. Let (V=C[0,1].) Consider S = {f € V : f(c) = d}.Then S is a subspace of V if and only if(a) c ∈ [0, 1] and (d=0)(b) d ∈ [0, 1] and (c=0)(c) c ∈ [0, 1] and d ∈ [0, 1].(d) (c=d=0)

The set S is a subspace of V if and only if the conditions for a subspace are met. These conditions are:

1. The zero vector is in S.
2. S is closed under vector addition.
3. S is closed under scalar multiplication.

Let's examine each option:

(a) c ∈ [0, 1] and (d=0) This condition means that for any function f in S, f(c) = 0. This satisfies the condition that the zero vector is in S (since the zero vector in V is the function that is zero everywhere). Also, if f and g are in S, then (f+g)(c) = f(c) + g(c) = 0 + 0 = 0, so S is closed under addition. Similarly, for any scalar k, (kf)(c) = kf(c) = k0 = 0, so S is closed under scalar multiplication. Therefore, S is a subspace of V under this condition.

(b) d ∈ [0, 1] and (c=0) This condition means that for any function f in S, f(0) = d. This does not necessarily satisfy the condition that the zero vector is in S, because the zero vector in V is the function that is zero everywhere, but here we have f(0) = d, not necessarily zero. Therefore, S is not necessarily a subspace of V under this condition.

(c) c ∈ [0, 1] and d ∈ [0, 1] This condition means that for any function f in S, f(c) = d. This does not necessarily satisfy the condition that the zero vector is in S, because the zero vector in V is the function that is zero everywhere, but here we have f(c) = d, not necessarily zero. Therefore, S is not necessarily a subspace of V under this condition.

(d) (c=d=0) This condition is the same as condition (a), where c is in [0,1] and d is zero. Therefore, S is a subspace of V under this condition.

So, the correct answer is (a) c ∈ [0, 1] and (d=0) and (d) (c=d=0).