The number of real roots of the equation tan ⁻¹√x(x + 1) + sin ⁻¹√(x² + x + 1) = π/4 is :

Solution

Maths sir · 4 steps

Before any algebra, find where both inverse functions are defined. Here that is just two points.

1. Square root condition. For √x(x+1) to be real, you need x² + x ≥ 0.
2. Inverse sine condition. sin⁻¹ takes inputs only from [-1, 1], so √(x²+x+1) ≤ 1. Squaring gives x² + x + 1 ≤ 1, that is, x² + x ≤ 0.
3. Combine both. Both conditions together force x² + x = 0, so x = 0 or x = -1. These are the only points where the left side exists.

   tan⁻¹√0 + sin⁻¹√1 = 0 + π/2 = π/2

4. So. At both x = 0 and x = -1 the left side equals π/2, never π/4. The equation has 0 real roots, option (d).

The idea to keep. With inverse trig equations, write the domain conditions first. If they squeeze x down to a few points, check those points directly instead of doing algebra.

Answer from the NTA final answer key. Solution by PrepWiser.