(i) Let g(x)=f(x)+f(−x)2
=f(x)−f(x)2 [Since f is odd then f(−x)=−f(x)∀x]
=0, which is an even function.
So (i) is true.
(ii) Let h(x)=[|f(x)|+1].
Now h(−x)=[|f(−x)|+1]=[|f(x)|+1]=h(x)∀x.
So h(x) is an even function.
So (ii) is also true.
(ii)Let p(x)=f(x)−f(−x)2
=f(x)+f(x)2 [Since f is odd then f(−x)=−f(x)∀x]
=f(x), which is an odd function. [Given]
So (iii) is not true.
(iv) Let q(x)=f(x)+f(−x)=f(x)−f(x) [Since f is odd then f(−x)=−f(x)∀x]
=0, which is an even function.
So (iv) is not true.
So option (B) is the correct option.