\(\underline{\mathrm{a}}=\underline{\mathrm{u}}-\underline{\mathrm{v}}, \underline{\mathrm{b}}=\underline{\mathrm{u}}+\underline{\mathrm{v}},|\underline{\mathrm{u}}|=|\underline{\mathrm{u}}|^{2}\) and \(|\underline{\mathrm{u}}|=|\underline{\mathrm{v}}|=2\) find \(|\underline{a} \times \underline{b}|=\) (b) \(\left.\sqrt{[} 4-(\underline{\mathrm{u}} \cdot \underline{\mathrm{v}})^{2}\right]\) (c) \(\sqrt{\left[16-(\underline{\mathrm{u}} \cdot \underline{\mathrm{v}})^{2}\right]}\) (d) \(\sqrt{\left[4-(\underline{\mathrm{u}} \cdot \underline{\mathrm{v}})^{2}\right]}\)

We have \(\underline{\mathrm{a}}=\underline{\mathrm{u}}-\underline{\mathrm{v}}\) and \(\underline{\mathrm{b}}=\underline{\mathrm{u}}+\underline{\mathrm{v}}\).

We have \(\underline{a}\times\underline{b} = (\underline{u}-\underline{v})\times(\underline{u}+\underline{v})\). Expanding this expression using properties of cross product, we get \[ \underline{a}\times\underline{b} = \underline{u}\times\underline{u} + \underline{u}\times\underline{v} - \underline{v}\times\underline{u} - \underline{v}\times\underline{v}. \] Remember that the cross product of a vector with itself is zero. Also, the cross product is anti-commutative, which means \(\underline{v}\times\underline{u}=-\underline{u}\times\underline{v}\). So, we can simplify the expression as: \[ \underline{a}\times\underline{b} = 2(\underline{u}\times\underline{v}). \]

Now, we need to find the magnitude of the cross product, which will be \[ |\underline{a}\times\underline{b}| = 2|\underline{u}\times\underline{v}|. \] Recall that \(|\underline{u}\times\underline{v}|=|\underline{u}||\underline{v}|\sin\theta\). We are given that \(|\underline{\mathrm{u}}|=|\underline{\mathrm{v}}|=2\), so we get \[ |\underline{a}\times\underline{b}| = 2(2)(2)\sin\theta = 8\sin\theta. \]

Now let's compare our result with the given options (b), (c), and (d). Notice that each option contains a term of the form \((\underline{u}\cdot\underline{v})^2\). We can use the fact that \(|\underline{u}\times\underline{v}|^2=|\underline{u}|^2|\underline{v}|^2-\left(\underline{u}\cdot\underline{v}\right)^2\) to eliminate this term and find the correct answer. First, we compute \(|\underline{u}\times\underline{v}|^2\): \[ |\underline{u}\times\underline{v}|^2 = \left(8\sin\theta\right)^2=64\sin^2\theta. \] Plugging back the values of \(|\underline{u}|\) and \(|\underline{v}|\), we get \[ 64\sin^2\theta=4^2(2^2)-(\underline{u}\cdot\underline{v})^2. \] Comparing this expression with the given options, we see that it matches option (c): \[ |\underline{a}\times\underline{b}| = \sqrt{\left[16-(\underline{\mathrm{u}} \cdot \underline{\mathrm{v}})^{2}\right]}. \]