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  • If are two distinct roots of the Q.E. <img alt="\mathrm[a x^2+b x+c=0, then \lim _{x \rightarr

If \alpha, \betaare two distinct roots of the Q.E. \mathrm[a x^2+b x+c=0, then \lim _{x \rightarrow a}\left[\frac{1-\cos \left(a x^2+b x+c\right)}{(x-a)^2}\right]] is equal to
 

Option: 1

0


Option: 2

\frac{1}{2}(\alpha-\beta)^2
 


Option: 3

\frac{1}{2} a^2(\alpha-\beta)^2

 


Option: 4

-\frac{1}{2} a^2(\alpha-\beta)^2


Answers (1)

best_answer

\mathrm{\alpha+\beta=-b / a }.............(1)

\mathrm{\alpha \beta=c / a }...................(2)

By L'1Hospital's Rule,

\mathrm{ L=\lim _{x \rightarrow a}\left\{\frac{\left[\sin \left(a x^2+b x+c\right)\right](2 a x+b)}{2(x-a)}\right\} }

   \mathrm{ =\lim _{x \rightarrow a}\left\{\begin{array}{l} \sin \left(a x^2+b x+c\right) \times 2 a+ \\ \frac{\cos \left(a x^2+b x+c\right)(2 a x+b)^2}{2} \end{array}\right\} }

   \mathrm{=\frac{1}{2}(2 a \alpha+b)^2 }             \mathrm{ {[\because b=-a \alpha-a \beta]} }

   \mathrm{ =\frac{1}{2}[2 a \alpha-a \alpha-a \beta]^2 \text {, } }        [From (1)]

   \mathrm{ =\frac{1}{2} a^2(\alpha-\beta)^2 }

Hence option 3 is correct.





 

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manish

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