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1. 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.
2. The solution to x2+6x+5=0x^2 + 6x + 5 = 0x2+6x+5=0 is
A. x=−1,−5x = -1, -5x=−1,−5
B. x=1,−5x = 1, -5x=1,−5
C. x=−1,5x = -1, 5x=−1,5
D. x=1,5x = 1, 5x=1,5
Answer: A) x=−1,−5x = -1, -5x=−1,−5
Rationale: The quadratic x2+6x+5=0x^2 + 6x + 5 = 0x2+6x+5=0
factors as (x+1)(x+5)=0(x + 1)(x + 5) = 0(x+1)(x+5)=0, so x=−1x = -
1x=−1 and x=−5x = -5x=−5.
,3. 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.
4. The value of limx→01−cos(x)x2\lim_{x \to 0} \frac{1 -
\cos(x)}{x^2}limx→0x21−cos(x) is
A. 0
B. 1
C. 12\frac{1}{2}21
D. ∞\infty∞
Answer: C) 12\frac{1}{2}21
Rationale: Using the Taylor series for cos(x)\cos(x)cos(x), we find
limx→01−cos(x)x2=12\lim_{x \to 0} \frac{1 - \cos(x)}{x^2} =
\frac{1}{2}limx→0x21−cos(x)=21.
5. 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.
6. If f(x)=2x+1f(x) = 2x + 1f(x)=2x+1 and g(x)=x2−1g(x) = x^2 -
1g(x)=x2−1, what is (f∘g)(x)(f \circ g)(x)(f∘g)(x)?
A. 2x2−12x^2 - 12x2−1
B. 2x2+12x^2 + 12x2+1
C. 2x2+32x^2 + 32x2+3
D. 2x2+22x^2 + 22x2+2
Answer: C) 2x2+32x^2 + 32x2+3
Rationale: The composition
(f∘g)(x)=f(g(x))=2(x2−1)+1=2x2−2+1=2x2+3(f \circ g)(x) = f(g(x)) =
2(x^2 - 1) + 1 = 2x^2 - 2 + 1 = 2x^2 +
3(f∘g)(x)=f(g(x))=2(x2−1)+1=2x2−2+1=2x2+3.
7. If tan(θ)=34\tan(\theta) = \frac{3}{4}tan(θ)=43, what is
sin(θ)\sin(\theta)sin(θ)?
A. 35\frac{3}{5}53
B. 45\frac{4}{5}54
1. 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.
2. The solution to x2+6x+5=0x^2 + 6x + 5 = 0x2+6x+5=0 is
A. x=−1,−5x = -1, -5x=−1,−5
B. x=1,−5x = 1, -5x=1,−5
C. x=−1,5x = -1, 5x=−1,5
D. x=1,5x = 1, 5x=1,5
Answer: A) x=−1,−5x = -1, -5x=−1,−5
Rationale: The quadratic x2+6x+5=0x^2 + 6x + 5 = 0x2+6x+5=0
factors as (x+1)(x+5)=0(x + 1)(x + 5) = 0(x+1)(x+5)=0, so x=−1x = -
1x=−1 and x=−5x = -5x=−5.
,3. 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.
4. The value of limx→01−cos(x)x2\lim_{x \to 0} \frac{1 -
\cos(x)}{x^2}limx→0x21−cos(x) is
A. 0
B. 1
C. 12\frac{1}{2}21
D. ∞\infty∞
Answer: C) 12\frac{1}{2}21
Rationale: Using the Taylor series for cos(x)\cos(x)cos(x), we find
limx→01−cos(x)x2=12\lim_{x \to 0} \frac{1 - \cos(x)}{x^2} =
\frac{1}{2}limx→0x21−cos(x)=21.
5. 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.
6. If f(x)=2x+1f(x) = 2x + 1f(x)=2x+1 and g(x)=x2−1g(x) = x^2 -
1g(x)=x2−1, what is (f∘g)(x)(f \circ g)(x)(f∘g)(x)?
A. 2x2−12x^2 - 12x2−1
B. 2x2+12x^2 + 12x2+1
C. 2x2+32x^2 + 32x2+3
D. 2x2+22x^2 + 22x2+2
Answer: C) 2x2+32x^2 + 32x2+3
Rationale: The composition
(f∘g)(x)=f(g(x))=2(x2−1)+1=2x2−2+1=2x2+3(f \circ g)(x) = f(g(x)) =
2(x^2 - 1) + 1 = 2x^2 - 2 + 1 = 2x^2 +
3(f∘g)(x)=f(g(x))=2(x2−1)+1=2x2−2+1=2x2+3.
7. If tan(θ)=34\tan(\theta) = \frac{3}{4}tan(θ)=43, what is
sin(θ)\sin(\theta)sin(θ)?
A. 35\frac{3}{5}53
B. 45\frac{4}{5}54
