Solutions to EF Exam
Texas A&M High School Math Contest
10 November, 2012
1. The average is
20(70) + 30(60)
= 64%.
50
2. The desired sum is 1 + 11 + 21 + 31 + · · · + 2001 + 2011. There are a total of 202 numbers which
can form 101 pairs of numbers which add to 2012. Therefore, the desired sum is (2012)(101) =
203, 212.
3. Let v be the man’s usual rowing speed, and let u be the rate of the stream’s current. Then
15
15
15
15
=
− 5 and
=
− 1. Clearing the fractions yields the system of
v+u
v−u
2v + u
2v − u
equations
15(v − u) = 15(v + u) − 5(v 2 − u2 )
15(2v − u) = 15(2v − u) − (4v 2 − u2 )
or 6u = v 2 − u2 , 30u = 4v 2 − u2 . Eliminating v from this system yields 6u = 3u2 , Since u 6= 0,
we must have u = 2 mph.
4.
360
180(n − 2)
=9·
, so n − 2 = 18, or n = 20 sides.
n
n
_
5. Let x = DE. Then 40 =
1 _
1
(x − (360 − 3x)), or x = 110. Therefore, ∠DBE = DE = 55◦ .
2
2
6. The equation can be rewritten as (5x )2 + 2(5x ) − 15 = 0, or (5x + 5)(5x − 3) = 0. This occurs
when 5x = −5 (no real solution) or 5x = 3 and x = log5 3.
7. In base 10, N = b3 + 1 = (b + 1)(b2 − b + 1). Therefore (d) must be true.
3
4
1
8. Since cos x = , sin x = − (since x lies outside quadrant 1). Similarly, since sin y = , cos y =
5
5
2
√
√ !
3
3
3
−
. Using the angle addition identity, cos(x + y) = cos x cos y − sin x sin y =
−
−
2
5
2
√
4
1
−3 3 + 4
−
=
.
5
2
10
32π
2π
9. Note that sin 16x = sin
= sin
= sin x. Multiply the given expression by 16 sin x:
5
5
16 sin x cos x cos 2x cos 4x cos 8x = 8 sin 2x cos 2x cos 4x cos 8x = 4 sin 4x cos 4x cos 8x = 2 sin 8x cos 8x =
1
.
sin 16x(= sin x) Therefore, the given expression is equal to
16
10. This is equivalent to finding the line tangent to y = −x2 − 8x − 7 which passes through the
point (−9, 0). If the line is tangent at (a, −a2 − 8a − 7), then equating slopes of the line yield
−a2 − 8a − 7
= −2a − 8, or a2 + 18a + 65 = 0 and a = −5 or a = −13. Since the point we desire
a+9
is between the x-intercepts (−7, 0) and (−1, 0), a = −5 and the slope of the line is 2. Therefore,
the angle must be arctan(2).
11.
d
(sin4 x+cos4 x) = 4 sin3 x cos x−4 cos3 x sin x = 4 sin x cos x(sin2 x−cos2 x) = −2 sin(2x) cos(2x)
dx
= − sin(4x). Since the 2011th derivative of sin x = − cos x (by pattern), The 2012th derivative
of f is 42011 cos(4x).
1
√
√
x
1
1
12. 2
=√
. Let u = x. Then du = √ dx, so
x +x
x(x + 1)
2 x
π π π
√
3√
−1
= 2 tan (u)|1/ 3 = 2
= .
−
3
6
3
ˆ
3
1/3
1
√
dx =
x(x + 1)
ˆ
√
3
√
1/ 3
u2
2
du
+1
13. For the student’s answer to be correct, we must have (x2 g(x))0 = 2xg 0 (x). Applying the Product
g 0 (x)
2x ∗
=
. After simplifying
Rule (correctly!) we obtain 2xg(x) + x2 g 0 (x) = 2xg 0 (x), or
2
2x − x
g(x)
A
eC
, or
and integrating both sides, we obtain −2 ln(2−x)+C = ln(g(x)), so g(x) =
(2 − x)2
(2 − x)2
(∗ NOTE that when x = 0 the original equation is trivially true no matter what g is).
2
1
= x4 + 10x2 y 2 + 5y 4 and
= 5x4 +
x
y
10x2 y 2 + y 4 . Clearing the fractions and adding the resulting equations
together yields 3 =
√
x5 + 5x4 y + 10x3 y 2 + 10x2 y 3 + 5xy 4 + y 5 = (x + y)5 , so x + y = 5 3.
14. Adding and subtracting the two equations gives us
15. Let x and y be the side lengths of the larger and smaller square, respectively. The smallest
rectangle containing these squares must have side lengths x and x + y, hence area x2 + xy.
ThereforehA is ithe largest possible value of x2 + xy given that x2 + y 2 = 1. Let x = cos t and y =
π
sin t (t ∈ 0, ). We now maximize cos2 t + cos t sin t on the given interval, which occurs when
4
π
−2 cos t sin t + cos2 t − sin2 t = cos(2t) − sin(2t) = 0, or t = (second derivative test shows this
8
π
π 1 + √2
2 π
.
critical value is indeed a maximum). Therefore, A = cos
+ cos
sin
=
8
8
8
2
16. Let t and u be the tens and units digit of the number, respectively. Then N = 10t+u = tu+t+u,
or 9t = tu. Since t 6= 0, we must have u = 9.
√
√
(1 + 5)2
5−1
(sin2 18◦ )(sin2 54◦ )
17. cos 72◦ = 2 cos 36◦ −1 =
−1 =
. Then (tan2 18◦ )(tan2 54◦ ) =
8
4
(cos2 18◦ )(cos2 54◦ )
√
√
!2
2
2
5+1
− 5−1
cos 36◦ − cos 72◦
2
1
4
4
√
√
=
= .
=
by Product-to-Sum formulas. = √
◦
◦
5+1
5−1
cos 36 + cos 72
5
2 5
+ 4
4
Therefore, p + q = 6
2x
1
x−1
−
=
, which is > 0 on the specified domain. Therefore, F is increasing
2 ln x ln x
ln x
for x > 1 and M = lim F (x) if such limit exists. In the integral, make the substitution t = ev .
x→1+
ˆ 2 ln x v
ˆ 2 ln x ln x
ˆ 2 ln x 2 ln x
e
dv
e
dv
e dv
. Now
≤ F (x) ≤
.
Then dt = ev dv and F (x) =
v
v
v
ln x
ln x
ln x
Evaluating the integrals yields eln x (ln(2 ln x) − ln(ln x)) ≤ F (x) ≤ e2 ln x (ln(2 ln x) − ln(ln x)),
or x ln 2 ≤ F (x) ≤ x2 ln 2. Therefore, by the Squeeze Theorem, M = lim+ F (x) = ln 2.
18. F 0 (x) =
x→1
2