Question

If $f\left(x\right)$ is an odd function then-
(i)$\frac{f(-x)+f\left(x\right)}{2}$ is an even function
(ii)$[\mid f\left(x\right)\mid +1]$ is even where [.] denotes greatest integer function.
(iii)$\frac{f\left(x\right)-f(-x)}{2}$ is neither even nor odd
(iv)$f\left(x\right)+f(-x)$ is neither even nor odd
Which of these statements are correct

• A. (i) & (ii) only
• B. (i) & (iv) only
• C. (iii) & (iv) only
• D. all of these

Answer 1

Answer: (A)

(i) Let $g\left(x\right)=\frac{f\left(x\right)+f(-x)}{2}$

$=\frac{f\left(x\right)-f\left(x\right)}{2}$ [Since $f$ is odd then $f(-x)=-f\left(x\right)\forall x$]

$=0$, which is an even function.

So (i) is true.

(ii) Let $h\left(x\right)=\left[|f\right(x)|+1]$.

Now $h(-x)=\left[|f\right(-x)|+1]=\left[|f\right(x)|+1]=h\left(x\right)\forall x$.

So $h\left(x\right)$ is an even function.

So (ii) is also true.

(ii)Let $p\left(x\right)=\frac{f\left(x\right)-f(-x)}{2}$

$=\frac{f\left(x\right)+f\left(x\right)}{2}$ [Since $f$ is odd then $f(-x)=-f\left(x\right)\forall x$]

$=f\left(x\right)$, which is an odd function. [Given]

So (iii) is not true.

(iv) Let $q\left(x\right)=f\left(x\right)+f(-x)$

$=f\left(x\right)-f\left(x\right)$ [Since $f$ is odd then $f(-x)=-f\left(x\right)\forall x$]

$=0$, which is an even function.

So (iv) is not true.

So option (B) is the correct option.