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TS EAMCET · Maths · Functions

Let \(f: \mathrm{A} \rightarrow \mathrm{B}\) be defined as \(f(x)=\frac{1}{2}-\tan \left(\frac{\pi x}{2}\right)\) and \(g: \mathrm{B} \rightarrow \mathrm{C}\) be defined \(g(x)=\sqrt{3+4 x-4 x^2}\). If A, B, C are subsets of \(\mathbb{R}\) and \(f\) is an onto function then the range of the function \(f(x)\) is

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

Answer & Solution

Correct Answer

(C) \(\left[-\frac{1}{2}, \frac{3}{2}\right]\)

Step-by-step Solution

Detailed explanation

Since \(f\) is an onto function Thus domain of \(\mathrm{g}(\mathrm{x})\) will be range of \(\mathrm{f}(\mathrm{x})\) Now, \(3+4 \mathrm{x}-4 \mathrm{x}^2 \geq 0\) to define \(\mathrm{g}(\mathrm{x})\) \[ \Rightarrow 4 \mathrm{x}^2-4 \mathrm{x}-3 \mathrm{x} 0 \] On solving we get…
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