Question

If $f\left( x+ay,x-ay \right)=axy$ then $f\left( x,y \right)$ is equal to:
(a) $\dfrac{{{x}^{2}}-{{y}^{2}}}{4}$
(b) $\dfrac{{{x}^{2}}+{{y}^{2}}}{4}$
(c) $4xy$
(d) none

Answer 1

Firstly, we need to substitute $x+ay=p$ and $x-ay=q$. On solving these two equations, we will obtain the values of x and y in terms of p and q. We can then substitute these values of x and y to obtain the value of $f\left( p,q \right)$. Finally, on replacing p by x and q by y we will obtain the required value of $f\left( x,y \right)$.

Complete step-by-step answer:
The given identity in the above question is written as
$\Rightarrow f\left( x+ay,x-ay \right)=axy......\left( i \right)$
According to the question, we need to find the value of $f\left( x,y \right)$. For this, le us assume
$\begin{aligned} & \Rightarrow x+ay=p.......\left( ii \right) \\\ & \Rightarrow x-ay=q.......\left( iii \right) \\\ \end{aligned}$
Adding the equations (ii) and (iii) we get

& \Rightarrow x+ay+x-ay=p+q \\\ & \Rightarrow 2x=p+q \\\ \end{aligned}$$ Dividing by $$2$$ we get $\Rightarrow x=\dfrac{p+q}{2}.......\left( iv \right)$ Putting this in the equation (ii) we get $\Rightarrow \dfrac{p+q}{2}+ay=p$ Subtracting $\dfrac{p+q}{2}$ from both the sides $$\begin{aligned} & \Rightarrow \dfrac{p+q}{2}+ay-\dfrac{p+q}{2}=p-\dfrac{p+q}{2} \\\ & \Rightarrow ay=\dfrac{2p-\left( p+q \right)}{2} \\\ & \Rightarrow ay=\dfrac{2p-p-q}{2} \\\ & \Rightarrow ay=\dfrac{p-q}{2} \\\ \end{aligned}$$ Dividing both the sides by a we get $\Rightarrow y=\dfrac{p-q}{2a}.......\left( v \right)$ On putting the equations (iv) and (v) into the equation (i) we get $$\begin{aligned} & \Rightarrow f\left( p,q \right)=a\left( \dfrac{p+q}{2} \right)\left( \dfrac{p-q}{2a} \right) \\\ & \Rightarrow f\left( p,q \right)=a\dfrac{\left( p+q \right)\left( p-q \right)}{4a} \\\ & \Rightarrow f\left( p,q \right)=\dfrac{\left( p+q \right)\left( p-q \right)}{4} \\\ \end{aligned}$$ Now, using the distributive rule of the algebraic identity, which is given by $a\left( b+c \right)$, we can write the above expression as $\begin{aligned} & \Rightarrow f\left( p,q \right)=\dfrac{\left( p+q \right)p+\left( p+q \right)\left( -q \right)}{4} \\\ & \Rightarrow f\left( p,q \right)=\dfrac{p\left( p+q \right)-q\left( p+q \right)}{4} \\\ \end{aligned}$ Now, again using the distributive law, we can write the above equation as $$\begin{aligned} & \Rightarrow f\left( p,q \right)=\dfrac{{{p}^{2}}+pq-pq-{{q}^{2}}}{4} \\\ & \Rightarrow f\left( p,q \right)=\dfrac{{{p}^{2}}-{{q}^{2}}}{4} \\\ \end{aligned}$$ Finally replacing p by x and q by y in the above equation, we get $$\Rightarrow f\left( x,y \right)=\dfrac{{{x}^{2}}-{{y}^{2}}}{4}$$ **So, the correct answer is “Option A”.** **Note:** Do not try to put $x+ay=x$ and $x-ay=y$ as that would give absurd results. For this reason only, we included dummy variables p and q which were finally replaced with x and y respectively. We can also use the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ to directly expand the expression $$\dfrac{\left( p+q \right)\left( p-q \right)}{4}$$.