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The range of the function f(x)=\sqrt{3-x}+\sqrt{2+x} is:

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

best_answer

3-x \geq 0 \quad \quad 2+x \geq 0
x \leq 3 \quad \quad x \geq-2
x \in[-2,3]

Now, f(-2)=\sqrt{3+2}=\sqrt{5}

           f(3)=\sqrt{2+3}=\sqrt{5}
           f(x)=\sqrt{3-x}+\sqrt{2+x}
         f^{\prime}(x)=\frac{1}{2 \sqrt{3-x}} x-1+\frac{1}{2 \sqrt{2+x}}=0
\Rightarrow \frac{1}{\sqrt{3-x}}=\frac{1}{\sqrt{2+x}}
\Rightarrow 3-x=2+x
\Rightarrow 2 \mathrm{x}=1
x=1 / 2

f\left(\frac{1}{2}\right)=\sqrt{3-1 / 2}+\sqrt{2+1 / 2}
                  =\sqrt{\frac{5}{2}}+\sqrt{\frac{5}{2}} \Rightarrow 2 \sqrt{\frac{5}{2}} \Rightarrow \sqrt{10}
        Range =[\sqrt{5}, \sqrt{10}]

 

Posted by

Divya Prakash Singh

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