Question

The value of ${lim}_{x\to \beta }\frac{1-\cos (a{x}^2+bx+c)}{(x^2-{\beta }^2)}$ where $\alpha ,\beta$ are the distinct roots of $a{x}^2+bx+c=0$ is

• A. ${(\alpha -\beta )}^2$
• B. $\frac{{(\alpha -\beta )}^2}{2}$
• C. ${\left(\frac{a(\alpha -\beta )}{2}\right)}^2$
• D. none of these

Answer 1

The correct option is A ${(\alpha -\beta )}^2$

${lim}_{x\to \beta }\frac{1-\cos (a{x}^2+bx+c)}{(x^2-{\beta }^2)}$

This is in $\frac{0}{0}$ form. Apply L'Hospital's rule.

$\Rightarrow {lim}_{x\to \beta }\frac{\sin (a{x}^2+bx+c)(2ax+b)}{2x}$

This is in $\frac{0}{0}$ form. Apply L'Hospital's rule.

$\Rightarrow {lim}_{x\to \beta }\frac{\sin (a{x}^2+bx+c)\left(2a\right)+(2ax+b{)}^2\cos (a{x}^2+bx+c)}{2}$

Substitute $x=\beta$

$\Rightarrow \frac{(2a+\beta {)}^2}{2}$

$\Rightarrow \frac{a^2(2\beta +\frac{b}{a}{)}^2}{2}$

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

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

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

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

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