If the function $$f(x) = \begin{cases} \frac{1}{x}\log_e\left(\frac{1+\frac{x}{b}}{1-\frac{x}{b}}\right), & x < 0 \\ k, & x = 0 \\ \frac{\cos^2 x - \sin^2 x - 1}{\sqrt{x^2+1}-1}, & x > 0 \end{cases}$$ is continuous at $$x = 0$$, then $$\frac{1}{a} + \frac{1}{b} + \frac{4}{k}$$ is equal to:

We have been given a piece-wise definition

$$ f(x)= \begin{cases} \dfrac1x\;\log_e\!\left(\dfrac{1+\dfrac{x}{a}}{1-\dfrac{x}{b}}\right), & x<0\\[10pt] k, & x=0\\[8pt] \dfrac{\cos ^2x-\sin ^2x-1}{\sqrt{x^{2}+1}-1}, & x>0 \end{cases} $$

and are told that the function is continuous at the point $$x=0$$. For continuity we must have

$$ \lim_{x\to 0^-}f(x)=f(0)=\lim_{x\to 0^+}f(x)=k . $$

We shall therefore compute the left-hand limit (LHL) and the right-hand limit (RHL) separately and then equate them to $$k$$.

Left-hand limit

The expression to be handled for $$x<0$$ is

$$ \frac{1}{x}\, \log_e\!\left(\frac{1+\dfrac{x}{a}}{1-\dfrac{x}{b}}\right) . $$

For small $$x,$$ use the series expansion

$$ \log_e(1+u)=u-\frac{u^{2}}{2}+\frac{u^{3}}{3}-\cdots ,\qquad |u|<1 . $$

Put

$$ u=\frac{x}{a}\quad\text{and}\quad v=\frac{x}{b}. $$

Then

$$ \log_e(1+u)=\frac{x}{a}-\frac{x^{2}}{2a^{2}}+O(x^{3}),\qquad \log_e(1-v)=-(v+\frac{v^{2}}{2}+O(v^{3}))=-\frac{x}{b}-\frac{x^{2}}{2b^{2}}+O(x^{3}). $$

Therefore

$$ \log_e\!\Bigl(\frac{1+\tfrac{x}{a}}{1-\tfrac{x}{b}}\Bigr) =\Bigl(\frac{x}{a}-\frac{x^{2}}{2a^{2}}\Bigr) -\Bigl(-\frac{x}{b}-\frac{x^{2}}{2b^{2}}\Bigr) +O(x^{3}) =x\!\left(\frac1a+\frac1b\right)+O(x^{2}). $$

Dividing by $$x$$ and letting $$x\to 0^-$$ gives

$$ \lim_{x\to 0^-}f(x)=\frac1a+\frac1b . $$

Right-hand limit

For $$x>0$$ we have

$$ \frac{\cos ^2x-\sin ^2x-1}{\sqrt{x^{2}+1}-1}. $$

Recall the double-angle identity

$$ \cos ^2x-\sin ^2x=\cos 2x . $$

Hence the numerator is $$\cos 2x-1$$. Using the Maclaurin series

$$ \cos 2x=1-\frac{(2x)^{2}}{2}+\frac{(2x)^{4}}{24}-\cdots =1-2x^{2}+\frac{2x^{4}}{3}-\cdots , $$

we obtain

$$ \cos 2x-1=-2x^{2}+O(x^{4}). $$

For the denominator employ the binomial expansion of the square root:

$$ \sqrt{1+x^{2}}=1+\frac{x^{2}}{2}-\frac{x^{4}}{8}+O(x^{6}), $$

so

$$ \sqrt{1+x^{2}}-1=\frac{x^{2}}{2}-\frac{x^{4}}{8}+O(x^{6}). $$

Now form the quotient:

$$ \frac{\cos 2x-1}{\sqrt{1+x^{2}}-1} =\frac{-2x^{2}+O(x^{4})}{\dfrac{x^{2}}{2}+O(x^{4})} =\frac{-2+O(x^{2})}{\tfrac12+O(x^{2})}\;. $$

Letting $$x\to 0^+$$ we get

$$ \lim_{x\to 0^+}f(x)=-4 . $$

Equating the limits to $$k$$

$$ k=\frac1a+\frac1b=-4. $$

Required expression

$$ \frac1a+\frac1b+\frac4k =k+\frac4k =-4+\frac4{-4} =-4-1 =-5 . $$

That value corresponds to Option D.

Hence, the correct answer is Option D.