Question:
The sum of two odd functions
The sum of two odd functions
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The sum of two odd functions preserves the symmetry: \( f(-x) + g(-x) = -[f(x) + g(x)] \)
Updated On: Aug 6, 2025
- is always an even function
- is always an odd function
- is sometimes odd and sometimes even
- may be neither odd nor even
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The Correct Option is B
Solution and Explanation
Let \( f(x) \) and \( g(x) \) be two odd functions. By definition:
\[
f(-x) = -f(x), \quad g(-x) = -g(x)
\]
Now consider \( h(x) = f(x) + g(x) \).
Then: \[ h(-x) = f(-x) + g(-x) = -f(x) - g(x) = -[f(x) + g(x)] = -h(x) \] Therefore, \( h(x) \) is also odd.
So, the sum of two odd functions is always an \boxed{\text{odd function}}.
Then: \[ h(-x) = f(-x) + g(-x) = -f(x) - g(x) = -[f(x) + g(x)] = -h(x) \] Therefore, \( h(x) \) is also odd.
So, the sum of two odd functions is always an \boxed{\text{odd function}}.
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Top Questions on Relations
- A classroom teacher is keen to assess the learning of her students the concept of "relations" taught to them. She writes the following five relations each defined on the set \( A = \{1, 2, 3\} \): \[ R_1 = \{(2, 3), (3, 2)\}, \quad R_2 = \{(1, 2), (1, 3), (3, 2)\}, \quad R_3 = \{(1, 2), (2, 1), (1, 1)\}, \] \[ R_4 = \{(1, 1), (1, 2), (3, 3), (2, 2)\}, \quad R_5 = \{(1, 1), (1, 2), (3, 3), (2, 2), (2, 1), (2, 3), (3, 2)\}. \] The students are asked to answer the following questions about the above relations:
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