MHT CET · Maths · Functions
If \(f: \mathrm{R} \rightarrow \mathrm{R}, \mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}\) are two functions defined by \(f(x)=2 x-3, \mathrm{~g}(x)=x^{3}+5\)
then \((\operatorname{fog})^{-1}(x)=\)
- A \(\left(\frac{2 x+3}{2}\right)^{\frac{1}{2}}\)
- B \(\left(\frac{x-7}{2}\right)^{\frac{1}{3}}\)
- C \(\left(\frac{x-7}{2}\right)^{\frac{1}{2}}\)
- D \(\left(\frac{x+7}{2}\right)^{\frac{1}{3}}\)
Answer & Solution
Correct Answer
(B) \(\left(\frac{x-7}{2}\right)^{\frac{1}{3}}\)
Step-by-step Solution
Detailed explanation
Given \(\mathrm{f}(\mathrm{x})=2 \mathrm{x}-3, \quad \mathrm{~g}(\mathrm{x})=\mathrm{x}^{3}+5\)
\(\begin{aligned} \therefore \quad(\mathrm{fog})(\mathrm{x}) &=\mathrm{f}[\mathrm{g}(\mathrm{x})]=\mathrm{f}\left(\mathrm{x}^{3}+5\right) \\ &=2\left(\mathrm{x}^{3}+5\right)-3=2 \mathrm{x}^{3}+7 \end{aligned}\)
\(\begin{aligned} & \text { Let } y=2 x^{3}+7 \Rightarrow \frac{y-7}{2}=x^{3} \\ \therefore &\left(\frac{y-7}{2}\right)^{\frac{1}{3}}=x \Rightarrow f^{-1}(y)=\left(\frac{y-7}{2}\right)^{\frac{1}{3}} \\ &(f \circ g)^{-1}(x)=\left(\frac{x-7}{2}\right)^{\frac{1}{3}} \end{aligned}\)
\(\begin{aligned} \therefore \quad(\mathrm{fog})(\mathrm{x}) &=\mathrm{f}[\mathrm{g}(\mathrm{x})]=\mathrm{f}\left(\mathrm{x}^{3}+5\right) \\ &=2\left(\mathrm{x}^{3}+5\right)-3=2 \mathrm{x}^{3}+7 \end{aligned}\)
\(\begin{aligned} & \text { Let } y=2 x^{3}+7 \Rightarrow \frac{y-7}{2}=x^{3} \\ \therefore &\left(\frac{y-7}{2}\right)^{\frac{1}{3}}=x \Rightarrow f^{-1}(y)=\left(\frac{y-7}{2}\right)^{\frac{1}{3}} \\ &(f \circ g)^{-1}(x)=\left(\frac{x-7}{2}\right)^{\frac{1}{3}} \end{aligned}\)
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