Question

If \(sin^{-1}x=2\,tan^{-1}x\) , then the number of integral values of x is equal to:

• A. 0
• B. 1
• C. 2
• D. More than 2

Answer 1

The Correct Option is D

To solve the given equation \( \sin^{-1}x = 2\tan^{-1}x \), we need to find the integral values of \( x \).

1. Start by expressing \( y = \tan^{-1}x \). Thus, we have \( \sin^{-1}x = 2y \). Consequently, \( \sin(2y) = x \).
2. Using the double angle formula for sine, we have: \[ \sin(2y) = 2\sin(y)\cos(y) \]. Since \( y = \tan^{-1}x \), then: - \( \sin(y) = \frac{x}{\sqrt{1+x^2}} \) - \( \cos(y) = \frac{1}{\sqrt{1+x^2}} \) Thus, \[ \sin(2y) = 2 \cdot \frac{x}{\sqrt{1+x^2}} \cdot \frac{1}{\sqrt{1+x^2}} = \frac{2x}{1+x^2} \].
3. Equating this to \( x \), we get: \[ \frac{2x}{1+x^2} = x. \]
4. Clear the fraction by multiplying both sides by \( 1+x^2 \): \[ 2x = x(1+x^2). \]
5. Simplify and rearrange: \[ 2x = x + x^3 \].
6. Bringing all terms to one side gives: \[ x^3 - x = 0. \]
7. Factor out \( x \): \[ x(x^2 - 1) = 0. \]
8. Further factor the quadratic term: \[ x(x - 1)(x + 1) = 0. \]
9. This equation implies three potential values for \( x \): - \( x = 0 \) - \( x = 1 \) - \( x = -1 \)
10. Thus, there are 3 integral solutions for the equation.

Therefore, the number of integral values of \( x \) is more than 2.