InterviewSolution
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InterviewSolution
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The range of the function `f(x) = sqrt(2-x)+sqrt( 1+x)`A. `[sqrt3, sqrt6]`B. `[0, sqrt6]`C. `(sqrt3, sqrt6)`D. none of these |
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Answer» Correct Answer - A 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)]` |
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