The number of real solution of `cot^(-1)sqrt(x(x+3))+sin^(-1)sqrt(x^(2)+3x+1)=(pi)/(2)` is /are

Options

A. 0
B. 1
C. 2
D. infinite

Answer

To solve the equation

\[ \cot^{-1}(\sqrt{x(x+3)}) + \sin^{-1}(\sqrt{x^2 + 3x + 1}) = \frac{\pi}{2}, \]

we can use the property of inverse trigonometric functions that states:

\[ \cot^{-1}(y) + \tan^{-1}(y) = \frac{\pi}{2}. \]

This implies that:

\[ \cot^{-1}(\sqrt{x(x+3)}) + \tan^{-1}(\sqrt{x^2 + 3x + 1}) = \frac{\pi}{2}. \]

From this, we can deduce that:

\[ \sqrt{x(x+3)} = \sqrt{x^2 + 3x + 1}. \]

Now, we will square both sides to eliminate the square roots:

\[ x(x + 3) = x^2 + 3x + 1. \]

Expanding the left side gives:

\[ x^2 + 3x = x^2 + 3x + 1. \]

Now, we can simplify this equation by subtracting \(x^2 + 3x\) from both sides:

\[ 0 = 1. \]

This is a contradiction, indicating that there are no values of \(x\) that satisfy the original equation. Therefore, the number of real solutions is:

\[ \boxed{0}. \]