If f : R → R, f(x) = 2x – 3; g : R → R, g(x) = x3 + 5, then the value of (fog)-1 (x):
Answer is (d)
Given,
f : R → R, f(x) = 2x – 3
g : R → R, g(x)= x3 + 5
∴ (fog)(x) = f(g(x)]
= f(x3 +5)
= 2(x3 + 5) – 3 = 2x3 + 10 – 3
= 2x3 + 7
Let y = (fog)(x) = 2x3 + 7
∴ (fog)-1(y) = x