If lim x→1^+ (x-1)(6+λcos(x-1))+μsin(1-x)/(x-1)^3 = -1, where λ, μ ∈ R, then λ+μ is equal to

Question

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

(1) 18

(2) 20

(3) 19

(4) 17

Answer 1

Correct option is: (1) 18  

Put \( \mathrm{x}=1+\mathrm{h}\)

\(\lim\limits _{h \rightarrow 0} \frac{h(6+\lambda \cosh )-\mu \sinh }{h^{3}}=-1\)

\(\lim\limits _{h \rightarrow 0} \frac{h\left(6+\lambda\left(1-\frac{h^{2}}{2!}\right)\right)-\mu\left(h-\frac{h^{3}}{3!}\right)}{h^{3}}=-1\)

\(6+\lambda-\mu=0\) and \(-\frac{\lambda}{2}+\frac{\mu}{6}=-1\)

\(\lambda+\mu=18\)