Let f:R→R,g:R→R be two functions given by f(x)=2x−3,g(x)=x3+5. Then (f∘g)−1(x) is equal to

f(x)=2x−3,g(x)=x3+5 are bijections and hence f−1 and g−1 both exist. y=2x−3⇒x=y+32 ∴ f−1(y)=y+32∴ f−1(x)=x+32→(1) y=x3+5 ⇒x=(y−5)1/3=g−1(y)∴ g−1(x)=(x−5)1/3 ∴ (fog)−1x=g−1o f-1) x =g−1f−1(x)=g−1x+32 by (1)=x+32−51/3=x−721/3

Let f:R→R,g:R→R be two functions given by f(x)=2x−3,g(x)=x3+5. Then (f∘g)−1(x) is equal to

f(x)=2x−3,g(x)=x3+5 are bijections and hence f−1 and g−1 both exist. y=2x−3⇒x=y+32 ∴ f−1(y)=y+32∴ f−1(x)=x+32→(1) y=x3+5 ⇒x=(y−5)1/3=g−1(y)∴ g−1(x)=(x−5)1/3 ∴ (fog)−1x=g−1o f-1) x =g−1f−1(x)=g−1x+32 by (1)=x+32−51/3=x−721/3

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Let $f : R \to R, g : R \to R$ be two functions given by $f (x) = 2 x - 3, g (x) = x^{3} + 5$ . Then $(f \circ g)^{- 1} (x)$ is equal to

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${(\frac{x - 2}{7})}^{1 / 8}$

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

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

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

answer is D.

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Detailed Solution

$f (x) = 2 x - 3, g (x) = x^{3} + 5$ are bijections and hence

$f^{- 1} and g^{- 1}$ both exist.

$\begin{array}{l} y = 2 x - 3 \Rightarrow x = \frac{y + 3}{2} \\ ∴ f^{- 1} (y) = \frac{y + 3}{2} \\ ∴ f^{- 1} (x) = \frac{x + 3}{2} \to (1) \\ y = x^{3} + 5 \\ \Rightarrow x = (y - 5)^{1 / 3} = g^{- 1} (y) \\ ∴ g^{- 1} (x) = (x - 5)^{1 / 3} \\ ∴ (f o g)^{- 1} x = (g^{- 1} {o f}^{- 1}) x \\ = g^{- 1} [f^{- 1} (x)] = g^{- 1} [\frac{x + 3}{2}] by (1) \\ = {(\frac{x + 3}{2} - 5)}^{1 / 3} = {(\frac{x - 7}{2})}^{1 / 3} \end{array}$