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Which of the following relations are eve...

Which of the following relations are even?
I. y=2
II. `f(x)=x`
III. `x^(2)+y^(2)=1`

A

Only I

B

Only I and II

C

Only II and III

D

only I and III

Text Solution

AI Generated Solution

The correct Answer is:
To determine which of the given relations are even, we need to recall the definition of an even function. A function \( f(x) \) is considered even if it satisfies the condition: \[ f(-x) = f(x) \] Now, let's analyze each of the given relations one by one. ### Step 1: Analyze \( y = 2 \) This is a constant function. 1. Replace \( x \) with \( -x \): \[ y = 2 \] Since it does not depend on \( x \), we have: \[ f(-x) = 2 \quad \text{and} \quad f(x) = 2 \] Therefore, \( f(-x) = f(x) \). **Conclusion**: The relation \( y = 2 \) is even. ### Step 2: Analyze \( f(x) = x \) Now, let's check the function \( f(x) = x \). 1. Replace \( x \) with \( -x \): \[ f(-x) = -x \] We compare this with \( f(x) \): \[ f(x) = x \] Since \( f(-x) \neq f(x) \) (as \( -x \neq x \)), this function is not even. **Conclusion**: The relation \( f(x) = x \) is not even. ### Step 3: Analyze \( x^2 + y^2 = 1 \) This is the equation of a circle centered at the origin with a radius of 1. 1. Replace \( x \) with \( -x \): \[ (-x)^2 + y^2 = 1 \] This simplifies to: \[ x^2 + y^2 = 1 \] Since this is the same as the original equation, we find that: \[ f(-x) = f(x) \] **Conclusion**: The relation \( x^2 + y^2 = 1 \) is even. ### Final Conclusion From our analysis: - \( y = 2 \) is even. - \( f(x) = x \) is not even. - \( x^2 + y^2 = 1 \) is even. Thus, the even relations are \( I \) and \( III \). ### Answer: The correct option is "only one and three". ---
To determine which of the given relations are even, we need to recall the definition of an even function. A function \( f(x) \) is considered even if it satisfies the condition: \[ f(-x) = f(x) \] Now, let's analyze each of the given relations one by one. ...
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