Mathematics_Today_-_October_2016

1. Suppose a quadratic function f(x) = ax^2 + bx + c
   (a, b, c ∈ R and a ≠ 0) satisfies the following
   conditions :
   (1) When x ∈ R, f(x – 4) = f(2 – x) and f(x) ≥ x.

(2) When x ∈ (0, 2), fx()≤⎛⎝⎜x+^1 ⎞⎠⎟

2

2

(3) The minimum value of f(x) on R is 0.

Find the maximal m(m > 1) such that there exists

t ∈ R, f(x + t) ≤ x holds so long as x ∈ [1, m].

2. Draw a tangent line of parabola y = x^2 at the point
   A(1, 1). Suppose the line intersects the x-axis and
   y-axis at D and B respectively. Let point C be on the
   parabola and point E on AC such that AE
   EC

=λ 1.

Let point F be on BC such that

BF

FC

=λ 2 and

λ 1 + λ 2 = 1. Assume that CD intersects EF at point

P. When point C moves along the parabola, find the

equation of the trail of P.

3. Suppose that α and β are different real roots of
   the equation 4x^2 – 4tx – 1 = 0 (t ∈ R). [α, β] is the
   domain of the function fx xt
   x

()= −

+

2

(^21)
.
(1) Find g(t) = max f(x) – min f(x).
(2) Prove that for ui∈⎛⎝⎜ 0 ⎞⎠⎟
2
,π (i = 1, 2, 3),
if sin u 1 + sin u 2 + sin u 3 = 1, then
1113
4
6
gu gu gu(tan 123 ) (tan ) (tan )
++<.

4. Suppose A, B, C are three non-collinear points
   corresponding to complex numbers z 0 = ai ,
   zbi 1 1
   2

=+, z 2 = 1 + ci (a, b and c being real

numbers), respectively. Prove that the curve

z = z 0 cos^4 t + 2z 1 cos^2 t · sin^2 t + z 2 sin^4 t (t ∈ R) shares

a single common point with the line bisecting AB

and parallel to AC in ΔABC and find this point.

5. A sequence is formed by the following rules :
   s 1 = a, s 2 = b and sn + 2 = sn + 1 + (–1)nSn for all n ≥ 1.
   If a = 3 and b is an integer less than 1000, what is the
   largest value of b for which 2015 is a member of the
   sequence? Justify your answer.
   SOLUTIONS


6. Since f(x – 4) = f(2 –x) for x ∈ R, it is known that
   the quadratic function f(x) has x = –1 as its axis
   of symmetry. By condition (3), we know that f(x)
   opens upward, that is, a > 0.
   Hence, f(x) = a(x + 1)^2 (a > 0).
   By condition (1), we get f(1) ≥ 1 and by (2),

f()1^11

2

1

2

≤⎝⎜⎛ + ⎞⎠⎟ =. It follows that f(1) = 1,

i.e., a(1 + 1)^2 = 1. So a=^14.

Thereby, fx()=+^1 (x )

4

12.

Since the graph of the parabola fx()=+(x )

1

4

(^12)
opens upward and a graph of y = f(x + t) can be
obtained by translating that of f(x) by t units. If we
want the graph of y = f(x + t) to lie under the graph of
y = x when x ∈ [1, m] and m to be maximal, then
1 and m should be two roots of an equation with
respect to x
1
4
().xt++ = 12 x ...(i)
Substituting x = 1 into (i), we get t = 0 or t = –4.
When t = 0, substituting it into (i), we get x 1 = x 2 = 1
(in contradiction with m > 1).
When t = –4, substituting it into (i), we get x 1 = 1
and x 2 = 9 and so m = 9.