If alpha, beta are the roots of the equation | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$, we can write $ax^2 + bx + c = a(x - \alpha)(x - \beta)$.We need

Vedclass · Accepted Answer

$\frac{a^2}{2}(\alpha - \beta)^2$

Vedclass · Answer

(D) Given the quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$, we can write $ax^2 + bx + c = a(x - \alpha)(x - \beta)$.We need to evaluate the limit $L = \lim_{x ightarrow \beta} \frac{1 - \cos(a(x - \alpha)(x - \beta))}{(x - \beta)^2}$.Using the identity $1 - \cos(	heta) = 2 \sin^2(\frac{	heta}{2})$, we get:$L = \lim_{x ightarrow \beta} \frac{2 \sin^2(\frac{a(x - \alpha)(x - \beta)}{2})}{(x - \beta)^2}$.Multiplying and dividing by $(\frac{a(x - \alpha)}{2})^2$, we have:$L = \lim_{x}$ ${ightarrow \beta} 2 \left[ \frac{\sin(\frac{a(x - \alpha)(x - \beta)}{2})}{\frac{a(x - \alpha)(x - \beta)}{2}} ight]^2 	imes \frac{a^2(x - \alpha)^2}{4}$.As $x ightarrow \beta$, the term in the bracket approaches $1$.Thus, $L = 2 	imes 1^2 	imes \frac{a^2(\beta - \alpha)^2}{4} = \frac{a^2(\alpha - \beta)^2}{2}$.

If $\alpha, \beta$ are the roots of the equation $ax^2 + bx + c = 0$, then $\lim_{x \rightarrow \beta} \frac{1 - \cos(ax^2 + bx + c)}{(x - \beta)^2}$ is

Similar Questions

If $\lim _{x \rightarrow 0} \frac{e^{a x}-\cos (b x)-\frac{c x e^{-c x}}{2}}{1-\cos (2 x)}=17$,then $5 a^2+b^2$ is equal to

If $f(x) = \begin{cases} 3ax - 2b, & x > 1 \\ ax + b + 1, & x < 1 \end{cases}$ and $\lim_{x \rightarrow 1} f(x)$ exists,then the relation between $a$ and $b$ is

If $\lim _{x \rightarrow \infty}\left(\frac{x^2+1}{x+1}-a x-b\right)=0$, where $a, b \in R$, then:

If $\lim _{x \rightarrow 0}\left\{1+x \log \left(1+a^2\right)\right\}^{1 / x}=2 a \sin ^2 \theta$, where $a>0$ and $\theta \in R$, then:

If $\lim _{x \rightarrow 1} \frac{x^2-ax+b}{x-1}=5$,then $(a+b)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module