The range of the function f ( x ) = 3 - x + 2 + x is [2023]

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Solution

Q.
The range of the function $f (x) = \sqrt{3 - x} + \sqrt{2 + x}$ is [2023]

1 $[2 \sqrt{2}, \sqrt{11}]$

2 $[\sqrt{5}, \sqrt{10}]$

3 $[\sqrt{5}, \sqrt{13}]$

4 $[\sqrt{2}, \sqrt{7}]$

Ans.
(2)

Given, $f (x) = \sqrt{3 - x} + \sqrt{2 + x}$

Let $y = \sqrt{3 - x} + \sqrt{2 + x}$

$\Rightarrow y^{2} = 3 - x + 2 + x + 2 \sqrt{3 - x} \sqrt{2 + x}$

$= 5 + 2 \sqrt{6 + x - x^{2}}$

$= 5 + 2 \sqrt{\frac{25}{4} - {(x - \frac{1}{2})}^{2}}$

$y_{m a x}^{2}$ when ${(x - \frac{1}{2})}^{2} = 0$

$∴$ $y_{\max}^{2} = 5 + 2 \times \frac{5}{2} = 10 \Rightarrow y_{\max} = \sqrt{10}$

$y_{m i n}^{2}$ when $\sqrt{\frac{25}{4} - {(x - \frac{1}{2})}^{2}} = 0 \Rightarrow y_{\min} = \sqrt{5}$

$Range of f (x) = [\sqrt{5}, \sqrt{10}]$

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