If from the point  $(h,2-5h),h\in R,h\ne 1$, two distinct tangents are drawn to the curve $y=x^3-3x^2-ax+b$  then  $a+b$ is equal to

Ans. 0003
Thus is based on the concept of points of inflection. For any cubic curve, if we draw tangent at the point of inflection, let say P. If we take any general point Q on this tangent, other than P, 
 then we can always draw two distinct tangents. 
Here, given points $(h, 2 – 5h)$ lies on the line $y=2-5x$ where $h\ne 1$ given
Hence, for point of inflection $P; x = 1; y = -3$       $(\because y=2-5x)$       
Curve must satisfy point of inflection,
$$\begin{gathered} y=x^3-3x^2-ax+b \\ -3=1-3-a+b;a-b=1 \end{gathered}$$  
Slope of tangent and curve, will be same at  $P(1,-3)$
$$\begin{gathered} \frac{dy}{dx}=3x^2-6x-a \\ {\frac{dy}{dx}|}_{atx=1}=3-6-a=-3-a \\ y=2-5x \\ {\frac{dy}{dx}|}_{atx=1}=-5 \\ -3-a=-5 \\ \Rightarrow a=2 \\ \therefore b=1 [\because a-b=1] \end{gathered}$$