Discussion

Question:

If (a + b) varies directly as (a – b), then (a² + b²) will vary as

Explanation:

(a + b) α (a - b)

⇒ (a + b) = k(a - b), when k is a constant

⇒ (a + b)/(a - b) = k

Applying componendo and dividendo,

a/b = (k + 1)/(k - 1) = x (say)

Squaring both sides; a²/b² = x²

Applying componendo and dividendo

$\frac{a^{2} + b^{2}}{a^{2} - b^{2}} = \frac{x^{2} + 1}{x^{2} - 1}$

⇒ $a^{2} + b^{2} = (a^{2} - b^{2}) \times \frac{x^{2} + 1}{x^{2} - 1}$

⇒ (a²+ b²) = (a² - b²) × (a constant, let's say t)

∴ (a²+ b²) = t × (a² - b²)

Also, $\frac{a^{2} + b^{2}}{a b} = \frac{x^{2} b^{2} + b^{2}}{x b \times b} = \frac{b^{2} (x^{2} + 1)}{b^{2} x}$ = $\frac{1 + x^{2}}{x}$ = constant

∴ (a²+ b²) α ab

∴ Both (A) and (B) are true.

Hence, option (c).

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