MCQMediumJEE Main 2025 · 4 April, Shift 1Limits

Mathematics Question from JEE Main 2025 · 4 April, Shift 1

If lim⁡x→1+(x−1)(6+λcos⁡(x−1))+μsin⁡(1−x)(x−1)3=−1\lim_{x \to 1^{+}} \frac{(x-1)(6+\lambda \cos(x-1)) + \mu \sin(1-x)}{(x-1)^3} = -1, where λ,μ∈R\lambda, \mu \in \mathbb{R}, then λ+μ\lambda + \mu is equal to

  • A

    1818

  • B

    2020

  • C

    1919

  • D

    1717

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