Question

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

• A. 2
• B. 1
• C. $-1$
• D. 0

Answer 1

Answer: (D)

$\underset{x\to p^{+}}{\lim }\left(\frac{1-\cos \left(x^2-4px+q^2+8q+16\right)}{{\left(x^2-4px+q^2+8q+16\right)}^2}\right)\left(\frac{{\left(x^2-4px+q^2+8q+16\right)}^2}{(x-2p{)}^2}\right)$
$\underset{h\to 0}{\lim }\frac{1}{2}{\left(\frac{(2p+h{)}^2-4p(2p+h)+q^2+82+16}{h^2}\right)}^2=\frac{1}{2}$
Using L'Hospital's
$\underset{x\to 2p^{+}}{\lim }[f(x)]=0$