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Step-by-Step Solution
Step 1: Recognize the Capacitor Configuration
We have a parallel plate capacitor of plate area $A$ and separation $d$. The capacitor is filled with a dielectric whose permittivity varies with position $x$. Such a setup can be treated as infinitesimal layers of capacitance in series.
Step 2: Set Up the Expression for Total Capacitance
For a parallel plate capacitor with a small thickness $dx$, if the local permittivity is $\varepsilon(x)$, the infinitesimal capacitance $dC$ is given by
$$
dC = \frac{\varepsilon(x)\,A}{dx}.
$$
Since these infinitesimal capacitors are in series, the total reciprocal capacitance is found by integrating over $0$ to $d$:
$$
\frac{1}{C} = \int_0^d \frac{dx}{\varepsilon(x)\,A}.
$$
Step 3: Apply the Given Permittivity Function
The problem states:
For $0 \le x \le \tfrac{d}{2}$:
$$
\varepsilon(x) = \varepsilon_0 + kx,
$$
and for $\tfrac{d}{2} \le x \le d$:
$$
\varepsilon(x) = \varepsilon_0 + k(d - x).
$$
This arrangement splits the integral from $0$ to $d$ into two parts, but notice the symmetry when we do the substitution $y = d - x$ for the second half. This results effectively in doubling the integral from $0$ to $d/2$ for $\frac{1}{\varepsilon(x)}$.
Hence,
$$
\frac{1}{C}
= \frac{2}{A}\int_0^{\frac{d}{2}} \frac{dx}{\varepsilon_0 + kx}.
$$
Step 4: Integrate Over the Specified Range
We now compute the integral:
$$
\int_0^{\frac{d}{2}} \frac{dx}{\varepsilon_0 + kx}.
$$
Notice that
$$
\int \frac{dx}{\varepsilon_0 + kx}
= \frac{1}{k}\ln(\varepsilon_0 + kx).
$$
So,
$$
\int_0^{\frac{d}{2}} \frac{dx}{\varepsilon_0 + kx}
= \left.\frac{1}{k}\ln(\varepsilon_0 + kx)\right|_0^{\frac{d}{2}}
= \frac{1}{k}\biggl[\ln\bigl(\varepsilon_0 + k\frac{d}{2}\bigr)
- \ln(\varepsilon_0)\biggr].
$$
Step 5: Combine Results to Get the Capacitance
Putting it all together,
$$
\frac{1}{C}
= \frac{2}{A}\times\frac{1}{k}\Bigl[\ln\Bigl(\varepsilon_0 + \frac{kd}{2}\Bigr) - \ln(\varepsilon_0)\Bigr]
= \frac{2}{A\,k}\ln\Bigl(\frac{\varepsilon_0 + \frac{kd}{2}}{\varepsilon_0}\Bigr).
$$
Invert to find $C$:
$$
C
= \frac{A\,k}{2\,\ln\Bigl(\frac{\varepsilon_0 + \frac{kd}{2}}{\varepsilon_0}\Bigr)}
= \frac{kA}{2\,\ln\Bigl(\frac{2\,\varepsilon_0 + kd}{2\,\varepsilon_0}\Bigr)}.
$$
This matches the given correct answer.
Final Answer
$$
C = \frac{kA}{2\,\ln\Bigl(\frac{2\,\varepsilon_0 + kd}{2\,\varepsilon_0}\Bigr)}.
$$
