If $f(x)=\frac{4 x+7}{7 x-4}$, then the value of $f\{f[f(2)]\}=$

If $f(x)=\frac{4 x+7}{7 x-4}$, then the value of $f\{f[f(2)]\}=$
  1. $\frac{3}{2}$
  2. $\frac{2}{3}$
  3. $\frac{35}{39}$
  4. $\frac{39}{35}$

Solution

$f(x)=\frac{4 x+7}{7 x-4}$ $f(2)=\frac{8+7}{14-4}=\frac{15}{10}=\frac{3}{2}$ $f[f(2)]=f\left(\frac{3}{2}\right)=\frac{\left(4 \times \frac{3}{2}\right)+7}{\left(7 \times \frac{3}{2}\right)-4}=\frac{6+7}{\left(\frac{21-8}{2}\right)}=\frac{13 \times 2}{13}=2$ $f\{f[f(2)]\}=f(2)=\frac{3}{2}$ This problem can also be solved as follows : $\begin{aligned} \mathrm{f}(\mathrm{x}) \quad=\frac{4 \mathrm{x}+7}{7 \mathrm{x}-4} & \Rightarrow \mathrm{f}[\mathrm{f}(\mathrm{x})]=\mathrm{f}\left[\frac{4 \mathrm{x}+7}{7 \mathrm{x}-4}\right] \\ \therefore \mathrm{f}\left[\frac{4 \mathrm{x}+7}{7 \mathrm{x}-4}\right] \quad &=\frac{4\left(\frac{4 \mathrm{x}+7}{7 \mathrm{x}-4}\right)+7}{7\left(\frac{4 \mathrm{x}+7}{7 \mathrm{x}-4}\right)-4} \\ &=\frac{16 \mathrm{x}+28+49 \mathrm{x}-28}{28 \mathrm{x}+49-28 \mathrm{x}+16}=\frac{65 \mathrm{x}}{65}=\mathrm{x} \\ \therefore \quad \mathrm{f}\{\mathrm{f}[\mathrm{f}(\mathrm{x})]\} &=\mathrm{f}\{\mathrm{x}\}=\frac{4 \mathrm{x}+7}{7 \mathrm{x}-4} \\ \therefore \quad \mathrm{f}\{\mathrm{f}[\mathrm{f}(2)]\} &=\frac{4(2)+7}{7(2)-4}=\frac{15}{10}=\frac{3}{2} \end{aligned}$

Asked in: MHT CET 2020 (12 Oct Shift 2)

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