MA1102R Calculus AY 2009/2010 Sem 1 NATIONAL UNIVERSITY OF SINGAPORE MATHEMATICS SOCIETY PAST YEAR PAPER SOLUTIONS solutions prepared by Boyan,Tay Jun Jie MA1102R Calculus AY 2009/2010 Sem 1 Question 1 (a) By L’Hˆopital’s Rule, limπ x→ 4 sec2 x − 2 tan x 2 sec2 x tan x − 2 sec2 x = limπ 1 + cos 4x −4 sin 4x x→ 4 2 sec4 x + 4 tan2 x sec2 x − 4 tan x sec2 x −16 cos 4x x→ 4 8+8−8 = −16 1 =− 2 = limπ (b) −x ≤ x cos √1x ≤ x Therefore, we have 1 lim −x ≤ lim x cos √ ≤ lim x x→0 x→0 x x→0 1 0 ≤ lim x cos √ ≤ 0 x→0 x By Squeeze Theorem, 1 lim x cos √ = 0 x x→0 Question 2 (a) Given > 0, let δ = 3 , such that |x − 2| < δ implies, 2 2 3x − 3x − 6 3x − x − 4 = 3 |x − 2| < 3δ = − 2 = x+1 x+1 (b) √ √ dy ((ex + 1) x2 + 2)0 (x − 8)5 − ((x − 8)5 )0 (ex + 1) x2 + 2 = dx ((x − 8)5 )2 √ √ ex x2 + 2 + (ex + 1) 2√2x (x − 8)5 − 5(x − 8)4 (ex + 1) x2 + 2 x2 +2 = (x − 8)10 √ √ ex x2 + 2 + (ex + 1) √xx2 +2 (x − 8) − 5(ex + 1) x2 + 2 = (x − 8)6 NUS Math LaTeXify Proj Team Page: 1 of 6 NUS Mathematics Society MA1102R Calculus AY 2009/2010 Sem 1 Question 3 (a) f 0 (x) = 3x2 − 18x + 24. let f 0 (x) > 0, we obtain that x > 4 or x < 2. let f 0 (x) < 0, we have 2 < x < 4. Therefore, f (x) is increasing on (−∞, 2) ∪ (4, ∞), and decreasing on (2, 4). (b) x ∈ R for f 0 (x), let f 0 (x) = 0, we can find out all the critical points at x = 2, and x = 4. Since f 0 (x) > 0 on (−∞, 2) ∪ (4, ∞), and f 0 (x) < 0 on (2, 4). By First Derivative Test, we obtain that f (x) have a local maximum at x = 2, and a local minimum at x = 4. (c) f 00 (x) = 6x − 18. let f 00 (x) > 0, then x > 3. let f 00 (x) < 0, then x < 3. Thus, f (x) is concave up on (3, ∞), and concave down on (−∞, 3). (d) x ∈ R for f 00 (x), let f 00 (x) = 0, we obtain its unique inflection point at x = 3. f (3) = 33 − 9 × 32 + 24 × 3 − 7 = 11. Hence, the coordinates of its inflection point is (3, 11). Question 4 Let x denotes the length of the side facing the main road in meters, f (x) denotes the total cost. Hence, x > 0, 1200 f (x) = 6x + 3 x + ×2 . x 7200 f (x) = 9x + . x 7200 f 0 (x) = 9 − 2 x √ 2. let f 0 (x) = 0, we have x = 20 √ √ In addition, f 0 (x) > 0 on (0, 20 2), and f 0 (x) < 0 on√(20 2, ∞). Hence, f (x) attains its absolute minimum at x = 20 2. Hence, √in order to minimize the cost of the fence, the length of the side facing the main road should be 20 2 meters. Question 5 (a) Z Z cos x ln(sin x) dx = sin x ln(sin x) − sin x 1 sin x cos x dx Z = sin x ln(sin x) − cos x dx = sin x ln(sin x) − sin x + C where C is a constant NUS Math LaTeXify Proj Team Page: 2 of 6 NUS Mathematics Society MA1102R (b) Let a = Calculus √ AY 2009/2010 Sem 1 2 − x, x = 2 − a2 , then a ∈ (0, 1), and dx = −2a da. Z 2 Z 1 √ x 2 − x dx = (2 − a2 )a(−2a) da 1 0 Z 1 2a4 − 4a2 da = 0 2 5 4 3 1 = a − a 5 3 0 2 4 = − 5 3 14 =− 15 Question 6 (a) 3 Z s s= 1+ 1 3 Z s = 1+ 1 3 dy dx 2 dx x2 1 − 2 2 2x 2 dx r 1 1 x4 + 4 − dx 1+ 4 4x 2 1 s Z 3 2 x 1 2 = + 2 dx 2 2x 1 Z 3 2 x 1 = + 2 dx 2 2x 1 3 3 x 1 + = 6 2x 1 Z = =4 (b) Z 2 (2x2 )2 dx a 4 5 2 =π x 5 a 128π 4π 5 = − a 5 5 r 2 Z 2a2 y 2 V2 = π a − dy 2 0 2 2a2 2 2a2 y =π a y 0 −π 4 0 V1 = π = 2a4 π − a4 π = a4 π NUS Math LaTeXify Proj Team Page: 3 of 6 NUS Mathematics Society MA1102R Calculus AY 2009/2010 Sem 1 thus, V = V1 + V2 4π 128π = − a5 + a4 π + 5 5 V 0 = −4a4 π + 4a3 π, a ∈ (0, 2), then V 0 = 0 at a = 1. In