Question

Let $\alpha$ and $\beta$ be the distinct roots of $\sf{ax^2\ +\ bx\ +\ c\ =\ 0}$ then,

$\displaystyle \sf{\lim_{x\ \to\ \alpha} \dfrac{1\ -\ \cos(ax^2\ +\ bx\ +\ c)}{(x\ -\ \alpha)^2}\ is\ :\ ?}$


• Give Proper Explanation!

• Don't Spam!
___________________

​Answer is $\sf{\dfrac{a^2}{2} ( \alpha\ -\ \beta)^2}$

Answer 1

Formula Applied :

$\bigstar\ \; \sf 1-cos\theta=2sin^2\dfrac{\theta}{2}$

$\bigstar\ \; \sf ax^2+bx+c=a(x-\alpha)(x-\beta)$

when  α , β are zeroes of polynomial .

$\bigstar\ \; \sf \displaystyle{\lim_{x\to 0}}\dfrac{sin\ x}{x}=1$

Explanation :

$\displaystyle \sf \lim_{x \to \alpha} \dfrac{1-cos(ax^2+bx+c)}{(x-\alpha)^2}\\\\\to \sf \lim_{x \to \alpha}\dfrac{2sin^2\left( \frac{ax^2+bx+c}{2}\right)}{(x-\alpha)^2}\\\\\to \sf \lim_{x\to \alpha} \dfrac{2sin^2 \left(\frac{a(x-\alpha)(x-\beta)}{2}\right)}{(x-\alpha)^2}$

$\to \displaystyle \sf \lim_{x\to \alpha} \dfrac{2sin^2 \left(\frac{a(x-\alpha)(x-\beta)}{2}\right)}{(x-\alpha)^2}\times \dfrac{\frac{a^2(x-\beta)^2}{4}}{\frac{a^2(x-\beta)^2}{4}}\\\\\to \sf \lim_{x\to \alpha} \dfrac{sin^2\left( \frac{a(x-\alpha)(x-\beta)}{2}\right)}{\left(\frac{a(x-\alpha )(x- \beta)}{2}\right)^2}\times \dfrac{2a^2(x-\beta)^2}{4}$

$\to \sf \displaystyle (1)^2\times \dfrac{a^2(\alpha - \beta)^2}{2}$

$\to \sf \pink{\dfrac{a^2}{2} (\alpha - \beta)^2}\ \; \bigstar$

Answer 2

Formula Applied :

$\bigstar\ \; \sf 1-cos\theta=2sin^2\dfrac{\theta}{2}★$

$\bigstar\ \; \sf ax^2+bx+c=a(x-\alpha)(x-\beta)★$

when α , β are zeroes of polynomial .

$\bigstar\ \; \sf \displaystyle{\lim_{x\to 0}}\dfrac{sin\ x}{x}=1★ </p><p>$

Explanation :

$\begin{gathered}\displaystyle \sf \lim_{x \to \alpha} \dfrac{1-cos(ax^2+bx+c)}{(x-\alpha)^2}\\\\\to \sf \lim_{x \to \alpha}\dfrac{2sin^2( \frac{ax^2+bx+c}{2})}{(x-\alpha)^2}\\\\\to \sf \lim_{x\to \alpha} \dfrac{2sin^2 (\frac{a(x-\alpha)(x-\beta)}{2})}{(x-\alpha)^2}\end{gathered}$

$\begin{gathered}\to \displaystyle \sf \lim_{x\to \alpha} \dfrac{2sin^2 (\frac{a(x-\alpha)(x-\beta)}{2})}{(x-\alpha)^2}\times \dfrac{\frac{a^2(x-\beta)^2}{4}}{\frac{a^2(x-\beta)^2}{4}}\\\\\to \sf \</p><p>lim_{x\to \alpha} \dfrac{sin^2( \frac{a(x-\alpha)(x-\beta)}{2})}{(\frac{a(x-\alpha )(x- \beta)}{2})^2}\times \dfrac{2a^2(x-\beta)^2}{4}\end{gathered}$

$\to \sf \displaystyle (1)^2\times \dfrac{a^2(\alpha - \beta)^2}{2}$

$\to \sf \pink{\dfrac{a^2}{2} (\alpha - \beta)^2}\ \; \bigstar$