The correct option is D both A and B
In oxidising flame, copper forms copper meta borate which is blue in colour.
Na2B4O7+CuSO4→2NaBO2+Cu(BO2)2+SO3
(Blue)
While in a reducing flame, copper meta borate is reduced to metallic copper, which is red and opaque.
2Cu(BO2)3+4NaBO2+2CΔ−→2Cu+2Na2B4O7+2CO
(Red opaque)
B2O3+CuSO4→Cu(BO2)2+SO3
(Blue)
(X) can be borax or microcosmic salt.