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1. If f(x)=3x2−4x+5f(x) = 3x^2 - 4x + 5f(x)=3x2−4x+5, what is
f′(x)f'(x)f′(x)?
A) 6x−46x - 46x−4
B) 6x−36x - 36x−3
C) 6x−56x - 56x−5
D) 6x+56x + 56x+5
Answer: A) 6x−46x - 46x−4
Rationale: Differentiating f(x)=3x2−4x+5f(x) = 3x^2 - 4x +
5f(x)=3x2−4x+5 with respect to xxx gives f′(x)=6x−4f'(x) = 6x -
4f′(x)=6x−4.
2. Solve for xxx in log2(x−3)=4\log_2 (x - 3) = 4log2(x−3)=4.
A. x=19x = 19x=19
B. x=17x = 17x=17
C. x=16x = 16x=16
D. x=12x = 12x=12
Answer: B) x=17x = 17x=17
Rationale: log2(x−3)=4\log_2 (x - 3) = 4log2(x−3)=4 means
x−3=24x - 3 = 2^4x−3=24, so x−3=16x - 3 = 16x−3=16, and x=17x
= 17x=17.
3. The general solution to sin(x)=12\sin(x) = \frac{1}{2}sin(x)=21
is
A. x=π6+2nπx = \frac{\pi}{6} + 2n\pix=6π+2nπ
B. x=π6+nπx = \frac{\pi}{6} + n\pix=6π+nπ
C. x=π6+2nπx = \frac{\pi}{6} + 2n\pix=6π+2nπ or x=5π6+2nπx =
\frac{5\pi}{6} + 2n\pix=65π+2nπ
D. x=π6+nπx = \frac{\pi}{6} + n\pix=6π+nπ or x=5π6+nπx =
\frac{5\pi}{6} + n\pix=65π+nπ
Answer: C) x=π6+2nπx = \frac{\pi}{6} + 2n\pix=6π+2nπ or
x=5π6+2nπx = \frac{5\pi}{6} + 2n\pix=65π+2nπ
Rationale: The sine function equals 12\frac{1}{2}21 at
π6\frac{\pi}{6}6π and 5π6\frac{5\pi}{6}65π plus integer multiples of
2π2\pi2π.
,4. Find the sum of the series 1+2+3+⋯+501 + 2 + 3 + \cdots +
501+2+3+⋯+50.
A. 1250
B. 1275
C. 2550
D. 2555
Answer: B) 1275
Rationale: The sum of the first nnn integers is given by
Sn=n(n+1)2S_n = \frac{n(n + 1)}{2}Sn=2n(n+1). For n=50n =
50n=50, S50=50(50+1)2=1275S_{50} = \frac{50(50 + 1)}{2} =
1275S50=250(50+1)=1275.
5. If cos(2θ)=0\cos(2\theta) = 0cos(2θ)=0, the general solution is
A. θ=π2+nπ\theta = \frac{\pi}{2} + n\piθ=2π+nπ
B. θ=π4+nπ\theta = \frac{\pi}{4} + n\piθ=4π+nπ
C. θ=π2+nπ\theta = \frac{\pi}{2} + n\piθ=2π+nπ or
θ=3π2+nπ\theta = \frac{3\pi}{2} + n\piθ=23π+nπ
D. θ=π4+nπ\theta = \frac{\pi}{4} + n\piθ=4π+nπ or
θ=5π4+nπ\theta = \frac{5\pi}{4} + n\piθ=45π+nπ
Answer: C) θ=π2+nπ\theta = \frac{\pi}{2} + n\piθ=2π+nπ or
θ=3π2+nπ\theta = \frac{3\pi}{2} + n\piθ=23π+nπ
Rationale: The cosine function equals zero at π2+nπ\frac{\pi}{2} +
n\pi2π+nπ and 3π2+nπ\frac{3\pi}{2} + n\pi23π+nπ.
