Lets Solve it by substitution.
T(n) = T(n-1) + T(n/2) + n
= T(n-2) + T(n/4) + (n-1) + n
= T(n-3) + T(n/8) + (n-2) + (n-1) + n
= T(n-4) + T(n/16) + (n-3) + (n-2) + (n-1) + n
= ------------
= ------------
= -------------
= T(n-k) + T(n/(2^k)) + (n-(k-1)) + (n-(k-2)) + ------------------------------ + (n-2) + (n-1) + n
Now consider n = 2^k, then k = log n
= T((2^k) - k) + T((2^k)/(2^k)) + (n - k + 1) + (n - k + 2) + ---------------------- + (n - 1) + n
Now Put value of K
= (2^(log n) - log n) + 1 + (n - log n + 1) + (n - log n + 2) + ------------------------ + n
Here series {(n - log n + 1) + (n - log n + 2) + ---------------------- + (n - 1) + n} are in AP with (log n) element then
Hence Sn = ((log n)/2) *(n -log n + 1 + n) = O(n.log n)
Hence,
T(n) = 2^(log n)
From all the given answer, Option C is asymptotically close to T(n), Hence C will be the answer.