Let A = B = C = R, and let f : A -> B and g : B -> C be defined by f(a) = a – 1 and g(b) = b² find : (i) (fog) (2) (ii) (gof) (2) (iii) (fof) (y) (iv) (gog) (y).

The problem defines three variables, A, B, and C, all of which are equal to R. Additionally, two functions are defined: f : A -> B and g : B -> C. Function f takes an input a from A and returns a – 1 in B, while function g takes an input b from B and returns b² in C.

We’re asked to find the following:

(i) (fog) (2): This means we need to find the composition of f and g, and then evaluate it at 2. To do this, we first apply g to 2, which gives us g(2) = 2² = 4. We then apply f to 4, which gives us f(4) = 4 – 1 = 3. So (fog)(2) = 3.

(ii) (gof) (2): This means we need to find the composition of g and f, and then evaluate it at 2. To do this, we first apply f to 2, which gives us f(2) = 2 – 1 = 1. We then apply g to 1, which gives us g(1) = 1² = 1. So (gof)(2) = 1.

(iii) (fof) (y): This means we need to find the composition of f with itself, and then evaluate it at y. To do this, we simply apply f twice to y, which gives us f(f(y)) = f(y – 1) = (y – 1) – 1 = y – 2.

(iv) (gog) (y): This means we need to find the composition of g with itself, and then evaluate it at y. To do this, we simply apply g twice to y, which gives us g(g(y)) = g(y²) = (y²)² = y⁴