Question

Solutions of the equation \( \tan^{-1}\sqrt{x^2+x} + \sin^{-1}\sqrt{x^2+x+1} = \frac{\pi}{2} \) are

• A. 0, 1
• B. 1, -1
• C. 0, -1
• D. 0, -2

Hint:

When solving equations with inverse trigonometric functions and square roots, always start by determining the domain. Finding the set of valid \( x \) values can often drastically simplify the problem or even give you the solutions directly, as it did in this case.

Answer 1

The Correct Option is C

We are given the equation \( \tan^{-1}\sqrt{x^2+x} + \sin^{-1}\sqrt{x^2+x+1} = \frac{\pi}{2} \).
First, let's consider the domain of the functions. For the terms under the square root to be non-negative:
1) \( x^2+x \geq 0 \implies x(x+1) \geq 0 \). This is true for \( x \in (-\infty, -1] \cup [0, \infty) \).
2) \( x^2+x+1 \geq 0 \). The discriminant is \( 1^2 - 4(1)(1) = -3 < 0 \). Since the leading coefficient is positive, this quadratic is always positive for all real \( x \).
Also, for \( \sin^{-1}(u) \) to be defined, we need \( u \leq 1 \). So, \( \sqrt{x^2+x+1} \leq 1 \implies x^2+x+1 \leq 1 \implies x^2+x \leq 0 \). This is true for \( x \in [-1, 0] \).
The overall domain is the intersection of these conditions: \( ([-1, 0]) \cap ((-\infty, -1] \cup [0, \infty)) \). The only points in the intersection are \( x=0 \) and \( x=-1 \).
So, the only possible solutions are 0 and -1. Let's check them.
Check x = 0: LHS = \( \tan^{-1}\sqrt{0} + \sin^{-1}\sqrt{1} = \tan^{-1}(0) + \sin^{-1}(1) = 0 + \frac{\pi}{2} = \frac{\pi}{2} \). So, \( x=0 \) is a solution.
Check x = -1:
LHS = \( \tan^{-1}\sqrt{(-1)^2 + (-1)} + \sin^{-1}\sqrt{(-1)^2 + (-1) + 1} = \tan^{-1}\sqrt{1-1} + \sin^{-1}\sqrt{1-1+1} \)
LHS = \( \tan^{-1}\sqrt{0} + \sin^{-1}\sqrt{1} = 0 + \frac{\pi}{2} = \frac{\pi}{2} \). So, \( x=-1 \) is a solution.
Both 0 and -1 are solutions.