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Limits Continuity and Differentiability question

2023 · 29 Jan · Shift 1 · Q40
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  5. /2023 · 29 Jan · Shift 1 · Q40

Limits Continuity and Differentiability question

2023 · 29 Jan · Shift 1 · Q40

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
Let x=2x=2x=2 be a root of the equation x2+px+q=0x^2+px+q=0x2+px+q=0 and f(x)={1−cos⁡(x2−4px+q2+8q+16)(x−2p)4,xe2p0,x=2pf(x) = \left\{ {\begin{matrix} {{{1 - \cos ({x^2} - 4px + {q^2} + 8q + 16)} \over {{{(x - 2p)}^4}}},} & {x e 2p} \\ {0,} & {x = 2p} \\ \end{matrix} } \right.f(x)={(x−2p)41−cos(x2−4px+q2+8q+16)​,0,​xe2px=2p​ Then lim⁡x→2p+[f(x)]\mathop {\lim }\limits_{x \to 2{p^ + }} [f(x)]x→2p+lim​[f(x)], where [.]\left[ . \right][.] denotes greatest integer function, is
  1. A
    2
  2. B
    1
  3. C
    0
  4. D
    −1-1−1
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