1999 DE EXAM SOLUTIONS
1. Let x , y and z be the respective length, width and height of the rectangular solid. Then zy= 12, xz = 8 and xy = 6 . The volume V of the rectangular solid is
V =xyz =p(xyz)2 =p(zy)(xz)(xy) =p(12)(8)(6) =√576 = 24
cubic inches.
2.
loga32−loga4 =−3
loga(32/4) =−3
loga8 =−3
aloga8 =a−3 8 =a−3
a = 1 2
3. If the length of the rectangle is x meters, then the width of the rectangle is 16 x meters. The perimeter P in meters is
P = 2l+ 2w= 2x+ 2
16 x
= 2x+ 32 x .
4. If 2 +x >0 , then
x
2 +x <2
x <2(2 +x) x <4 + 2x x >−4 . Combining 2 +x >0 and x >−4 gives x >−2 . If 2 +x <0 , then
x
2 +x <2
The solution set is (−∞,−4)∪(−2,∞) .
5.
x=f(f(x))
x= cf(x) 2f(x) + 3 x(2f(x) + 3) = cf(x)
2xf(x) + 3x=cf(x) (c−2x)f(x) = 3x
f(x) = −3x 2x−c cx
2x+ 3 = −3x 2x−c and we have c=−3 .
6.
sin−1(cosθ) = 71◦
sin(sin−1(cosθ)) = sin(71◦)
cosθ= sin(71◦)
sin(90◦
−θ) = sin(71◦)
90◦
−θ= 71◦
θ= 19◦ .
7. Substituting x = 3 and then x = 1/3 into the expression, 3f(1/3)−f(3) = 3
3f(3)−f(1/3) = 1/3 . Then
−f(3) + 3f(1/3) = 3 9f(3)−3f(1/3) = 1 , and we conclude that 8f(3) = 4 , or f(3) = 1/2 .
8. Now BC=3, so EC=4 andAB=8.
9. The triangle with (−1,0), (3,0) and (3,3) as vertices is a right triangle. The legs have length 4 and 3 , so the area of the triangle is 6 .
10.
cosπ
2 + 2 cos 2π
2 + 3 cos 3π
2 +· · ·+ 45 cos 45π
2 =−2 + 4−6 + 8− · · ·+ 44
11. The sum of the interior angles of a pentagon is 540◦. Thus
6x+ 4x+ (5x+ 5) + (6x−20) + (7x−5) = 540 28x−20 = 540 28x= 560 x= 20◦ .
12. Solving for y gives y = 47−(3/4)x. Consequently, x must be a multiple of 4 in order for y to be an integer. Let x= 4j, then y= 47−3j and y−x = 47−3j−4j = 47−7j. The smallest positive value occurs when j = 6 . Thus, x = 4j = 24 and y= 47−3j = 29 .
13. If (0, y) is equidistant from (5,−5) and (1,1) , then
p
(5−0)2+ (
−5−y)2 =p(1−0)2+ (1−y)2 25 + (5 +y)2 = 1 + (y−1)2
25 +y2+ 10y+ 25 = 1 +y2−2y+ 1 y2+ 10y+ 50 =y2−2y+ 2
12y=−48 y=−4 .
14. Let z be a common real solution. Then
z2 −az+ 1 =z2−z+a= 0 (1−a)z =a−1
a= 1 or z =−1
If a= 1, then x2
−x+ 1 = 0 has no real solutions. So z =−1 and
(−1)2−a(−1) + 1 = 0 1 +a+ 1 = 0
a=−2 .
15. Let x and y be the lengths of the legs and z the length of the hypotenuse. Then 1
2xy = 1
2(z)(12) = 6z x+y+z = 60
Since xy= 12z,
x2+ 2xy+y2 = (x2+y2) + 24z =z2+ 24z
(x+y)2 =z2+ 24z
(60−z)2 =z2+ 24z
z2
−120z+ 3600 =z2+ 24z 144z = 3600
z = 25 .
Now
x+y= 60−z = 60−25 = 35 xy= 12z = 12(25) = 300
y= 300 x , so
x+y=x+ 300 x = 35 x2
−35x+ 300 = 0 (x−15)(x−20) = 0 .
The solutions are x= 15 and x= 20 . If x= 15 , then y= (300/x) = 20 . If x= 20 , then y= (300/x) = 15 . The triangle has sides 15, 20, 25 .
16. The graph of f is a line with slope 1/4 and y-intercept 3 . Now f(x) = (1/4)x+ 3 and f(x) = 9 when
1
4x+ 3 = 9 1 4x = 6
x = 24 , and f−1(9) = 24 .
17. The largest such constrained value of 5x+4y must occur at either B=(4,0), C=(3 2,
15 4 )
or D=(0,5) as shown.
(4,0) (5,0)
(3/2,15/4)
x y
3x + 2y = 12
5x + 6y = 30
A B=
Now 5x+ 4y= 20 at both B and D and 5x+ 4y = 45/2 at C. The largest value of 5x+ 4y is 45/2 .
18. For nonzero solutions we need
det
4−p 2 −1 7−p
= 0
(4−p)(7−p) + 2 = 0 p2−11p+ 28 + 2 = 0 p2−11p+ 30 = 0 (p−5)(p−6) = 0
So p= 5 is the smallest value.
19.
log10cotA+ log10sinA = log10(cotAsinA)
= log10cosA
= log100.1
= log10(10)−1
