In $\triangle A B C$, with usual notations, $a \cos \mathrm{~B}=\mathrm{b} \cos \mathrm{A}, a \cos \mathrm{C} \neq \mathrm{c} \cos \mathrm{A}$ then $\mathrm{A}(\triangle \mathrm{ABC})$ _____ sq. units.

The area of triangle $\triangle ABC$ is determined by analyzing the given conditions: $a \cos B = b \cos A$ and $a \cos C \ne c \cos A$.

From the equation $a \cos B = b \cos A$, substituting the cosine rule $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$ and $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ yields:

$a \left( \frac{a^2 + c^2 - b^2}{2ac} \right) = b \left( \frac{b^2 + c^2 - a^2}{2bc} \right)$

Simplifying gives $a^2 + c^2 - b^2 = b^2 + c^2 - a^2$, leading to $a^2 = b^2$. Since side lengths are positive, $a = b$, so the triangle is isosceles with sides $a$ and $b$ equal.

The inequality $a \cos C \ne c \cos A$ is examined with $a = b$. Substituting $\cos C = \frac{2b^2 - c^2}{2b^2}$ and $\cos A = \frac{c}{2b}$ gives $b \left( \frac{2b^2 - c^2}{2b^2} \right) \ne c \left( \frac{c}{2b} \right)$, which simplifies to $2b^2 - c^2 \ne c^2$ and thus $b^2 \ne c^2$, so $b \ne c$.

To compute the area, let $c$ be the base and $h$ be the altitude from vertex $C$ to base $AB$, which bisects $AB$ into segments of length $\frac{c}{2}$. By the Pythagorean theorem in $\triangle AMC$, $h^2 + \left( \frac{c}{2} \right)^2 = b^2$, so $h = \frac{\sqrt{4b^2 - c^2}}{2}$.

The area is $\frac{1}{2} \cdot c \cdot h = \frac{1}{2} \cdot c \cdot \frac{\sqrt{4b^2 - c^2}}{2} = \frac{c}{4} \sqrt{4b^2 - c^2}$. Since $a = b$, the expression becomes $\frac{c}{4} \sqrt{4a^2 - c^2}$.

This matches option B, so the answer is $\boxed{B}$.