Question

If $\sin^{-1}(x^2 - 4x + 5) + \cos^{-1}(y^2 - 2y + 2) = \pi / 2$, then:

• A. $x + y = 3$
• B. $xy = 2$
• C. $x - y = 1$
• D. $x - y = 0$

Answer 1

Answer: (A), (B), (C)

A, B, C

Answer 2

The given equation is $\sin^{-1}(x^2 - 4x + 5) + \cos^{-1}(y^2 - 2y + 2) = \pi / 2$.

For $\sin^{-1}(A)$ and $\cos^{-1}(B)$ to be defined, their arguments must be in the interval $[-1, 1]$. Let's analyze the arguments:

1. $x^2 - 4x + 5$: This expression can be rewritten by completing the square: $x^2 - 4x + 5 = (x^2 - 4x + 4) + 1 = (x-2)^2 + 1$. Since $(x-2)^2 \ge 0$ for all real $x$, we have $(x-2)^2 + 1 \ge 1$. For $\sin^{-1}((x-2)^2 + 1)$ to be defined, we must have $(x-2)^2 + 1 \in [-1, 1]$. The only value in $[-1, 1]$ that is also $\ge 1$ is $1$. Therefore, $(x-2)^2 + 1 = 1$, which implies $(x-2)^2 = 0$. This gives $x-2 = 0$, so $x=2$.

2. $y^2 - 2y + 2$: This expression can be rewritten by completing the square: $y^2 - 2y + 2 = (y^2 - 2y + 1) + 1 = (y-1)^2 + 1$. Since $(y-1)^2 \ge 0$ for all real $y$, we have $(y-1)^2 + 1 \ge 1$. For $\cos^{-1}((y-1)^2 + 1)$ to be defined, we must have $(y-1)^2 + 1 \in [-1, 1]$. The only value in $[-1, 1]$ that is also $\ge 1$ is $1$. Therefore, $(y-1)^2 + 1 = 1$, which implies $(y-1)^2 = 0$. This gives $y-1 = 0$, so $y=1$.

Thus, the only possible values are $x=2$ and $y=1$. Now we check the given options with $x=2$ and $y=1$: (A) $x + y = 2 + 1 = 3$. This statement is true. (B) $xy = 2 \times 1 = 2$. This statement is true. (C) $x - y = 2 - 1 = 1$. This statement is true. (D) $x - y = 0$. This statement is false, as $2-1=1 \neq 0$.

Since the question asks "then:", it implies which of the given statements are true. All options (A), (B), and (C) are true.