If $\mathrm{f}: \mathbb{R} \backslash\{0\} \rightarrow \mathbb{R}$ is defined by…

If $\mathrm{f}: \mathbb{R} \backslash\{0\} \rightarrow \mathbb{R}$ is defined by $\mathrm{f}(\mathrm{x})=\mathrm{x}+\frac{1}{\mathrm{x}}$, then the value of $(\mathrm{f}(\mathrm{x}))^2=$
  1. $\mathrm{f}(\mathrm{x})+\mathrm{f}(0)$
  2. $f\left(x^2\right)+f(2)$
  3. $f\left(x^3\right)+f(0)$
  4. $f\left(x^2\right)+f(1)$

Solution

$\begin{aligned} & \text { Given : } \mathrm{f}(\mathrm{x})=\mathrm{x}+\frac{1}{\mathrm{x}} \\ & \begin{array}{l}\Rightarrow(\mathrm{f}(\mathrm{x}))^2=\mathrm{x}^2+\frac{1}{\mathrm{x}^2}+2 \cdot \mathrm{x} \cdot \frac{1}{\mathrm{x}} \\ \Rightarrow(\mathrm{f}(\mathrm{x}))^2=\mathrm{x}^2+\frac{1}{\mathrm{x}^2}+2 \\ \Rightarrow(\mathrm{f}(\mathrm{x}))^2=\mathrm{f}\left(\mathrm{x}^2\right)+\mathrm{f}(1) \quad[\because \mathrm{f}(1)=2]\end{array}\end{aligned}$

Asked in: AP EAMCET 2023 (18 May Shift 2)

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