I want to know if I've become a practitioner of voodoo mathematics (i.e. if my solution is nonsensical).
Let a and b be two numbers such that b≥a>0.Consider the equation ∫abx2dx=(∫abxdx)2.i) Which value of b satisfies this equation when a=0?∫0bx2dx=3b3=(∫0bxdx)2=4b4.
As
b=0 we can divide both sides by
b3 and multiply both sides by 12 to get
b=34.
ii. In the case a=1, derive the equation 3b3−b2−7b−7=0 and with the aid of a sketch, show that there is only one solution to the equation and that it lies between 2 and 3.∫1bx2dx=3b3−1=(∫1bxdx)2=(2b2−1)2.
3b3−1=3(b−1)(b2+b+1) and
(2b2−1)2=4(b+1)2(b−1)2.
Now cancel
(b−1) from both sides to get
3b2+b+1=4(b+1)2(b−1).
We get
4b2+4b+4=3b3+3b2−3b−3 which leads to the required equation of
3b3−b2−7b−7=0.
The y-intercept is (0,-7) and we can find the x-coordinates of the turning points by differentiating the cubic to give
9b2−2b−7=(9b+7)(b−1). The x-coordinates of the turning points are thus
b=1,b=−97.
f(1)=−12 and
f′′(1)=16 which is a minimum so
there is a maximum at
b=−97 and
f)(−97)<0. This all becomes clearer if I can actually show my sketch here! f(2)=-1<0 and f(3)=44>0; the sign changes so there is a root between b=2 and b=3.
Let p=b+a and q=b−a. Show that 3p2+q2=3p2q and describe p2 in terms of q. Deduce that 0<q≤34. ∫abx2dx=3b3−a3=3(b−a)(a2+b2+ab)=3q(a2+b2+ab).
(∫abxdx)2=4b2−a2=4(b+a)2(b−a)2=4p2q2.
Cancelling q from both sides and multiplying both sides by 12,
4a2+4b2+4=3p2q.
3(b+a)2+(b−a)2=3p2+q2=3p2q.
Thus subtracting both sides by
3p2 we obtain
q2=3p2(q−1) to obtain
p2=3(q−1)q2.
As
p2=(b+a)2>0,
3(q−1)>0 for the equality to hold as we know that
q2 is positive so 3(q-1) can't be negative (neither can it be zero as it is not possible to divide by zero). So
q>1. We also know that
(b+a)2≥(b−a)2, with equality achieved when a=0. Thus
p2≥q2 so
3(q−1)≤1 so
q≤34.