For this problem, we just take an example which satisfies all the conditions which are given in the Qs.

For example let A={1,2} and B={ a,b,c}. Let f(1) =a, f(2)=b. Now,

g({})={}

g({1})={a}

g({2})={b}

g({1,2})={a,b}

h({})={}

h({a})={1}

h({b})={2}

h({c})={}

h({a,b})={1,2}

h({a,c})={1}

h({b,c})={2}

h({a,b,c})={1,2}

we can see that h is not One-to-One.

Now, we find out g(h(D)).

D =2^{B} = {{},{a},{b},{c} ,{a,b} ,{a,c},{b,c},{a,b,c} }

h(D)= { {},{1},{2},{1,2} }

g({})={}

g({1})={a}

g({2})={b}

g({1,2})={a,b}

g(h(D)) = {{},{a},{b},{a,b}}

Now we can see that for any D⊆B,g(h(D)) ⊆ D. Had the function f been bijective (one-one and onto or one-one and co-domain = range) , then we would have got g(h(D))=D.

## The correct answer is,(A) g(h(D))⊆D