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Theory of Equations: Roots Equation Roots Equation

JEE Maths question with a full step-by-step solution.

Question
If roots of the equation a(x−ℓ)(x−m)+n=0a(x-\ell)(x-m) + n = 0 are α,β\alpha, \beta, then roots of the equation a ⁣(xx−k−α) ⁣(xx−k−β)−n=0a\!\left(\dfrac{x}{x-k} - \alpha\right)\!\left(\dfrac{x}{x-k} - \beta\right) - n = 0 are
Aℓℓ+1, mm+1\dfrac{\ell}{\ell+1},\, \dfrac{m}{m+1}
Bk(ℓ+1)ℓ, k(m+1)m\dfrac{k(\ell+1)}{\ell},\, \dfrac{k(m+1)}{m}
Ckℓℓ−1, kmm−1\dfrac{k\ell}{\ell-1},\, \dfrac{km}{m-1}correct
Dℓ−1ℓ, m−1m\dfrac{\ell-1}{\ell},\, \dfrac{m-1}{m}
Solution
Step 1: Since α,β\alpha, \beta are roots of a(t−ℓ)(t−m)+n=0a(t-\ell)(t-m)+n = 0, we have a(t−ℓ)(t−m)+n=a(t−α)(t−β)a(t-\ell)(t-m)+n = a(t-\alpha)(t-\beta). Therefore a(t−α)(t−β)−n=a(t−ℓ)(t−m)a(t-\alpha)(t-\beta) - n = a(t-\ell)(t-m). Step 2: Set t=xx−kt = \dfrac{x}{x-k}. The equation becomes:
a ⁣(xx−k−α) ⁣(xx−k−β)−n=0  ⟹  a ⁣(xx−k−ℓ) ⁣(xx−k−m)=0a\!\left(\frac{x}{x-k}-\alpha\right)\!\left(\frac{x}{x-k}-\beta\right) - n = 0 \implies a\!\left(\frac{x}{x-k}-\ell\right)\!\left(\frac{x}{x-k}-m\right) = 0
Step 3: xx−k=ℓ  ⟹  x=ℓ(x−k)  ⟹  x=kℓℓ−1\dfrac{x}{x-k} = \ell \implies x = \ell(x-k) \implies x = \dfrac{k\ell}{\ell-1}. Similarly: xx−k=m  ⟹  x=kmm−1\dfrac{x}{x-k} = m \implies x = \dfrac{km}{m-1}. Correct answer: (3)
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