Find the range of `f(x)sqrt(x-1)+sqrt(5-1)`
Correct Answer – `[2,2sqrt(2)]`
Let `y=sqrt(x-1)=sqrt(5-x)`
or `y^(2)=x-1+5-x+2 sqrt((x-1)(5-x))`
or `y^(2)=4+2 sqrt(-x^(2)-5+6x)`
or `y^(2)=4+2 sqrt(4-(x-3)^(2))`
then `y^(2)` has minimum value 4 [when `4-(x-3)^(2)=0`] and maximum value 8 when `x=3`.
Therefore, `y in [2,2sqrt(2)]`.
Correct Answer – `[2,2sqrt(2)]`
Let `y = sqrt(x – 1) + sqrt(5 -x)`
`rArr y^(2) = x – 1 + 5 – x + 2 sqrt((x – 1)(5 – x))`
`= 4 + 2 sqrt(-x^(2)-5 +6x)`
`= 4 + 2 sqrt( 4 – ( x – 3)^(2))`
Then `y^(2)` has minimum value 4 [when `4 – (x – 3)^(2) = 0]` and maximum value 8 when x = 3. They, y `in[2,2,sqrt(2)]` .