Functions

Algebra Level 2

Let f ( x , y ) = y + x f(x,y) = y + x , f ( x , y ) = 2 a f(x,y) = \sqrt{2a} and x 2 + y 2 = − 2 b x^2 + y^2 = -2b . Find f ( 1 x , 1 y ) f \left( \dfrac1x , \dfrac1y\right) in terms of a a and b b .

2 a a + b \frac {\sqrt{2a}}{a+b} 2 a 2 a + 2 b \frac {\sqrt{2a}}{2a+2b} 2 a a − b \frac {\sqrt{2a}}{a-b} 2 a 2 a − 2 b \frac {\sqrt{2a}}{2a-2b}

Abdelrahman Samy by Abdelrahman Samy , 24, Egypt

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1 solution

Hung Woei Neoh
Jun 30, 2016

f ( x , y ) = y + x = x + y ⟹ f(x,y)= y+x = x+y \implies Eq.(1)

f ( x , y ) = 2 a ⟹ f(x,y) = \sqrt{2a} \implies Eq.(2)

x 2 + y 2 = − 2 b ⟹ x^2+y^2 = -2b \implies Eq.(3)

Eq.(1) = Eq.(2):

x + y = 2 a ⟹ x+y=\sqrt{2a} \implies Eq.(4)

Now, we are looking for the value of f ( 1 x , 1 y ) f\left(\dfrac{1}{x},\dfrac{1}{y}\right) . We have:

f ( 1 x , 1 y ) = 1 x + 1 y = x + y x y f\left(\dfrac{1}{x},\dfrac{1}{y}\right)=\dfrac{1}{x}+\dfrac{1}{y} =\dfrac{x+y}{xy}

Substitute Eq.(4) in here:

f ( 1 x , 1 y ) = x + y x y = 2 a x y f\left(\dfrac{1}{x},\dfrac{1}{y}\right) =\dfrac{x+y}{xy}= \dfrac{\sqrt{2a}}{xy}

From Eq.(3):

x 2 + y 2 = − 2 b x 2 + 2 x y + y 2 = 2 x y − 2 b ( x + y ) 2 = 2 x y − 2 b x^2+y^2 = -2b\\ x^2+2xy+y^2=2xy-2b\\ (x+y)^2 = 2xy-2b

Substitute Eq.(4) here too:

( 2 a ) 2 = 2 x y − 2 b 2 a = 2 x y − 2 b x y = a + b (\sqrt{2a})^2 =2xy-2b\\ 2a=2xy-2b\\ xy=a+b

Therefore,

f ( 1 x , 1 y ) = 2 a x y = 2 a a + b f\left(\dfrac{1}{x},\dfrac{1}{y}\right) = \dfrac{\sqrt{2a}}{xy}=\boxed{\dfrac{\sqrt{2a}}{a+b}}

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