addition, V 0 > 0 in (0, 1), V 0 < 0 in (1, 2). By First Derivative Test, for a ∈ (0, 2), V attains its maximum value at a = 1. Question 7 (a) dy 1 + 2y = dx x + x3 dy 2 1 + y= 2 dx x x + x4 x Therefore, p(x) = x2 , the integrating factor is e R p(x) dx =e R 2 x dx = e2lnx = x2 . x2 2 x2 dy + yx2 = 2 dx x x + x4 d 2 1 x y= dx 1 + x2 R 1 2 dx + C y = 1+x 2 , x arctan x + C y= , x2 C∈R C∈R (b) dP = 0.0008P (100 − P ) dt 1 dP = dt 0.0008P (100 − P ) Z Z 1250 dP = dt 100P − P 2 Z 1 dP = t + C1 1250 502 − (P − 50)2 1250 P 1250 P ln = ln = t + C1 100 |P − 100| 100 100 − P P e0.08t+C2 = 100 − P 100 P (t) = 1 + C3 e−0.08t NUS Math LaTeXify Proj Team Page: 4 of 6 NUS Mathematics Society MA1102R Calculus AY 2009/2010 Sem 1 where C1 , C2 , C3 are some real numbers. Since P (0) = 20, we obtain C3 = 4. Hence, P (t) = 100 1 + 4e−0.08t Question 8 (a) Assume g(c) = 0 for some c ∈ (a, b) By Mean Value Theorem, there exist two values m ∈ (a, c), and n ∈ (c, b) and g(c) − g(a) =0 c−a g(b) − g(c) g 0 (n) = =0 b−c g 0 (m) = By Mean Value Theorem, there exists a value x ∈ (m, n) such that g 00 (x) = contradicts with g 00 (x) 6= 0. Thus, g(x) 6= 0 for all x ∈ (a, b). g 0 (n)−g 0 (m) n−m = 0 which (b) Let h(x) = f (x)g 0 (x) − f 0 (x)g(x) Thus, we obtain that h(a) = 0 and h(b) = 0. By Mean Value Theorem, there exists a value c ∈ (a, b) such that h0 (c) = 0. Hence, h0 (c) = f (c)g 00 (c) − f 00 (c)g(c) = 0 Since c ∈ (a, b), we have g(c) 6= 0 and g 00 (c) 6= 0, (c) we obtain that there exists c ∈ (a, b) such that fg(c) = f 00 (c) g 00 (c) . Question 9 Consider xg(x) for x ∈ R. Z 1 Z x xf (xt) dt = f (u) du where u = xt 0 0 Rx f (u) du ⇒ g(x) = 0 for x ∈ R \ {0} x xg(x) = Hence g(x) is differentiable for all x ∈ R \ {0} and Rx xf (x) − 0 f (u) du f (x) − g(x) 0 g (x) = = 2 x x Now, let limx→0 f (x) x for x ∈ R \ {0}. = M. f (x) ⇒ lim f (x) = lim ·x= x→0 x→0 x NUS Math LaTeXify Proj Team f (x) lim x→0 x Page: 5 of 6 lim x = M · 0 = 0 x→0 NUS Mathematics Society MA1102R Calculus Hence, by the continuity of f , f (0) = 0. Furthermore, notice that g(0) = AY 2009/2010 Sem 1 R1 0 f (0) dt = f (0) = 0. g(x) − g(0) x→0 Rx x f (u) du = lim 0 x→0 x2 f (x) = lim by L’Hˆopital’s rule x→0 2x M = 2 f (x) − g(x) lim g 0 (x) = lim x→0 x→0 x f (x) g(x) = lim − lim x→0 x x→0 x M M =M− = 2 2 ⇒ g 0 (0) = lim Therefore, g 0 is continuous at 0. NUS Math LaTeXify Proj Team Page: 6 of 6 NUS Mathematics Society MA1102R Errata: Calculus AY 2009/2010 Sem 1 NATIONAL UNIVERSITY OF SINGAPORE MATHEMATICS SOCIETY ERRATA FOR PAST YEAR PAPER SOLUTIONS MA1102R Calculus AY 2009/2010 Sem 1 Compiled November 22, 2014 Written by Lee Kee Wei Question 1a 8+8−8 = 12 . 16 Audited by Henry Morco The original solution evaluated the limit to be 8+8−8 −16 = − 21 . It should have been R1 Question 5b The original solution gave the definite intergal to be 0 (2 − a2 )a(−2a)da after the R0 substitution. The substitued intergal should have been 1 (2 − a2 )a(−2a)da. This would lead the answer 14 14 to be 15 instead of the given − 15 . Question 6a The original solution evaluated the integral as: R3 2 3 1 3 have been: 1 x2 + 2x12 dx = [ x6 − 2x ]1 = 14 3 . R3 1 x2 2 3 1 3 + 2x12 dx = [ x6 + 2x ]1 = 4. It should END OF ERRATA NUS Math LATEXify Proj Team Page 1 of ?? NUS Mathematics Society

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