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JEE Main 20254 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
4 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Limits

JEE Main 2025 · 4 April · Morning Shift · Mathematics · Limits

If _ x arrow 1^ + ((x-1)(6+λ (x-1))+μ (1-x))/((x-1)^3)=-1 , where...

If lim⁡x→1+(x−1)(6+λcos⁡(x−1))+μsin⁡(1−x)(x−1)3=−1\quad \lim _{x \rightarrow 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

Options

  1. A18
  2. B20
  3. C19
  4. D17

Correct answer

A. 18

Step-by-step solution

 Put x=1+hlim⁡h→0h(6+λcosh⁡)−μsinh⁡h3=−1lim⁡h→0h(6+λ(1−h22!))−μ(h−h33!)h3=−16+λ−μ=0 and  −λ2+μ6=−1λ+μ=18\begin{aligned} & \text{ Put } x=1+h \\ & \lim _{h \rightarrow 0} \frac{h(6+\lambda \cosh )-\mu \sinh }{h^3}=-1 \\ & \lim _{h \rightarrow 0} \frac{h\left(6+\lambda\left(1-\frac{h^2}{2!}\right)\right)-\mu\left(h-\frac{h^3}{3!}\right)}{h^3}=-1 \\ & 6+\lambda-\mu=0 \text{\text{ and } }-\frac{\lambda}{2}+\frac{\mu}{6}=-1 \\ & \lambda+\mu=18\end{aligned}
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