If the probability distribution function of a random variable X is given as

Then F(0) is equal to

The particular solution of the differential equation $$ \frac{d y}{d x}=\frac{x+y+1}{x+y-1} $$ when $$ \mathrm{x}=\frac{2}{3} $$ and $$ y=\frac{1}{3} $$ is

If $$a \sin \theta=b \cos \theta$$, where $$a, b \neq 0$$, then $$a\cos 2 \theta+b \sin 2 \theta=$$

$$\vec{a}=4 \hat{i}+13 \hat{j}-18 \hat{k}, \vec{b}=\hat{i}-2 \hat{j}+3 \hat{k}$$ and $$\vec{c}=2 \hat{i}+3 \hat{j}-4 \hat{k}$$ are three vectors such that $$\vec{a}=x \vec{b}+y \vec{c}$$, then $$x+y=$$