Question

# Let f:R→R,g:R→R, be two functions,  such that f(x) =2x – 3, g (x) = x3 + 5. The function (fog)−1 (x) is equal to  (x−72)13(x−72)13(x+72)13(x−27)13

Solution

## The correct option is B (x−72)13We have, f : R → R, g: R → R defined by f(x) = 2x - 3 and g (x) = x3 + 5 It can be checked that f(x) and g(x) are bijective functions ∴ fo g is also bijective and (fog) = f(g(x)) = f (x3+5)=2(x3+5)−3=2x3+7 (fog)(x)=y⇒2x3+7=y⇒x=(y−72)13 ∴ (fog)−1(x)=(x−72)13,x ϵ R ∴ The correct answer is (d).

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