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JEE Main 2005
Exam
JEE Main
Year
2005
Subject
Mathematics
Chapter
Limits

JEE Main 2005 · Mathematics · Limits

Let α and β be the distinct roots of a x^2+b x+c=0 , then _ x arrow...

Let αand\alpha and\betabethedistinctrootsofbe the distinct roots ofa x^2+b x+c=0,then, then\lim _{x \rightarrow \alpha} \frac{1-\cos \left(a x^2+b x+c\right)}{(x-\alpha)^2} is equal to

Options

  1. Aa22(α−β)2\frac{a^2}{2}(\alpha-\beta)^2
  2. B0
  3. C−a22(α−β)2-\frac{a^2}{2}(\alpha-\beta)^2
  4. D12(α−β)2\frac{1}{2}(\alpha-\beta)^2

Correct answer

A. a22(α−β)2\frac{a^2}{2}(\alpha-\beta)^2

Step-by-step solution

 Given limit =lim⁡x→α1−cos⁡a(x−α)(x−β)(x−α)2=lim⁡x→α2sin⁡2(a(x−α)(x−β)2)(x−α)2=lim⁡x→α2(x−α)2×sin⁡2(a(x−α)(x−β)2)a2(x−α)2(x−β)2×a2(x−α)2(x−β)24=a2(α−β)22.\begin{aligned}& \text{ Given limit }=\lim _{x \rightarrow \alpha} \frac{1-\cos a(x-\alpha)(x-\beta)}{(x-\alpha)^2}=\lim _{x \rightarrow \alpha} \frac{2 \sin ^2\left(a \frac{(x-\alpha)(x-\beta)}{2}\right)}{(x-\alpha)^2}\\ & =\lim _{x \rightarrow \alpha} \frac{2}{(x-\alpha)^2} \times \frac{\sin ^2\left(a \frac{(x-\alpha)(x-\beta)}{2}\right)}{a^2(x-\alpha)^2(x-\beta)^2} \times \frac{a^2(x-\alpha)^2(x-\beta)^2}{4}\\ & =\frac{a^2(\alpha-\beta)^2}{2} . \end{aligned}
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