1. If (a+b) is directly proportional to (a-b), then $$(a^2-b^2)$$ is directly proportional to ab.
2. If a is directly proportional to b, then $$(a^2-b^2)$$ is directly proportional to ab.
From 1. $$(a+b) ∝ (a-b)$$
=> $$(a+b)=k(a-b)$$, where k is constant
squaring we get,
=>$$(a+b)^2=k^2(a-b)^2$$
=>$$\frac{(a+b)^2}{(a-b)^2}=k2$$
=>$$\frac{a^2+2ab+b^2}{a^2-2ab+b^2}=k^2$$
Using Componendo and Dividendo, $$\frac{x}{y}=\frac{x+y}{x-y},$$
=>$$\frac{a^2+2ab+b^2+a^2-2ab+b^2}{a^2+2ab+b^2-(a^2-2ab+b^2)}=\frac{k^2+1}{k^2-1}$$
=> $$\frac{2(a^2+b^2)}{4ab}=\frac{k^2+1}{k^2-1}=c$$(let)
=>$$(a^2+b^2) ∝ ab$$
Hence, Statement 1 is correct.
From 2. $$a ∝ b$$
=> $$a=kb$$
=> $$\frac{a}{b}=k$$
Using Componendo and Dividendo,
=>$$\frac{ a+b}{a-b}=\frac{k+1}{k-1}$$
Squaring both sides,
=>$$ (\frac{a+b}{a-b})^2=(\frac{k+1}{k-1})^2$$
=>$$\frac{a^2+b^2+2ab}{a^2+b^2-2ab}=\frac{k^2+2k+1}{k^2-2k+1}$$
Again using C/D,
=>$$\frac{a^2+b^2+2ab+a^2+b^2-2ab}{a^2+b^2+2ab-(a^2+b^2-2ab)}$$=$$\frac{k^2+2k+1+k^2-2k+1}{k^2+2k+1-(k^2-2k+1)}$$
=>$$\frac{2(a^2+b^2)}{4ab}=\frac{2(k^2+1)}{4k}$$
=>$$(a^2+b^2) ∝ ab$$
Hence, statement 2 is incorrect.
Answer (A) – 1 only