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TS EAMCET · Maths · Functions

The domain and range of \(f(x)=\frac{1}{\sqrt{|x|-x^2}}\) are A and B respectively. Then \(\mathrm{A} \cup \mathrm{B}=\)

  1. A \(\mathbb{R}-\{-1,0,1\}\)
  2. B \((-1, \infty)-\{0,1\}\)
  3. C \((-1,0) \cup(0,1) \cup[2, \infty)\)
  4. D \((-1,1) \cup[2, \infty)\)
Verified Solution

Answer & Solution

Correct Answer

(C) \((-1,0) \cup(0,1) \cup[2, \infty)\)

Step-by-step Solution

Detailed explanation

\(|x|-x^2 > 0\) If \(x \ge 0\): \(x-x^2 > 0 \Rightarrow x(1-x) > 0 \Rightarrow 0 If \(x 0 \Rightarrow -x(1+x) > 0 \Rightarrow x(1+x) \(\mathrm{A} = (-1, 0) \cup (0, 1)\) For \(x \in A\), the maximum value of \(|x|-x^2\) is \(1/4\) (at \(x=\pm 1/2\)) and approaches…