If f x is a function whose domain is symmetric about the origin- then f x - f - x isA- EvenB- One-oneC- OddD- Both even and odd

The correct option is A Even(a, b) g(x) = f(x) + f(–x) g(–x) = f(–x) + f(x) = g(x) therefore g(x) is even ...

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Standard XII

Mathematics

Domain

If f x is a f...

Question

If f(x) is a function whose domain is symmetric about the origin, then f(x) + f(–x) is

One-one

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Even

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Odd

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Both even and odd

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Solution

The correct option is B Even
(a, b)
g(x) = f(x) + f(–x)
g(–x) = f(–x) + f(x) = g(x)
therefore g(x) is even

Suggest Corrections

Similar questions

Q. Assertion :

f

is even,

g

is odd then

\frac{f}{g} (g \neq 0)

is an odd function. Reason: If

f (- x) = - f (x)

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f (x)

is called odd function and if

f (- x) = f (x)

for every

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Q. If

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is an invertible function such that

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are symmetric about the line

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If $f (x) = (h_{1} (x) - h_{1} (- x)) (h_{2} (x) - h_{2} (- x)) \dots (h_{2 n + 1} (x) - h_{2 n + 1} (- x)),$ where $h_{1} (x), h_{2} (x), \dots \dots h_{n} (x)$ are defined everywhere and $f (200) = 0$ , then $f (x)$ is

If f(x) is a function whose domain is symmetric about the origin, then f(x) + f(–x) is

Q. Identify the true statements in the following :
(i) If a curve is symmetrical about the origin, then it is symmetrical about both the axes.
(ii) If a curve is symmetrical about both the axes, then it is symmetrical about the origin.
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f (x, y) = 0

is symmetrical about the line

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.
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f (x, y) = f (- y, - x)

then it is symmetrical about the origin.

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If f(x) is a function whose domain is symmetric about the origin, then f(x) + f(–x) is

The correct option is B Even(a, b) g(x) = f(x) + f(–x) g(–x) = f(–x) + f(x) = g(x) therefore g(x) is even

The correct option is B Even
(a, b)
g(x) = f(x) + f(–x)
g(–x) = f(–x) + f(x) = g(x)
therefore g(x) is even