If $$\displaystyle\lim_{x \to 1} \dfrac{(x - 1)(6 + \lambda \cos(x - 1)) + \mu \sin(1 - x)}{(x - 1)^3} = -1$$, where $$\lambda, \mu \in \mathbb{R}$$, then $$\lambda + \mu$$ is equal to

Let $$x - 1 = h$$ so that $$h \rightarrow 0$$ when $$x \rightarrow 1$$.
Rewrite every term in powers of $$h$$.

Because $$\cos(-h)=\cos h$$ and $$\sin(1-x)=\sin(-h)=-\sin h$$, the numerator becomes

$$N(h)=h\bigl(6+\lambda\cos h\bigr)-\mu\sin h$$.

Use the standard Maclaurin expansions valid for $$h \rightarrow 0$$:
$$\cos h = 1-\dfrac{h^{2}}{2}+O(h^{4}), \qquad \sin h = h-\dfrac{h^{3}}{6}+O(h^{5}).$$

Substituting these in $$N(h)$$ gives

$$\begin{aligned} N(h) &= h\Bigl[6+\lambda\Bigl(1-\dfrac{h^{2}}{2}+O(h^{4})\Bigr)\Bigr] -\mu\Bigl(h-\dfrac{h^{3}}{6}+O(h^{5})\Bigr) \\[4pt] &= 6h+\lambda h-\dfrac{\lambda}{2}h^{3}-\mu h+\dfrac{\mu}{6}h^{3}+O(h^{5}). \end{aligned}$$

Collect like powers of $$h$$:

$$N(h)=\underbrace{\bigl(6+\lambda-\mu\bigr)h}_{\text{order }h} +\underbrace{\Bigl(-\dfrac{\lambda}{2}+\dfrac{\mu}{6}\Bigr)h^{3}}_{\text{order }h^{3}} +O(h^{5}).$$

For the limit $$\dfrac{N(h)}{h^{3}}$$ to exist and be finite, the lower-order term in $$h$$ must vanish. Hence

$$6+\lambda-\mu=0 \quad\Longrightarrow\quad \mu = 6+\lambda.$$(1)

Now the limit simplifies to

$$\lim_{h\to 0}\dfrac{N(h)}{h^{3}} = -\dfrac{\lambda}{2}+\dfrac{\mu}{6}.$$

The problem states that this limit equals $$-1$$, so

$$-\dfrac{\lambda}{2}+\dfrac{\mu}{6}=-1.$$ (2)

Substitute $$\mu=6+\lambda$$ from (1) into (2):

$$-\dfrac{\lambda}{2}+\dfrac{6+\lambda}{6}=-1.$$

Multiply every term by $$6$$ to clear denominators:

$$-3\lambda + 6 + \lambda = -6.$$

Simplify:

$$-2\lambda + 6 = -6 \;\;\Longrightarrow\;\; -2\lambda = -12 \;\;\Longrightarrow\;\; \lambda = 6.$$

Using (1), $$\mu = 6 + \lambda = 6 + 6 = 12.$$

Therefore, $$\lambda + \mu = 6 + 12 = 18.$$

Hence the required value is 18, which corresponds to Option A.