JEE Mains · Maths · STD 11- 2. Relation and Function
The range of the function \(f ( x )=\sqrt{3-x}+\sqrt{2+x}\) is
- A \([\sqrt{5}, \sqrt{10}]\)
- B \([2 \sqrt{2}, \sqrt{11}]\)
- C \([\sqrt{5}, \sqrt{13}]\)
- D \([\sqrt{2}, \sqrt{7}]\)
Answer & Solution
Correct Answer
(A) \([\sqrt{5}, \sqrt{10}]\)
Step-by-step Solution
Detailed explanation
\(y^2=3-x+2+x+2 \sqrt{(3-x)(2+x)}\) \(=5+2 \sqrt{6+x-x^2}\) \(y^2=5+2 \sqrt{\frac{25}{4}-\left(x-\frac{1}{2}\right)^2}\) \(y_{\max }=\sqrt{5+5}=\sqrt{10}\) \(y_{\min }=\sqrt{5}\)
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