JEE MAIN - Mathematics (2023 - 29th January Morning Shift - No. 19)
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
2
1
0
$$-1$$
Explanation
$\lim \limits_{x \rightarrow 2^{+}}\left(\frac{1-\cos \left(x^{2}-4 p x+q^{2}+8 q+16\right)}{\left(x^{2}-4 p x+q^{2}+8 q+16\right)^{2}}\right)\left(\frac{\left(x^{2}-4 p x+q^{2}+8 q+16\right)^{2}}{(x-2 p)^{2}}\right)$
$\lim \limits_{h \rightarrow 0} \frac{1}{2}\left(\frac{(2 p+h)^{2}-4 p(2 p+h)+q^{2}+82+16}{h^{2}}\right)^{2}=\frac{1}{2}$
Using L'Hospital's
$\lim _{x \rightarrow 2 p^{+}}[f(x)]=0$
$\lim \limits_{h \rightarrow 0} \frac{1}{2}\left(\frac{(2 p+h)^{2}-4 p(2 p+h)+q^{2}+82+16}{h^{2}}\right)^{2}=\frac{1}{2}$
Using L'Hospital's
$\lim _{x \rightarrow 2 p^{+}}[f(x)]=0$
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