A Vanishing Polynomial Problem (Posted on 2007-01-17)

	Submitted by K Sengupta
Rating: 2.6667 (3 votes)
Solution:	(Hide)
Since M(0) = 0; we can write: M(x) = (x+2)(a*x^2 + bx + c) Or, M'(x) = (x+2)(2ax+b) + (a*x^2 + bx +c) Now, since M(x)has a relative minimum/ maximum at x = -1 and x = 1/3; it follows that: M'(-1) = M'(1/3) = 0 Or, c-a = 0; 15a + 24b + 9c = 0 Or, c = a, b = -a; and, so: Integral M(x) dx; x = -1 to 1 = 14/3 gives: a* Integral (x^3 + x^2 -x +2) dx; x = -1 to 1 = 14/3 Or, a* Integral (x^2 + 2) dx; x = -1 to 1 = 14/3 ( as Integral (x^3 - x) dx; x = -1 to 1 = 0, since (x^3 - x) is an odd function) Or, a =1 Consequently, the required polynomial is given by: M(x) = (x+2)(x^2 - x + 1)

	Subject	Author	Date
	Solution	Bractals	2007-01-17 12:52:16
	solution	Charlie	2007-01-17 10:52:41