If $0<\mathrm{x}<\frac{\mathrm{\pi }}{4}\text{ and }\cos \mathrm{x}+\sin \mathrm{x}=\frac{5}{4}\text{, }$then the value of $16(\cos \mathrm{x}-\sin \mathrm{x}{)}^2$ is

$\begin{array}{l}\text{ Let }\mathrm{y}=\cos \mathrm{x}-\sin \mathrm{x}(\mathrm{y}>0)\\ {\mathrm{y}}^2=1-2\sin \mathrm{x}\cdot \cos \mathrm{x}\\ \cos \mathrm{x}+\sin \mathrm{x}=\frac{5}{4}\\ \therefore 1+2\sin \mathrm{xcos}\mathrm{x}=\frac{25}{16}\\ \text{ or }2\sin \mathrm{x}\cdot \cos \mathrm{x}=\frac{9}{16}\end{array}$

$\text{ Now }{y}^2=1-2\sin x\cos x=1-\frac{9}{16}=\frac{7}{16}$