Let f be a differentiable function satisfying f(x)=1-2 x+ ₀^ x e ^ (x-t) f(t) dt , x R and let g (x)= ₀^ x (f( t )+2)¹⁵( t -4)⁶( t +12)¹⁷ dt , x R . If p and q are respectively the points of local minima and local maxima of g, then the value of | p + q | is equal to _ _ _ _ .

Step-by-step solution

Given f(x) = 1 - 2x + ₀^x e^ (x-t) f(t) , dt . Differentiating and simplifying: e^ -x f'(x) - e^ -x f(x) = -2e^ -x + (1-2x)e^ -x (-1) + e^ -x f(x) This reduces to: f'(x) - 2f(x) = 2x - 3 Solving the linear ODE dy dx - 2y = 2x - 3 : y e^ -2x = e^ -2x (2x-3) ,dx On solving, we get f(x) = 1 - x . Now, g(x) = ₀^x (3-t)¹⁵(t-4)^6(t+12)¹⁷ , dt g'(x) = (3-x)¹⁵(x-4)^6(x+12)¹⁷ = -(x-3)¹⁵(x-4)^6(x+12)¹⁷ Sign analysis of g'(x) : g'(x) changes from + - at x = 3 → local maxima (q = 3) g'(x) changes from - + at x = -12 → local mi