MATHEMATICS PRACTICE PROBLEMS
1. 6 and –4 are roots of which polynomial equation?
(A) 2 x 2 + 3x − 24 = 0
(B) x 2 + 10 x − 24 = 0
(C) x 2 − 2 x − 24 = 0
(D) x 2 − 4 x − 32 = 0
Solution
Since 6 and −4 are roots, then x − 6 = 0 and x + 4 = 0 .
Multiply these two terms.
( x − 6 )( x + 4 ) = 0
x 2 − 2 x − 24 = 0
Answer is (C).
2. If log a 10 = 0.250 , then log10 a equals
(A) 0.25
(B) 0.50
(C) 2.0
(D) 4.0
Solution
log a 10 = 0.250
Take the base a antilogarithm of both sides.
10 = a 0.250
Take the base 10 logarithm of both sides.
log10 = log a 0.250
Simplify using simple logarithm identities.
Copyright 2001 Professional Publications, Inc.
Mathematics Practice Problems - 1
1 = 0.250 log a
Rearrange.
log a =
1
=4
0.250
Answer is (D).
3. What is the determinant of matrix D?
1 1 1
D = 2 −1 1
1 2 −1
(A) −5
(B) −3
(C) 3
(D) 7
Solution
Expand by the top row.
1 1 1
−1 1
2 1
2 −1
2 −1 1 = 1
−1
+1
2 −1
1 −1
1 2
1 2 −1
= (1)(1 − 2) − (1)( −2 − 1) + (1)(4 + 1)
= −1 + 3 + 5 = 7
Answer is (D).
4. Find
d2y
given that x = 2 + t and y = 1 + t2.
2
dx
(A) − 1 2
(B) 2
(C) 3
(D) 5
Copyright 2001 Professional Publications, Inc.
Mathematics Practice Problems - 2
Solution
Method 1:
Take the first derivatives of x and y with respect to t.
x = 2 + t and y = 1 + t 2
dx
dy
=1
= 2t
dt
dt
Divide the second derivative by the first.
dy dt dy 2t
=
=
= 2t
dx dt dx 1
t = x−2
Substitute into the equation for
dy
.
dx
dy
= 2( x − 2)
dx
Take the derivative with respect to x.
d2y
=2
dx 2
Answer is (B).
Method 2:
t = x−2
Substitute into the equation for y.
1 + ( x − 2) = 1 + x2 − 4x + 4 = x2 − 4x + 5
2
Take the first and second derivatives.
dy
= 2x − 4
dx
Copyright 2001 Professional Publications, Inc.
Mathematics Practice Problems - 3
d2y
=2
dx 2
Answer is (B).
5. What is the radius of a circle with equation x 2 − 6 x + y 2 − 4 y − 12 = 0 ?
(A) 4
(B) 5
(C) 6
(D) 7
Solution
Put the equation in standard form by rearranging and completing the square.
(x
2
− 6 x + _ ) + ( y 2 − 4 y + _ ) = 12 + _
The square for x is completed by adding 9 to both sides, and the square for y is completed by
adding 4 to both sides.
(x
2
− 6 x + 9 ) + ( y 2 − 4 y + 4 ) = ( x − 3) + ( y − 2 ) = 12 + 9 + 4 = 25
2
2
r 2 = 25
The radius is r = 25 = 5 .
Answer is (B).
6. Points (1, 2) and (4, 8) are on a straight line. Find the slope of a line perpendicular to this line.
(A) −2
(B) − 1 2
(C) 1 2
(D) 2
Solution
The slope of the line is
Copyright 2001 Professional Publications, Inc.
Mathematics Practice Problems - 4
m=
y 2 − y1 8 − 2 6
=
= =2
x 2 − x1 4 − 1 3
The perpendicular line has a slope of
−1
m
= − 12 .
Answer is (B).
7. Determine the value of the following limit:
lim 8x − 2 x
x → 0 4x
(A) 0
ln 2
(B)
4
ln 2
(C)
2
ln 8 − ln 2
(D)
2
Solution
Use L’Hôpital’s rule.
lim 8 x − 2 x
lim 8 x ln 8 − 2 x ln 2 ln 8 − ln 2 ln 23 − ln 2
=
=
=
x → 0 4x
x→0
4
4
4
3 ln 2 − ln 2 2 ln 2 ln 2
=
=
=
4
4
2
Answer is (C).
8. The area bounded by y = 1 , x = 1 , and y = e − x is most nearly
(A) 0.32
(B) 0.38
(C) 0.50
(D) 0.63
Solution
Calculate e− x at the corners of the region.
Copyright 2001 Professional Publications, Inc.
Mathematics Practice Problems - 5
x
0
0.5
1
e− x
1
0.61
0.37
Area = ∫ (1 − e − x ) dx = ∫ dx − ∫ e − x dx = [ x ]0 −  −e − x 
1
1
1
0
0
0
1
1
0
= (1 − 0 ) − ( −0.37 − 1) = 1 − 0.63 = 0.37
Answer is (B).
9. What is the general solution of the differential equation y′′ − y′ − 6 y = 0 ?
(A) C1e2 y + C2 e −3 y
(B) C1e−2 y + C2 e3 y
(C) C1e3 y + C2 e−2 y
(D) C1e−2 y + C2 e−3 y
Solution
Write the equation in standard form.
y′′ − 2 ( − 12 y′ ) − 6 y = 0
The roots of the characteristic equation are
2
−a ± a 2 − b =
1
1
1
±   + 6 = ± 2.5 = 3, −2
2
2
2
Answer is (B).
Copyright 2001 Professional Publications, Inc.
Mathematics Practice Problems - 6
10. Given y = 2 x 2 + 3 , x = 2 , and ∆x = 0.20 , find ∆y .
(A) 1.0
(B) 1.2
(C) 1.7
(D) 2.0
Solution
∆y = y ( x + ∆x ) − y ( x ) = y ( 2.20 ) − y ( 2 )
(
)
= 2 ( 2.2 ) + 3 − 2 ( 2 ) + 3 = 2 ( 2.2 ) − 8 = 9.68 − 8
2
= 1.68
2
2
Answer is (C).
Copyright 2001 Professional Publications, Inc.
Mathematics Practice Problems - 7