If $f: \mathrm{R} \rightarrow \mathrm{R}, \mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}$ are two 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 $(\operatorname{fog})^{-1}(x)=$
  1. $\left(\frac{2 x+3}{2}\right)^{\frac{1}{2}}$
  2. $\left(\frac{x-7}{2}\right)^{\frac{1}{3}}$
  3. $\left(\frac{x-7}{2}\right)^{\frac{1}{2}}$
  4. $\left(\frac{x+7}{2}\right)^{\frac{1}{3}}$

Solution

Given $\mathrm{f}(\mathrm{x})=2 \mathrm{x}-3, \quad \mathrm{~g}(\mathrm{x})=\mathrm{x}^{3}+5$ $\begin{aligned} \therefore \quad(\mathrm{fog})(\mathrm{x}) &=\mathrm{f}[\mathrm{g}(\mathrm{x})]=\mathrm{f}\left(\mathrm{x}^{3}+5\right) \\ &=2\left(\mathrm{x}^{3}+5\right)-3=2 \mathrm{x}^{3}+7 \end{aligned}$ $\begin{aligned} & \text { Let } y=2 x^{3}+7 \Rightarrow \frac{y-7}{2}=x^{3} \\ \therefore &\left(\frac{y-7}{2}\right)^{\frac{1}{3}}=x \Rightarrow f^{-1}(y)=\left(\frac{y-7}{2}\right)^{\frac{1}{3}} \\ &(f \circ g)^{-1}(x)=\left(\frac{x-7}{2}\right)^{\frac{1}{3}} \end{aligned}$

Asked in: MHT CET 2020 (13 Oct Shift 1)

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