Tangents

Here inflection point is x=1

so equation of tangent at inflection point is {x=1 ,y=(b-a-2) , y'=(-a-3)}

y=(b+1)-(a+3)x

which should coincide with the given straight line y=2-5x

which gives b=1 and a=2 hence ab+a+b=5

Let (X,Y) be a point on curve $y = x^3-3x^2 -ax+b$

$\frac{dy}{dx}$ = $3{ x }^{ 2 }-6x-a$

$\frac{dy}{dx}$ at (X,Y) = $3{ X }^{ 2 }-6X-a$

Equation of tangent to (X,Y)

$\frac{y-Y}{x-X}$ = $3{ X }^{ 2 }-6X-a$ .........Equation 1

Now this passes through (h,2-5h) So y= 2-5h and x=h in equation 1

So ${ 2X }^{ 3 }-3(h+1){ X }^{ 2 }+6hX+(a-5)h+2-b\quad =\quad 0$ , let it be f(X)

Now according to question there are exactly two values of x satisfying f(x)=0 , but f(X) is cubic equation means that one of roots is a repeated root, this implies that f(x) and f'(x) has a common root

f'(x) = 6{ x }^{ 2 }-6(h+1)x+6h

Thus f'(x) has two roots x =1 and x=h

Now for all h , one of these must be a root of f(x) so x = 1 must be a root of f(x) for all h

f (1) = (a-2)h + (b-1)

Hence if a = 2 and b = 1 then f(1) will be zero irrespective of h

Hence ab+a+1 = 2+2+1 = 5