If `f={(a ,1),(b ,-2),(c ,3)},g={(a,-2),(b,0),(c,1)}` then `f^2+g^2=`

If f={(a, 0), (b, -2), (c, 3)}, g={a, -2), (b, 0), (c,1)} then f//g=

If f={(a, 0), (b, 2), (c, -3)}, g={(a, -1), (b, 1), (c, 2)} then 2f-3g=

Let A={1,2,3},B={a,b,c),C={p,q,r}. If f:A to B, g: B to C are defined by f={(1,a),(2,c),(3,b)},g={a,q),(b,r),(c,p)} then show that f^(-1)og^(-1)=(gof)^(-1) .

A={a, b, c}, B={2, 1, 0}, f={(a, 2), (b, 0), (c, 2)}. Then f is

If f and g are two real functions defined by f={(a,1)(b,0),(c,8),(0,4)} and g={(a,5),(c,-3),(b,6)(0,2)} then dom ((sqrt(g))/(f)) is

If f={(1,a), (2, b), (1, b), (3,c), (1,c)}, g=(a, p), (b,r), (c,q), (c,p)} , then (gof)^(-1)=

Which of the above is one to many relation ? If A= {1,2,3},B={a,b,c},f={(1,a),(1,b),(1,c)},g={(1,a),(3,c),(2,b)},h={(1,a),(2,b),(3,a)},q={(1,a),(2,a),(3,b),(1,b),(3,a)} then

Let f={(1,a),(2,c),(4,d),(3,b)} and g^(-1) = {(2,a),(4,b),(1,c),(3,d)} then show that (gof)^(-1) = f^(-1) og^(-1) .