Question:
Let $\alpha$ and $\beta$ be the distinct roots of $ax^2 + bx + c = 0$ , then $\displaystyle \lim_{x \to\alpha} \frac{1- \cos \left(ax^{2} + bx + c\right)}{\left(x-\alpha\right)^{2}} $ is equal to
Let $\alpha$ and $\beta$ be the distinct roots of $ax^2 + bx + c = 0$ , then $\displaystyle \lim_{x \to\alpha} \frac{1- \cos \left(ax^{2} + bx + c\right)}{\left(x-\alpha\right)^{2}} $ is equal to
Updated On: Jul 5, 2022
- $\frac{a^2}{2} (\alpha - \beta)^2$
- 0
- $\frac{-a^2}{2} (\alpha - \beta )^2$
- $\frac{1}{2} (\alpha - \beta )^2$
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The Correct Option is A
Solution and Explanation
Given limit $= \displaystyle \lim_{x \to\alpha} \frac{1- \cos a\left(x -\alpha\right)\left(x-\beta\right)}{\left(x -\alpha\right)^{2}} $
$= \displaystyle \lim_{x \to\alpha} \frac{2\sin^{2} \left(a \frac{\left(x -\alpha\right)\left(x - \beta\right)}{2}\right)}{\left(x -\alpha\right)^{2}} $
$=\displaystyle \lim_{x \to\alpha } \frac{2}{\left(x -\alpha\right)^{2}} \times\frac{\sin^{2}\left(a \frac{\left(x-\alpha\right)\left(x -\beta\right)}{2}\right)}{\frac{a^{2}\left(x-\alpha\right)^{2}\left(x-\beta\right)^{2}}{4}} \times\frac{a^{2} \left(x-\alpha\right)^{2} \left(x -\beta\right)^{2}}{4}$
$ = \frac{a^{2}\left(\alpha - \beta\right)^{2}}{2} $.
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Concepts Used:
Limits And Derivatives
Mathematically, a limit is explained as a value that a function approaches as the input, and it produces some value. Limits are essential in calculus and mathematical analysis and are used to define derivatives, integrals, and continuity.
Limits Formula:
Derivatives of a Function:
A derivative is referred to the instantaneous rate of change of a quantity with response to the other. It helps to look into the moment-by-moment nature of an amount. The derivative of a function is shown in the below-given formula.
Properties of Derivatives:
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