Question
The range of the function $$f\left( x \right) = 2\sqrt {x - 2} + \sqrt {4 - x} $$ is :
A.
$$\left( {\sqrt 2 ,\,\sqrt {10} } \right)$$
B.
$$\left[ {\sqrt 2 ,\,\sqrt {10} } \right)$$
C.
$$\left( {\sqrt 2 ,\,\sqrt {10} } \right]$$
D.
$$\left[ {\sqrt 2 ,\,\sqrt {10} } \right]$$
Answer :
$$\left[ {\sqrt 2 ,\,\sqrt {10} } \right]$$
Solution :
Clearly, domain of the function is $$\left[ {2,\,4} \right].$$
$$\eqalign{
& {\text{Now, }}f'\left( x \right) = \frac{1}{{\sqrt {x - 2} }} - \frac{1}{{2\sqrt {4 - x} }} \cr
& {\text{or, }}f'\left( x \right) = 0 \cr
& {\text{or, }}\sqrt {x - 2} = 2\sqrt {4 - x} \cr
& {\text{or, }}x - 2 = 16 - 4x \cr
& {\text{or, }}x = \frac{{18}}{5} \cr
& {\text{Now, }}f\left( 2 \right) = \sqrt 2 , \cr
& f\left( {\frac{{18}}{5}} \right) = 2\sqrt {\frac{{18}}{5} - 2} + \sqrt {4 - \frac{{18}}{5}} = \sqrt {10} , \cr
& f\left( 4 \right) = 2\sqrt 2 \cr} $$
Hence, range of the function is $$\left[ {\sqrt 2 ,\,\sqrt {10} } \right].$$
Also, here $$x = \left( {\frac{{18}}{5}} \right)$$ is the point of global maxima.