Interpretation of the question

The expression in the prompt is truncated, but from the options it is the standard problem where the dielectric constant between the plates varies continuously from $1$ to $K$ across the separation $d$.

So take the permittivity at distance $x$ from one plate as $\varepsilon(x)=\varepsilon_0\,k(x)$ where $k(x)$ varies linearly from $1$ to $K$: $k(x)=1+\frac{(K-1)x}{d}.$

Treat the capacitor as series combination of thin slabs

Consider a thin slab of thickness $dx$ at position $x$.

Its capacitance is $dC=\frac{\varepsilon(x)A}{dx},$ but since these slabs are stacked along the field direction, they are in series, so we add their inverses: $d\left(\frac{1}{C}\right)=\frac{dx}{\varepsilon(x)A}.$

Here area of square plate is $A=a^2.$

Therefore,

Evaluate the integral

Let $u=1+\frac{(K-1)x}{d} \quad \Rightarrow \quad dx=\frac{d}{K-1}\,du.$

Limits:

• when $x=0$, $u=1$
• when $x=d$, $u=K$

So, $\frac{1}{C}=\frac{1}{\varepsilon_0 a^2}\cdot \frac{d}{K-1}\int_1^K \frac{du}{u}$ $\frac{1}{C}=\frac{d}{\varepsilon_0 a^2 (K-1)}\ln K.$

Hence, $C=\frac{\varepsilon_0 a^2 (K-1)}{d\ln K}.$

Compare with options

The derived capacitance is $C=\frac{\varepsilon_0 a^2 (K-1)}{d\ln K}.$

But option B is $\frac{K\varepsilon_0 a^2}{d(K-1)}\ln K,$ which is not equal to the derived expression.

In fact, option B resembles neither the exact result nor its reciprocal in the correct form.

Conclusion

The correct capacitance for a linearly varying dielectric constant from $1$ to $K$ across separation $d$ is $\boxed{\frac{\varepsilon_0 a^2 (K-1)}{d\ln K}}.$

Therefore, none of the given options matches this result exactly.