The equation of pair of lines $y=p x$ and $y=q x$ can be written as $(y-p x)(y-q x)=0$. Then the equation of the pair of the angle bisectors of the lines $x^2-4 x y-5 y^2=0$ is

The equation $(\operatorname{cosp}-1) x^2+(\operatorname{cosp}) x+\sin p=0$ in the variable $x$, has real roots. Then p can take any value in the interval

If $\overline{\mathrm{a}}=\frac{1}{\sqrt{10}}(4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+\hat{\mathrm{k}}), \overline{\mathrm{b}}=\frac{1}{5}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})$, then the value of $(2 \bar{a}-\bar{b}) \cdot\{(\bar{a} \times \bar{b}) \times(\bar{a}+2 \bar{b})\}$ is

Let a random variable X have a Binomial distribution with mean 8 and variance 4 . If $\mathrm{P}(x \leqslant 2)=\frac{\mathrm{k}}{2^{16}}$, then k is equal to