Bivariate range

Algebra Level pending

Let x , y ∈ [ 0 , 1 ] x,y\in [0,1] . If the ranges of

1 + x y 1 + x 2 + 1 − x y 1 + y 2 \sqrt{\frac{1+xy}{1+x^2}}+\sqrt{\frac{1-xy}{1+y^2}}

is [ a , b ] [a, b] , find the value of a + b a+b .


The answer is 3.

Danish Ahmed by Danish Ahmed , 24, India

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1 solution

Tom Engelsman
Nov 26, 2020

The minimum value of f ( x , y ) = 1 + x y 1 + x 2 + 1 − x y 1 + y 2 f(x,y) = \sqrt{\frac{1+xy}{1+x^2}} + \sqrt{\frac{1-xy}{1+y^2}} occurs when both denominators are at their largest value. That is, x = y = 1 ⇒ 1 + 1 2 = 2. x=y=1 \Rightarrow 1+1^2 = 2. Likewise, the maximum value occurs when both denominators are at their smallest value: x = y = 0 ⇒ 1 + 0 2 = 1. x=y=0 \Rightarrow 1 + 0^2 = 1. Hence, f M I N = f ( 1 , 1 ) = 1 , f M A X = f ( 0 , 0 ) = 2 ⇒ a = 1 , b = 2 ⇒ a + b = 3 . f_{MIN} = f(1,1) = 1, f_{MAX} = f(0,0) = 2 \Rightarrow a = 1, b = 2 \Rightarrow a+b=\boxed{3}.

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