🛠️ JEE➗ Maths

Let \(x=2\) be a root of the equation \({x}^{2}+px+q=0\) and \(f\left(x\right)=\left{\begin{matrix}\frac{1-\cos \left({x…

Q1

Let \(x=2\) be a root of the equation \({x}^{2}+px+q=0\) and

\(f\left(x\right)=\left{\begin{matrix}\frac{1-\cos \left({x}^{2}-4px+{q}^{2}+8q+16\right)}{(x-2p{)}^{4}}, & x\neq 2p \\ 0 & x=2p\end{matrix}\right.\)

Then \(\lim _{x \rightarrow 2 p^{+}}[f(x)]\) where [.] denotes greatest integer function, is

[JEE Main 2023, 29 Jan (Shift 1)]

a

2

b

1

c

0

d

-1

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