Problem:
Let f(x)=100x3−300x2+200x. For how many real numbers a does the graph of y=f(x−a) pass through the point (1,25)?
Answer Choices:
A. 1
B. 2
C. 3
D. 4
E. more than 4
Solution:
We need the graph of y=f(x−a) to pass through (1,25).
That means
f(1−a)=25.
Given
f(x)=100x3−300x2+200x=100x(x−1)(x−2),
we have
f(1−a)=100(1−a)(1−a−1)(1−a−2)=100(1−a)(−a)(−1−a)=100a(1−a)(1+a)=100a(1−a2)
So
100a(1−a2)=25⟹4a(1−a2)=1,
4a−4a3=1⟹4a3−4a+1=0.
Let g(a)=4a3−4a+1. Then
g′(a)=12a2−4=4(3a2−1),
so critical points at
a=±31.
We can evaluate
g(−31)>0,g(31)<0.
Since g(a)→−∞ as a→−∞ and g(a)→+∞ as a→+∞, the cubic crosses the a-axis once in each of the intervals
(−∞,−31),(−31,31),(31,∞),
giving 3 distinct real roots.
Therefore, there are
(C) 3
real numbers a such that the graph of y=f(x−a) passes through (1,25).
The problems on this page are the property of the MAA's American Mathematics Competitions