The range of the function `f(x) = sqrt(2-x)+sqrt( 1+x)`

Options

A. `[sqrt3, sqrt6]`
B. `[0, sqrt6]`
C. `(sqrt3, sqrt6)`
D. none of these

Answer

A

Explanation

We have, `f(x) = sqrt(2-x) + sqrt(1+x)` is
Clearly, `f(x)` is defined for
`2-x le 0 and 1+ x ge 0 rArr x le 2 and x ge - 1 rArr x in [-1,2]`
So, domain `(f) = [-1,2]`
Let `y = sqrt(2-x) + sqrt(1+x)" "….(i)`
` rArr y^(2) = 3+2 sqrt(2+x-x^(2))" "......(ii)`
`rArr ((y^(2) -3)/(2))^(3) = 2+ x-x^(2)`
`rArr x^(2) - x = 2 -((y^(2)-3)/(2))^(2)`
`rArr (x-(1)/(2))^(2) = (9)/(4) -((y^(2)-3)/(2))^(2)`
`rArr x -(1)/(2) = pm sqrt((9)/(4)-((y^(2)-3)/(2))^(2))`
`rArr x = (1)/(2) pm sqrt((9)/(4) - ((y^(2)-3)/(2))^(2))`
For x to be real, we must have
`(9)/(4) -((y^(2)-3)/(2))^(2) ge 0`
`rArr ((y^(2)-3)/(2))^(2) -((3)/(2))^(2) le 0`
`-(3)/(2) le (y^(2) -3)/(2) le (3)/(2)`
`rArr 0 ley^(2) le 6 rAra - sqrt(6) le y le sqrt(6) rArr y in [-sqrt(6),sqrt(6)]`
Alos, from (i) and (ii), we have
`y^(2) ge 3 and y ge 0 y ge sqrt(3)`
Form (iii) and (iv) ,we have
`y in [sqrt(3),sqrt(6)]`
Hence range `(f) = [ sqrt(3), sqrt(6)]`