The correct option is C g(2n)=0,n∈N
g(x)=x∫0f(t) dt
g(−x)=−x∫0f(t) dt=−x∫0f(−t) dt⇒g(−x)=x∫0f(t) dt[∵f(−t)=−f(t)]
⇒g(x)=g(−x)
Thus, g(x) is even function.
Also,
g(x+2)=x+2∫0f(t) dt⇒g(x+2)=2∫0f(t) dt+x+2∫2f(t) dt⇒g(x+2)=g(2)+x∫0f(t+2) dt⇒g(x+2)=g(2)+x∫0f(t) dt⇒g(x+2)=g(2)+g(x)
Now,
g(2)=2∫0f(t) dt⇒g(2)=1∫0f(t) dt+2∫1f(t) dt⇒g(2)=1∫0f(t) dt+0∫−1f(t+2) dt⇒g(2)=1∫0f(t) dt+0∫−1f(t) dt⇒g(2)=1∫−1f(t) dt=0
As f(t) is odd function.
So,
g(2)=0⇒g(x+2)=g(x)
i.e. g(x) is periodic with period 2.
∴g(4)=0 or f(6)=0, ⇒g(2n)=0,n∈N.