Practice Test 2027 Edition | Updated
2026–27 Questions, Correct Answers &
Detailed Rationales
1. A function is defined by f(x)=x3−6x2+9x+4f(x)=x^3-6x^2+9x+4. At which
value of xx does the function have a local maximum?
A. 1
B. 2
C. 3
D. 4
Answer: x=1x=1
Rationale: The derivative is f′(x)=3x2−12x+9=3(x−1)(x−3)f'(x)=3x^2-12x+9=3(x-
1)(x-3). The derivative changes from positive to negative at x=1x=1, so x=1x=1 is
the local maximum.
2. Evaluate the definite integral
,∫02(3x2−4x+5) dx\int_0^2 (3x^2-4x+5)\,dx
A. 8
B. 10
C. 12
D. 14
Answer: 10
Rationale: The antiderivative is x3−2x2+5xx^3-2x^2+5x. Evaluating from 0 to 2
gives 8−8+10=108-8+10=10.
3. A civil engineering model uses y=Ce0.15xy=Ce^{0.15x}. If y=20y=20 when
x=0x=0, what is yy when x=10x=10?
A. 67.2
B. 78.4
C. 89.6
D. 94.8
Answer: 89.6
Rationale: Since C=20C=20,
y=20e0.15(10)=20e1.5≈89.6y=20e^{0.15(10)}=20e^{1.5}\approx89.6.
4. Find the solution to
ln(x−2)+ln(x+2)=ln21\ln(x-2)+\ln(x+2)=\ln 21
A. 3
B. 4
C. 5
D. 6
Answer: x=5x=5
,Rationale: Combining logarithms gives ln(x2−4)=ln21\ln(x^2-4)=\ln21, so
x2=25x^2=25. Because x>2x>2, the valid solution is x=5x=5.
5. A matrix equation is
[2134][xy]=[718]\begin{bmatrix} 2&1\\ 3&4 \end{bmatrix}
\begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}7\\18\end{bmatrix}
What is xx?
A. 1
B. 2
C. 3
D. 4
Answer: 2
Rationale: The equations are 2x+y=72x+y=7 and 3x+4y=183x+4y=18. From the
first equation, y=7−2xy=7-2x. Substitution gives 3x+28−8x=183x+28-8x=18,
resulting in x=2x=2.
6. What is the magnitude of the vector v=6i−8j+3k\mathbf{v}=6\mathbf{i}-
8\mathbf{j}+3\mathbf{k}?
A. 9.22
B. 10.44
C. 11.00
D. 12.37
Answer: 10.44
Rationale: The magnitude is 62+(−8)2+32=109≈10.44\sqrt{6^2+(-
8)^2+3^2}=\sqrt{109}\approx10.44.
, 7. Two vectors are a=(3,−2,4)\mathbf{a}=(3,-2,4) and
b=(1,5,−2)\mathbf{b}=(1,5,-2). What is a⋅b\mathbf{a}\cdot\mathbf{b}?
A. -15
B. -5
C. 7
D. 15
Answer: -15
Rationale: The dot product is 3(1)+(−2)(5)+4(−2)=3−10−8=−153(1)+(-2)(5)+4(-
2)=3-10-8=-15.
8. Determine the eigenvalues of
A=[4123]A= \begin{bmatrix} 4&1\\ 2&3 \end{bmatrix}
A. 1 and 6
B. 2 and 5
C. 3 and 4
D. 2 and 7
Answer: 2 and 5
Rationale: The characteristic equation is (4−λ)(3−λ)−2=0(4-\lambda)(3-\lambda)-
2=0, which simplifies to λ2−7λ+10=0\lambda^2-7\lambda+10=0. Thus
λ=2,5\lambda=2,5.
9. A quantity is modeled by Q(t)=100(1−e−0.2t)Q(t)=100(1-e^{-0.2t}). What is
dQ/dtdQ/dt at t=5t=5?
A. 0.368
B. 7.358