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Limits Continuity and Differentiability question

2025 · 4 Apr · Shift 1 · Q42
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  5. /2025 · 4 Apr · Shift 1 · Q42

Limits Continuity and Differentiability question

2025 · 4 Apr · Shift 1 · Q42

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
If lim⁡x→1+(x−1)(6+λcos⁡(x−1))+μsin⁡(1−x)(x−1)3=−1\lim_{x \rightarrow 1^{+}} \frac{(x-1)(6+\lambda \cos (x-1))+\mu \sin (1-x)}{(x-1)^3}=-1limx→1+​(x−1)3(x−1)(6+λcos(x−1))+μsin(1−x)​=−1, where λ,μ∈R\lambda, \mu \in \mathbb{R}λ,μ∈R, then λ+μ\lambda+\muλ+μ is equal to
  1. A
    20
  2. B
    19
  3. C
    18
  4. D
    17
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