Question

Let $x=2$ be a root of the equation $x^2+p x+q=0$ and $f(x)=\begin{cases}\frac{1-\cos \left(x^2-4 p x+q^2+8 q+16\right)}{(x-2 p)^4}, & x \neq 2 p \\ 0, & , x=2 p\end{cases}$ Then $\displaystyle\lim _{x \rightarrow 2 p^{+}}[f(x)]$, where [.] denotes greatest integer function, is

• A. 2
• B. 1
• C. $-1$
• D. 0

Answer 1

Answer: (D)

$\displaystyle \lim _{x \rightarrow p ^{+}}\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) $
$\displaystyle\lim _{h \rightarrow 0} \frac{1}{2}\left(\frac{(2 p+h)^2-4 p(2 p+h)+q^2+82+16}{h^2}\right)^2=\frac{1}{2}$
Using L'Hospital's
$\displaystyle\lim _{x \rightarrow 2 p^{+}}[f(x)]=0$