The function \(f(x)=\operatorname{sech}(x)\) on \(\mathbf{R}\) has the range

The function \(f(x)=\operatorname{sech}(x)\) on \(\mathbf{R}\) has the range
  1. \((0, \infty)\)
  2. \((0,1]\)
  3. \([1, \infty)\)
  4. \((1, \infty)\)

Solution

\(f(x)=\operatorname{sech} x\) \(=\frac{1}{\cosh x}=\frac{2}{e^x+e^{-x}}\) At \(\quad x=0, f(x)=\frac{2}{e^0+e^{-0}}=\frac{2}{2}=1\) At \(x \neq 0, f(x)\) is less than 1 or \(e^x+e^{-x}\) is greater than 2 . Also \(f(x) \neq 0\) at any value of \(x\). Hence, range \(\operatorname{sech} x\) \(=(0,1]\)

Asked in: AP EAMCET 2020 (17 Sep Shift 2)

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