Q92 / 180 · COMEDK 2024 (Evening Shift) · Mathematics
Let $\alpha$ and $\beta$ be the distinct roots of $a x^2+b x+c=0$, then $\lim _\limits{x \rightarrow \alpha} \dfrac{1-\cos \left(a x^2+b x+c\right)}{(x-\alpha)^2}$ is equal to
- A$\dfrac{(\alpha-\beta)^2}{2}$
- Correct answerB$\dfrac{a^2(\alpha-\beta)^2}{2}$
- C0
- D$\dfrac{-a^2(\alpha-\beta)^2}{2}$
Solution
The quadratic equation $ax^2 + bx + c = 0$ has roots $\alpha$ and $\beta$. Thus, $ax^2 + bx + c = a(x - \alpha)(x - \beta)$.
The limit is given by $L = \lim_{x \rightarrow \alpha} \dfrac{1 - \cos(a(x - \alpha)(x - \beta))}{(x - \alpha)^2}$.
Using the identity $1 - \cos \theta = 2 \sin^2(\theta/2)$, we have $L = \lim_{x \rightarrow \alpha} \dfrac{2 \sin^2(\dfrac{a(x - \alpha)(x - \beta)}{2})}{(x - \alpha)^2}$.
Using the standard limit $\lim_{u \rightarrow 0} \dfrac{\sin u}{u} = 1$, we multiply and divide by the square of the argument of the sine function:
$L = \lim_{x \rightarrow \alpha} 2 \left[ \dfrac{\sin(\dfrac{a(x - \alpha)(x - \beta)}{2})}{\dfrac{a(x - \alpha)(x - \beta)}{2}} \right]^2 \times \dfrac{a^2(x - \alpha)^2(x - \beta)^2}{4(x - \alpha)^2}$.
As $x \rightarrow \alpha$, the term in the square brackets approaches 1. The expression simplifies to:
$L = 2 \times 1^2 \times \dfrac{a^2(x - \beta)^2}{4} = \dfrac{a^2(\alpha - \beta)^2}{2}$.
Answer: $\dfrac{a^2(\alpha-\beta)^2}{2}$
The limit is given by $L = \lim_{x \rightarrow \alpha} \dfrac{1 - \cos(a(x - \alpha)(x - \beta))}{(x - \alpha)^2}$.
Using the identity $1 - \cos \theta = 2 \sin^2(\theta/2)$, we have $L = \lim_{x \rightarrow \alpha} \dfrac{2 \sin^2(\dfrac{a(x - \alpha)(x - \beta)}{2})}{(x - \alpha)^2}$.
Using the standard limit $\lim_{u \rightarrow 0} \dfrac{\sin u}{u} = 1$, we multiply and divide by the square of the argument of the sine function:
$L = \lim_{x \rightarrow \alpha} 2 \left[ \dfrac{\sin(\dfrac{a(x - \alpha)(x - \beta)}{2})}{\dfrac{a(x - \alpha)(x - \beta)}{2}} \right]^2 \times \dfrac{a^2(x - \alpha)^2(x - \beta)^2}{4(x - \alpha)^2}$.
As $x \rightarrow \alpha$, the term in the square brackets approaches 1. The expression simplifies to:
$L = 2 \times 1^2 \times \dfrac{a^2(x - \beta)^2}{4} = \dfrac{a^2(\alpha - \beta)^2}{2}$.
Answer: $\dfrac{a^2(\alpha-\beta)^2}{2}$
Hence, the correct answer is option: 'B'.