If the normal to the curve $y=\mathrm{f}(x)$ at the point $(3,4)$ makes an angle of $\left(\frac{3 \pi}{4}\right)$ with the positive $X$-axis, then the value of $f^{\prime}(3)$ is

$$ \int \frac{2 x+5}{\sqrt{7-6 x-x^2}} d x=A \sqrt{7-6 x-x^2}+B \sin ^{-1}\left(\frac{x+3}{4}\right)+\mathrm{c} $$ (where c is a constant of integration) then the value of $A+B$ is

A straight line L through the point $(3,-2)$ is inclined at an angle of $60^{\circ}$ to the line $\sqrt{3} x+y=1$. If L also intersects the X -axis, then the equation of $L$ is

If $y(x)$ is the solution of the differential equation $(x+2) \frac{\mathrm{d} y}{\mathrm{~d} x}=x^2+4 x-9, x \neq-2$ and $y(0)=0$, then $y(-4)$ is equal to