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Question Number 122834 by liberty last updated on 20/Nov/20

I f {a x}^{2} + b x + c = 0 h a s t h e r o o t s a r e

p a n d q t h e n t h e v a l u e o f \lim_{x \to p} \frac{1 - \cos ({a x}^{2} + b x + c)}{{(x - p)}^{2}}

i s___

Commented by bemath last updated on 20/Nov/20

\lim_{x - p \to 0} \frac{1 - \cos ({a x}^{2} + b x + c)}{{(x - p)}^{2}} =

\lim_{x - p \to 0} \frac{2 \sin^{2} (\frac{{a x}^{2} + b x + c}{2})}{{(x - p)}^{2}} =

2 \times \lim_{x - p \to 0} {(\frac{\sin (\frac{a (x - p) (x - q)}{2})}{x - p})}^{2} =

2 \times \frac{1}{4} a^{2} {(p - q)}^{2} = \frac{a^{2} {(p - q)}^{2}}{2}

Answered by $@y@m last updated on 20/Nov/20

\frac{{(p - q)}^{2}}{2}

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