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TS EAMCET · Maths · Functions

If \(\mathrm{D}\) is the domain and \(\mathrm{G}\) is the range of the real valued function \(f(x)=\sqrt{\frac{1-x^2}{1+x^2}}\), then \(\mathrm{D} \cap \mathrm{G}=\)

  1. A \([0, \infty)\)
  2. B \([0,1]\)
  3. C \(\left[0, \frac{1}{2}\right]\)
  4. D \([-1,1]\)
Verified Solution

Answer & Solution

Correct Answer

(B) \([0,1]\)

Step-by-step Solution

Detailed explanation

We have \(f(x)=\sqrt{\frac{1-x^2}{1+x^2}}\) For domain of \(\mathrm{f}(\mathrm{x}), \frac{1-\mathrm{x}^2}{1+\mathrm{x}^2} \geq 0\) \(\Rightarrow \mathrm{x} \in[-1,1]\) For range of \(\mathrm{f}(\mathrm{x})\) \[ y=\frac{1-x^2}{1+x^2} \] On solving we get \(\mathrm{y} \leq 1\) but…
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