If fx-y- x- y - xy- then fx- y- fy- x2 is

Assuming x + y = u, x y = v ⇒ nbsp; x= u+ v2, y =u v2 f(x+y, x y) = xy ⇒ f(u, v) = (u+ v2 ) × (u v2 ) ⇒ f (u, v) = u^2 v^24 ⇒ f (u, v) + f(v, u) = ...

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Standard XI

Mathematics

Constant Function

If fx+y, x- y...

Question

If $f (x + y, x - y) = x y$ , then $\frac{f (x, y) + f (y, x)}{2}$ is

x

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y

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0

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\frac{x + y}{2}

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Solution

The correct option is C $0$
Assuming $x + y = u, x - y = v$
$\Rightarrow x = \frac{u + v}{2}, y = \frac{u - v}{2}$

$f (x + y, x - y) = x y$
$\Rightarrow f (u, v) = (\frac{u + v}{2}) \times (\frac{u - v}{2})$
$\Rightarrow f (u, v) = \frac{u^{2} - v^{2}}{4}$
$\Rightarrow f (u, v) + f (v, u) = \frac{u^{2} - v^{2}}{4} + \frac{v^{2} - u^{2}}{4} \Rightarrow f (u, v) + f (v, u) = 0$
$\Rightarrow f (x, y) + f (y, x) = 0$

$∴ \frac{f (x, y) + f (y, x)}{2} = 0$

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Constant Function

Standard XI Mathematics

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If f(x+y,x−y)=xy, then f(x,y)+f(y,x)2 is

The correct option is C 0Assuming x+y=u, x−y=v ⇒x=u+v2, y=u−v2 f(x+y,x−y)=xy ⇒f(u,v)=(u+v2)×(u−v2) ⇒f(u,v)=u2−v24 ⇒f(u,v)+f(v,u)=u2−v24+v2−u24⇒f(u,v)+f(v,u)=0 ⇒f(x,y)+f(y,x)=0 ∴f(x,y)+f(y,x)2=0

If $f (x + y, x - y) = x y$ , then $\frac{f (x, y) + f (y, x)}{2}$ is