Let $x=2$ be a root of the equation $x^2+px+q=0$ and $$f(x) = \left\{ {\matrix{ {{{1 - \cos ({x^2} - 4px + {q^2} + 8q + 16)} \over {{{(x - 2p)}^4}}},} & {x \ne 2p} \cr {0,} & {x = 2p} \cr } } \right.$$
Then $\mathop {\lim }\limits_{x \to 2{p^ + }} [f(x)]$, where $\left[ . \right]$ denotes greatest integer function, is
Solution
$\lim \limits_{x \rightarrow 2^{+}}\left(\frac{1-\cos \left(x^{2}-4 p x+q^{2}+8 q+16\right)}{\left(x^{2}-4 p x+q^{2}+8 q+16\right)^{2}}\right)\left(\frac{\left(x^{2}-4 p x+q^{2}+8 q+16\right)^{2}}{(x-2 p)^{2}}\right)$
<br/><br/>
$\lim \limits_{h \rightarrow 0} \frac{1}{2}\left(\frac{(2 p+h)^{2}-4 p(2 p+h)+q^{2}+82+16}{h^{2}}\right)^{2}=\frac{1}{2}$
<br/><br/>
Using L'Hospital's
<br/><br/>
$\lim _{x \rightarrow 2 p^{+}}[f(x)]=0$
About this question
Subject: Mathematics · Chapter: Limits, Continuity and Differentiability · Topic: Limits and Standard Results
This question is part of PrepWiser's free JEE Main question bank. 162 more solved questions on Limits, Continuity and Differentiability are available — start with the harder ones if your accuracy is >70%.