Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

$\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} $
Can you answer this question?

1 Answer

0 votes
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App