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plugin-e3_prac_S2010.pdf5

# plugin-e3_prac_S2010.pdf5 - M408D Practice problems for...

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Unformatted text preview: M408D Practice problems for Exam #3 Directions. Indicate the correct answer for each problem by ﬁlling in the appropriate space, as in (A) (•) (C) (D) (E). 1) If f (t) and g (t) are as shown to the right, which ﬁgure below shows the parametric curve (x, y ) = (f (t), g (t)), 0 ≤ t ≤ 3? y f y g (A) (B) y (C) x y (D) x y (E) x x 1 2 3 t 1 2 3 t x 2) The second-degree Taylor polynomial for f (x) = 2 √ 3 + x at the reference point a = 1 is: (A) 2 + (x − 1) (x − 1)2 (x − 1) (x − 1)2 (x − 1) (x − 1) − . (B) 1 − + . (C) 2 − − . 4 64 2 32 2 16 (x − 1) (x − 1)2 (x − 1) (x − 1)2 + . (E) 2 + − . (D) 4 64 8 24 3) A series representation for f (x) = e3x at the reference point a = 0 is: ∞ 2 (A) k=0 (−1)k 3k xk . (k + 2)! ∞ (B) k=0 xk . 32k k ! ∞ (C) k=0 3k x2k . k! ∞ (D) k=0 3kxk+2 . k! ∞ (E) k=0 32k x2k . (2k )! y 1 t= π/2 C t=0 4x 4 S 4) A surface S is formed by rotating a quarter-ellipse C about the x-axis. Which integral below represents the volume enclosed by S ? 4 0 0 0 2 (A) 2 π sin3 t dt . (B) π/2 2π sin t cos2 t dt . (C) π/2 4π cos3 t dt . (D) π/2 2π sin3 t dt . (E) 2 π sin2 t dt . ∞ 5) The interval of convergence of the series k=1 k (x − 2)k is: (C) [−3, 1). (D) [1, 2]. (E) (−∞, ∞). (A) (−2, 1]. (B) (1, 3). 6) The slope dy of the curve x = s + ln(s), y = 2s ln(3s) at s = 1 is: dx 3 (A) 3 + ln 6. (B) 2 − ln 3. (C) 1 + ln . 2 (D) 3 − ln 6. (E) 1 + ln 3. 7) Which one of the following ﬁgures shows a polar graph of r = 3 cos θ, 0 ≤ θ ≤ π ? (A) y (B) x y y (C) x (D) y y (E) x x x 8) A series representation of f (x) = ∞ ∞ 2x2 with reference point a = 0 is: 1 − x2 ∞ ∞ ∞ (A) k=0 2k xk . (B) k=0 (−1)k 2xk+2 . (C) k=0 (−2)k xk+2 . (D) k=0 2x2k+2 . (E) k=0 2xk+2 . 9) If T1 (x) is the ﬁrst-degree Taylor polynomial for x1/2 at a = 1, then Taylor’s inequality implies that the error |x1/2 − T1 (x)| in the interval [1, 1.3] is less than or equal to: (A) 1 . 900 (B) 3 . 200 (C) 1 . 300 (D) 7 . 250 (E) 9 . 800 10) If f (x) = ∞ (−1)n xn , then n! n=0 (B) ∞ 1/2 f (x) dx = 0 (A) (−1)m−1 2m . (m + 1)! m=1 (−1)m . 2m (m + 1)! m=1 ∞ (C) (−1)m−1 . 2 m m! m=1 ∞ (D) (−1)m−1 2m−1 . m! m=1 ∞ (E) (−1)m . 2 m− 1 m ! m=1 ∞ y r = sin(2θ) 11) The area of the shaded region enclosed by the polar graph in the ﬁgure is: (A) π . (B) π . 3 (C) π . 4 (D) 3π . 4 (E) π . 8 x 12) If f (x) = 4x3 ex−1 , then by using the ﬁrst two non-zero terms of an appropriate Taylor series we ﬁnd f (1.1) ≈ (A) 28 . 5 (B) 14 . 3 (C) 17 . 4 (D) 29 . 4 (E) 31 . 5 ...
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