1. If (a+b) is directly proportional to (a-b), then (a^2-b^2)
2. If a is directly proportional to b, then (a^2-b^2)
From 1. (a+b) ∝ (a-b)
=> (a+b)=k(a-b)
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
=>(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{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