# $\large\frac{dy}{dx}$$=1 passes through \large[$$0,\large\frac{3}{2}]$ and $\large\frac{dy}{dx}$$=2 passes through [0,1] represents two curve which passes through a point whose coordinates are zeroes of a quadratic polynomial . The quadratic equation is gradient polynomial. Find sum of coefficients of x^3,x^2,x (a)\;\frac{3}{4}\\(b)\;\frac{1}{4}\\ (c)\;\frac{2}{3} \\ (d)\;\frac{1}{3} ## 1 Answer y=x+c y-x=\large\frac{3}{2}-----(1) y=2x+c_1 y-2x=1-----(2) Solving (1) ans (2) we get point (\large\frac{1}{2}$$,2)$
$\bigg (x -\large\frac{1}{2}\bigg)$$(x-2)=x^2-\large\frac{5}{2}$$x+1$
$\large\frac{dy}{dx}$$=x^2-\large\frac{5}{2}$$ x+1$
$y= \int (x^2- 5/2x+1)dx$
$y= x^3 -\large\frac{5x^2}{4}$$+x+c 1- \large\frac{5}{4}$$+1=\large\frac{3}{4}$
Hence a is the correct answer.
edited Mar 20, 2014