6. The equation of the straight line passing through (1,3)(1, 3)(1,3) with
slope 444 is
A. y=4x+3y = 4x + 3y=4x+3
B. y=4x−1y = 4x - 1y=4x−1
C. y=3x+1y = 3x + 1y=3x+1
D. y=4x+1y = 4x + 1y=4x+1
Answer: B) y=4x−1y = 4x - 1y=4x−1
Rationale: Using the point-slope form y−y1=m(x−x1)y - y_1 = m(x -
x_1)y−y1=m(x−x1) with m=4m = 4m=4, x1=1x_1 = 1x1=1, and
, y1=3y_1 = 3y1=3, we get y−3=4(x−1)y - 3 = 4(x - 1)y−3=4(x−1),
simplifying to y=4x−1y = 4x - 1y=4x−1.
7. The limit of limx→0sin(x)x\lim_{x \to 0}
\frac{\sin(x)}{x}limx→0xsin(x) is
A. 0
B. 1
C. -1
D. Does not exist
Answer: B) 1
Rationale: The well-known limit limx→0sin(x)x=1\lim_{x \to 0}
\frac{\sin(x)}{x} = 1limx→0xsin(x)=1.
8. If f(x)=ln(x)f(x) = \ln(x)f(x)=ln(x), what is f′(x)f'(x)f′(x)?
A. 1x\frac{1}{x}x1
B. xx\frac{x}{x}xx
C. xxx
D. ln(x)\ln(x)ln(x)
Answer: A) 1x\frac{1}{x}x1
Rationale: The derivative of f(x)=ln(x)f(x) = \ln(x)f(x)=ln(x) is
f′(x)=1xf'(x) = \frac{1}{x}f′(x)=x1.
9. The integral ∫x2dx\int x^2 dx∫x2dx is
A. x33+C\frac{x^3}{3} + C3x3+C
B. x32+C\frac{x^3}{2} + C2x3+C
C. x35+C\frac{x^3}{5} + C5x3+C
D. x3+Cx^3 + Cx3+C
Answer: A) x33+C\frac{x^3}{3} + C3x3+C
Rationale: The integral of x2x^2x2 with respect to xxx is
x33+C\frac{x^3}{3} + C3x3+C.
10. The solution to the quadratic equation x2−5x+6=0x^2 - 5x + 6 =
0x2−5x+6=0 is
A. x=1,6x = 1, 6x=1,6
B. x=−1,−6x = -1, -6x=−1,−6
1. If f(x)=3x2−4x+5f(x) = 3x^2 - 4x + 5f(x)=3x2−4x+5, what is
f′(x)f'(x)f′(x)?
A) 6x−46x - 46x−4
B) 6x−36x - 36x−3
C) 6x−56x - 56x−5
D) 6x+56x + 56x+5
Answer: A) 6x−46x - 46x−4
Rationale: Differentiating f(x)=3x2−4x+5f(x) = 3x^2 - 4x +
5f(x)=3x2−4x+5 with respect to xxx gives f′(x)=6x−4f'(x) = 6x -
4f′(x)=6x−4.
2. Solve for xxx in log2(x−3)=4\log_2 (x - 3) = 4log2(x−3)=4.
A. x=19x = 19x=19
B. x=17x = 17x=17
C. x=16x = 16x=16
D. x=12x = 12x=12
Answer: B) x=17x = 17x=17
Rationale: log2(x−3)=4\log_2 (x - 3) = 4log2(x−3)=4 means
x−3=24x - 3 = 2^4x−3=24, so x−3=16x - 3 = 16x−3=16, and x=17x
= 17x=17.
3. The general solution to sin(x)=12\sin(x) = \frac{1}{2}sin(x)=21
is
A. x=π6+2nπx = \frac{\pi}{6} + 2n\pix=6π+2nπ
B. x=π6+nπx = \frac{\pi}{6} + n\pix=6π+nπ
C. x=π6+2nπx = \frac{\pi}{6} + 2n\pix=6π+2nπ or x=5π6+2nπx =
\frac{5\pi}{6} + 2n\pix=65π+2nπ
D. x=π6+nπx = \frac{\pi}{6} + n\pix=6π+nπ or x=5π6+nπx =
\frac{5\pi}{6} + n\pix=65π+nπ
Answer: C) x=π6+2nπx = \frac{\pi}{6} + 2n\pix=6π+2nπ or
x=5π6+2nπx = \frac{5\pi}{6} + 2n\pix=65π+2nπ
Rationale: The sine function equals 12\frac{1}{2}21 at
π6\frac{\pi}{6}6π and 5π6\frac{5\pi}{6}65π plus integer multiples of
2π2\pi2π.
