Differentiation Question 494
Question: Let $ \alpha $ and $ \beta $ be the roots of $ ax^{2}+bx+c=0. $ Then $ \underset{x\to \alpha }{\mathop{\lim }},\frac{1-\cos (ax^{2}+bx+c)}{{{(x-\alpha )}^{2}}} $ is equal to:
Options:
A) 0
B) $ \frac{1}{2}{{(\alpha -\beta )}^{2}} $
C) $ \frac{a^{2}}{2}{{(\alpha -\beta )}^{2}} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ \underset{x\to \alpha }{\mathop{\lim }},\frac{1-\cos (ax^{2}+bx+c)}{{{(x-\alpha )}^{2}}} $ $ =\underset{x\to \alpha }{\mathop{\lim }},\frac{2{{\sin }^{2}}( \frac{ax^{2}+bx+c}{2} )}{{{(x-\alpha )}^{2}}} $ $ =\underset{x\to \alpha }{\mathop{\lim }},2{{[ \frac{\sin ,\frac{a(x-\alpha )(x-\beta )}{2}}{\frac{a(x-\alpha )(x-\beta )}{2}} ]}^{2}}\times \frac{a^{2}{{(x-\beta )}^{2}}}{4} $ $ =\frac{a^{2}}{2}{{(\alpha -\beta )}^{2}} $ [using $ ax^{2}+bx+c=a(x-\alpha )(x-\beta ) $ ]
BETA
