JEE Main 2025 — 4 April, Morning Shift — Question 33

$\lim _{x \rightarrow 1^{+}} \frac{(x-1)(6+\lambda \cos (x-1))+\mu \sin (1-x)}{(x-1)^{3}}=-1$

Let $x-1=t$

$\lim _{t \rightarrow 0^{+}} \frac{6 t+\lambda t \cos t-\mu \sin t}{t^{3}}=-1$

$=\lim _{t \rightarrow 0^{+}} \frac{6 t+\lambda t\left(1-\frac{t^{2}}{2!}+\frac{t^{4}}{4!}+\cdots\right)-\mu\left(t-\frac{t^{3}}{3!}+\cdots\right)}{t^{3}}=-1$

$=\lim _{t \rightarrow 0^{+}} \frac{t(6+\lambda-\mu)+t^{3}\left(-\frac{\lambda}{2}+\frac{\mu}{6}\right)+\cdots}{t^{3}}=-1$

$\lambda-\mu+6=0 …(i)$

$\frac{\mu}{6}-\frac{\lambda}{2}=-1 …(ii)$

Solving (i) and (ii)

$\lambda=6, \mu=12$

$\lambda+\mu=18$