Right answer is (a) True
To explain I would say: We know that if x(n) is a real signal, then xI(n)=0 and xR(n)=x(n)
We know that, xR(n)=x(n)=\(\frac{1}{2π}\int_0^{2π}\)[XR(ω) cosωn- XI(ω) sinωn] dω
Since both XR(ω) cosωn and XI(ω) sinωn are even, x(n) is also even
=> x(n)=\(\frac{1}{π} \int_0^π\)[XR(ω) cosωn- XI(ω) sinωn] dω