AP EAMCET · Maths · Functions
The range of \(f(x)=\sqrt{\frac{a-|x|}{(a+1)-|x|}},(a>0)\) is
- A \([0, a]\)
- B \([0, \infty)-\left[-\sqrt{\frac{a}{a+1}}, \sqrt{\frac{a}{a+1}}\right]\)
- C \(\left[0, \sqrt{\frac{a}{a+1}}\right] \cup(1, \infty)\)
- D \(\left[0, \sqrt{\frac{a}{a+1}}+1\right]\)
Answer & Solution
Correct Answer
(C) \(\left[0, \sqrt{\frac{a}{a+1}}\right] \cup(1, \infty)\)
Step-by-step Solution
Detailed explanation
Given function is \(f(x)=\sqrt{\frac{a-|x|}{(a+1)-|x|}},(a > 0)\) \(\because f(x) \geq 0, \forall x \in\) domain of \(f(x)\). Now, let \(\frac{a-|x|}{(a+1)-|x|}=y\)…
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