.
First part: \[ -1 \le 1 - 2x^2 \] \[ -2 \le -2x^2 \] \[ 1 \ge x^2 \] \[ -1 \le x \le 1 \] Second part: \[ 1 - 2x^2 \le 1 \] \[ -2x^2 \le 0 \] This is always true for all real \(x\).
Therefore the domain is \[ -1 \le x \le 1 \] Step 3: Find \(x\) when \(p(x)=\frac{\pi{6}\).}
Given \[ \sin^{-1}(1-2x^2)=\frac{\pi}{6} \] Apply sine on both sides: \[ 1 - 2x^2 = \sin\frac{\pi}{6} \] \[ 1 - 2x^2 = \frac{1}{2} \] \[ 2x^2 = \frac{1}{2} \] \[ x^2 = \frac{1}{4} \] \[ x = \pm \frac{1}{2} \] Step 4: Determine the range of \(p(x)\).
Since \[ p(x)=\sin^{-1}(1-2x^2) \] and \[ 1-2x^2 \in [-1,1] \] the range of \(\sin^{-1}\) is \[ \left[-\frac{\pi}{2},\frac{\pi}{2}\right] \] However the maximum value of \(1-2x^2\) is \(1\) at \(x=0\), giving \[ p(x)=\frac{\pi}{2} \] The minimum occurs when \[ x=\pm1 \] \[ 1-2(1)=-1 \] \[ p(x)=-\frac{\pi}{2} \] Thus \[ -\frac{\pi}{2} \le p(x) \le \frac{\pi}{2} \] Step 5: Find the range of \(2p(x)+\frac{\pi{2}\).}
Multiply the inequality by 2: \[ -\pi \le 2p(x) \le \pi \] Add \(\frac{\pi}{2}\): \[ -\frac{\pi}{2} \le 2p(x)+\frac{\pi}{2} \le \frac{3\pi}{2} \] Thus the range is \[ \left[-\frac{\pi}{2},\frac{3\pi}{2}\right] \] Final Answer:
Domain: \[ [-1,1] \] Values of \(x\): \[ x=\pm\frac{1}{2} \] Range of \(2p(x)+\frac{\pi}{2}\): \[ \boxed{\left[-\frac{\pi}{2},\frac{3\pi}{2}\right]} \]