If one zero of the quadratic polynomial $(\mathrm{k}-1) \mathrm{x}^{2}-\mathrm{kx}+1$ is $-4$, then the value of $\mathrm{k}$ is (a) $\frac{-5}{4}$ (b) $\frac{5}{4}$ (c) $\frac{-4}{3}$ (d) $\frac{4}{3}$

The correct option is option (b) $\frac{5}{4}$

Since $-4$ is a zero of $(k-1) x^{2}+k x+1$

$(k-1) \times(-4)^{2}+k \times(-4)+1=0$

$\Rightarrow 16 \mathrm{k}-16-4 \mathrm{k}+1=0$

$\Rightarrow 12 \mathrm{k}-15=0$

$\Rightarrow \mathrm{k}=\frac{-\frac{15}{\mathrm{Tz}}}{}$

$\Rightarrow \mathrm{k}=\frac{5}{4}$