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$\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} $
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$y=x+c$
$y-x=\large\frac{3}{2}$-----(1)
$y=2x+c_1$
$y-2x=1$-----(2)
Solving (1) ans (2) we get point $(\large\frac{1}{2}$$,2)$
So quadratic polynomial will be
$ \bigg (x -\large\frac{1}{2}\bigg)$$ (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.
answered Feb 7, 2014 by meena.p
edited Mar 20, 2014 by balaji.thirumalai
 

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