JEE Mains · Maths · STD 11 - 12. limits
If \(\lim _{x \rightarrow 1^{+}} \frac{(x-1)(6+\lambda \cos (x-1))+\mu \sin (1-x)}{(x-1)^3}=-1\), where \(\lambda, \mu \in \mathbb{R}\), then \(\lambda+\mu\) is equal to
- A 18
- B 20
- C 19
- D 17
Answer & Solution
Correct Answer
(A) 18
Step-by-step Solution
Detailed explanation
\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…
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