Consider the relations $R_1$ and $R_2$ defined as \[a R_1 b \iff a^2 + b^2 = 1 \quad \text{for all } a, b \in \mathbb{R},\]and \[(a, b) R_2 (c, d) \iff a + d = b + c \quad \text{for all } (a, b), (c, d) \in \mathbb{N} \times \mathbb{N}.\]Then:

Question

The maximum value of the function $ f(x) = -|x+1| + 3, \; x \in \mathbb{R} $ is _____

Answer 1

To determine if $ R_1 $ and $ R_2 $ are equivalence relations, we will verify reflexivity, symmetry, and transitivity.

Reflexivity: For $ a R_1 a $, we require $ a^2 + a^2 = 1 $, which simplifies to $ 2a^2 = 1 $. This condition is not met for all $ a \in \mathbb{R} $, thus $ R_1 $ is not reflexive.
Symmetry: Assume $ a R_1 b $, meaning $ a^2 + b^2 = 1 $. Since $ a^2 + b^2 = 1 $ implies $ b^2 + a^2 = 1 $, it follows that $ b R_1 a $. Therefore, $ R_1 $ is symmetric.
Transitivity: If $ a R_1 b $ and $ b R_1 c $, then $ a^2 + b^2 = 1 $ and $ b^2 + c^2 = 1 $. Summing these equations yields $ a^2 + 2b^2 + c^2 = 2 $. This does not necessarily imply $ a^2 + c^2 = 1 $ unless $ b^2 = 0 $, which is not universally true for all $ a, b, c $. Consequently, $ R_1 $ is not transitive.

As $ R_1 $ fails reflexivity and transitivity, it is not an equivalence relation.

Reflexivity: For $ (a, b) R_2 (a, b) $, we need $ a + b = b + a $. This equality holds for all $ (a, b) \in \mathbb{N} \times \mathbb{N} $, confirming $ R_2 $ is reflexive.
Symmetry: If $ (a, b) R_2 (c, d) $, then $ a + d = b + c $. Rearranging this gives $ c + b = d + a $, which means $ (c, d) R_2 (a, b) $. Thus, $ R_2 $ is symmetric.
Transitivity: Let $ (a, b) R_2 (c, d) $ and $ (c, d) R_2 (e, f) $. This implies $ a + d = b + c $ and $ c + f = d + e $. Adding these two equations results in $ a + f = b + e $. This shows that $ (a, b) R_2 (e, f) $, proving $ R_2 $ is transitive.

Since $ R_2 $ exhibits reflexivity, symmetry, and transitivity, it is an equivalence relation.

Based on these findings, the correct conclusion is: Only $ R_2 $ is an equivalence relation.

The Correct Option is B