Q39 / 75 · JEE Main 2025 (4 Apr Shift 1) · Mathematics

If $\quad \lim _{x \rightarrow 1^{+}} \frac{(x-1)(6+\lambda \cos (x-1))+\mu \sin (1-x)}{(x-1)^3}=-1$, where $\lambda, \mu \in \mathbb{R}$, then $\lambda+\mu$ is equal to

  • Correct answer
    A
    18
  • B
    20
  • C
    19
  • D
    17

Solution

$\begin{aligned} & \text { Put } x=1+h \\ & \lim _{h \rightarrow 0} \frac{h(6+\lambda \cosh )-\mu \sinh }{h^3}=-1 \\ & \lim _{h \rightarrow 0} \frac{h\left(6+\lambda\left(1-\frac{h^2}{2!}\right)\right)-\mu\left(h-\frac{h^3}{3!}\right)}{h^3}=-1 \\ & 6+\lambda-\mu=0 \text { and }-\frac{\lambda}{2}+\frac{\mu}{6}=-1 \\ & \lambda+\mu=18\end{aligned}$

Hence, the correct answer is option: 'A'.

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