,4. Find the sum of the series 1+2+3+⋯+501 + 2 + 3 + \cdots +
501+2+3+⋯+50.
A. 1250
B. 1275
C. 2550
D. 2555
Answer: B) 1275
Rationale: The sum of the first nnn integers is given by
Sn=n(n+1)2S_n = \frac{n(n + 1)}{2}Sn=2n(n+1). For n=50n =
50n=50, S50=50(50+1)2=1275S_{50} = \frac{50(50 + 1)}{2} =
1275S50=250(50+1)=1275.
5. If cos(2θ)=0\cos(2\theta) = 0cos(2θ)=0, the general solution is
A. θ=π2+nπ\theta = \frac{\pi}{2} + n\piθ=2π+nπ
B. θ=π4+nπ\theta = \frac{\pi}{4} + n\piθ=4π+nπ
C. θ=π2+nπ\theta = \frac{\pi}{2} + n\piθ=2π+nπ or
θ=3π2+nπ\theta = \frac{3\pi}{2} + n\piθ=23π+nπ
D. θ=π4+nπ\theta = \frac{\pi}{4} + n\piθ=4π+nπ or
θ=5π4+nπ\theta = \frac{5\pi}{4} + n\piθ=45π+nπ
Answer: C) θ=π2+nπ\theta = \frac{\pi}{2} + n\piθ=2π+nπ or
θ=3π2+nπ\theta = \frac{3\pi}{2} + n\piθ=23π+nπ
Rationale: The cosine function equals zero at π2+nπ\frac{\pi}{2} +
n\pi2π+nπ and 3π2+nπ\frac{3\pi}{2} + n\pi23π+nπ.
6. The equation of the straight line passing through (1,3)(1, 3)(1,3) with
slope 444 is
A. y=4x+3y = 4x + 3y=4x+3
B. y=4x−1y = 4x - 1y=4x−1
C. y=3x+1y = 3x + 1y=3x+1
D. y=4x+1y = 4x + 1y=4x+1
Answer: B) y=4x−1y = 4x - 1y=4x−1
Rationale: Using the point-slope form y−y1=m(x−x1)y - y_1 = m(x -
x_1)y−y1=m(x−x1) with m=4m = 4m=4, x1=1x_1 = 1x1=1, and
, y1=3y_1 = 3y1=3, we get y−3=4(x−1)y - 3 = 4(x - 1)y−3=4(x−1),
simplifying to y=4x−1y = 4x - 1y=4x−1.
7. The limit of limx→0sin(x)x\lim_{x \to 0}
\frac{\sin(x)}{x}limx→0xsin(x) is
A. 0
B. 1
C. -1
D. Does not exist
Answer: B) 1
Rationale: The well-known limit limx→0sin(x)x=1\lim_{x \to 0}
\frac{\sin(x)}{x} = 1limx→0xsin(x)=1.
8. If f(x)=ln(x)f(x) = \ln(x)f(x)=ln(x), what is f′(x)f'(x)f′(x)?
A. 1x\frac{1}{x}x1
B. xx\frac{x}{x}xx
C. xxx
D. ln(x)\ln(x)ln(x)
Answer: A) 1x\frac{1}{x}x1
Rationale: The derivative of f(x)=ln(x)f(x) = \ln(x)f(x)=ln(x) is
f′(x)=1xf'(x) = \frac{1}{x}f′(x)=x1.
9. The integral ∫x2dx\int x^2 dx∫x2dx is
A. x33+C\frac{x^3}{3} + C3x3+C
B. x32+C\frac{x^3}{2} + C2x3+C
C. x35+C\frac{x^3}{5} + C5x3+C
D. x3+Cx^3 + Cx3+C
Answer: A) x33+C\frac{x^3}{3} + C3x3+C
Rationale: The integral of x2x^2x2 with respect to xxx is
x33+C\frac{x^3}{3} + C3x3+C.
10. The solution to the quadratic equation x2−5x+6=0x^2 - 5x + 6 =
0x2−5x+6=0 is
A. x=1,6x = 1, 6x=1,6
B. x=−1,−6x = -1, -6x=−1,−6
