Question

If \( n_1 \) and \( n_2 \) are the number of real valued solutions of \( x = |\sin^{-1} x| \) \(\text{and}\) \( x = \sin(x) \text{ respectively, then the value of} \, n_2 - n_1 \text{ is:}\)

• A. 1
• B. 0
• C. 2
• D. 3

Hint:

For solving transcendental equations like \( x = \sin(x) \), graphical or numerical methods are often the best approach to finding the number of solutions.

Answer 1

The Correct Option is A

We are given two equations: 1. \( x = |\sin^{-1} x| \) 2. \( x = \sin(x) \) 

Step 1: Analyze the first equation \( x = |\sin^{-1} x| \) We know that \( \sin^{-1} x \) (the inverse sine function) gives values in the range \( [-\frac{\pi}{2}, \frac{\pi}{2}] \) for \( x \in [-1, 1] \). For the equation \( x = |\sin^{-1} x| \), we examine the two cases: 

- Case 1: When \( x \geq 0 \), \( |\sin^{-1} x| = \sin^{-1} x \). Thus, \( x = \sin^{-1} x \). 

- Case 2: When \( x < 0 \), \( |\sin^{-1} x| = -\sin^{-1} x \). Thus, \( x = -\sin^{-1} x \). 

These two cases give us real solutions in the interval \( [-1, 1] \). Through solving, we find that there are 2 real solutions for \( x = |\sin^{-1} x| \). Hence, \( n_1 = 2 \). 

Step 2: Analyze the second equation \( x = \sin(x) \) Now, we solve \( x = \sin(x) \). This is a transcendental equation, and we solve it by plotting or numerically approximating. The solutions occur within the interval \( [-1, 1] \). 

We find 3 real solutions for this equation. Thus, \( n_2 = 3 \). 

Step 3: Calculate \( n_2 - n_1 \) Now, we calculate the difference: \[ n_2 - n_1 = 3 - 2 = 1 \] Thus, the correct answer is \( \boxed{(a) \, 1} \).