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
A
(x+72)13
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B
(x−72)13
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C
(x−27)13
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D
(x−72)13
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Solution
The correct option is D(x−72)13 We 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,xϵR ∴The correct answer is (d).