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

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Mathematics

Graphical Interpretation of Continuity

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Even

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Odd

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Both even and odd

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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)

for every

x

of its domain then

f (x)

is called odd function and if

f (- x) = f (x)

for every

x

of its domain, then

f (x)

is called even function.

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

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

Q. If

f : R \to R

is an invertible function such that

f (x)

and

f^{- 1} x

are symmetric about the line

y = - x

, then

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.
(iii) A curve

f (x, y) = 0

is symmetrical about the line

y = x

f (x, y) = f (y, x)

.
(iv) For the curve

f (x, y) = 0

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

then it is symmetrical about the origin.

Join BYJU'S Learning Program

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