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Question

Consider the following statements: Which of the statements given above is/are correct?

  1. 1 only
  2. 2 only
  3. Both 1 and 2
  4. Neither 1 nor 2
Answer

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