Let f : R → R , g : R → R , be two functions, such that f(x) =2x – 3, g (x) = x 3 + 5. The function ( f o g ) − 1 (x) is equal to Options: A . &nbsp ( x + 7 2 ) 1 3 B . &nbsp ( x − 7 2 ) 1 3 C . &nbsp ( x − 2 7 ) 1 3 D . &nbsp ( x − 7 2 ) 1 3

Let f:R→R,g:R→R, be two functions, such that f(x) =2x

Question -

Let $f : R \to R, g : R \to R$ , be two functions, such that f(x) =2x – 3, g (x) = $x^{3}$ + 5.
The function $(f o g)^{- 1}$ (x) is equal to

Options:

A .

{(\frac{x + 7}{2})}^{\frac{1}{3}}

B .

{(x - \frac{7}{2})}^{\frac{1}{3}}

C .

{(\frac{x - 2}{7})}^{\frac{1}{3}}

D .

{(\frac{x - 7}{2})}^{\frac{1}{3}}

Answer: Option D
:
D
We have, f : R

\to

R, g: R

\to

R defined by f(x) = 2x - 3 and g (x) =

x^{3}

+ 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

(x^{3} + 5) = 2 (x^{3} + 5) - 3 = 2 x^{3} + 7

(f o g) (x) = y \Rightarrow 2 x^{3} + 7 = y \Rightarrow x = {(\frac{y - 7}{2})}^{\frac{1}{3}}

∴ (f o g)^{- 1} (x) = {(\frac{x - 7}{2})}^{\frac{1}{3}}, x ϵ R

∴

The correct answer is (d).

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