Let x=2 be a root of the equation x²+px+q=0 and f(x) = \ (1 - cos (x² - 4px + q² + 8q + 16))/((x - 2p)⁴), x ≠ 2p 0, x = 2p. Then lim x → 2p⁺ [f(x)], where [. ] denotes greatest integer function, is

Limit, Continuity and Differentiability: JEE Main 2023 Mathematics question

JEE Main 2023 · 29 Jan, Morning shift · Q4297

JEE Main 202329 Jan, Morning shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002

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