AP EAMCET · Maths · Functions
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 \((\mathrm{f}(\mathrm{x}))^2=\)
- A \(\mathrm{f}(\mathrm{x})+\mathrm{f}(0)\)
- B \(f\left(x^2\right)+f(2)\)
- C \(f\left(x^3\right)+f(0)\)
- D \(f\left(x^2\right)+f(1)\)
Answer & Solution
Correct Answer
(D) \(f\left(x^2\right)+f(1)\)
Step-by-step Solution
Detailed explanation
\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}} \\…
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