Final examination semester 2 academic year 2019-2020 course name Calculus 1 - ĐH Sư phạm Kỹ thuật

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HCMC UNIVERSITY OF TECHNOLOGY AND FINAL EXAM, SEMESTER 2, 2019-2020 EDUCATION Subject: Calculus 1 HIGH QUALITY TRAINING FACULTY Course code: MATH141601E GROUP OF MATHEMATICS Number of pages: 02 pages. ------------------------- Duration: 90 minutes. Date of exam: 31/07/2020 Materials are allowed during the exam. Question 1 (1 point) Given f ( x ) = x 3 + 3, g ( x) = cos (2 x ) . a) f ( x) and g ( x) are even, odd or neither? b) Find ( f g )( x) and ( g f )( x) . Question 2 (1.5 points) a) Find the value of the constant m such that the following piecewise - defined function is continuous everywhere.  e3 x − 1  , x≠0 f ( x) =  x . m , x=0  b) With m found in question a), find f ' ( x ) , ∀x. Question 3 (1 point) Let y be an implicit function of x satisfying the equation: x 2 + xy − y 3 = 1 Find the tangent line to the graph of the equation at the point M (1; −1) . Question 4 (1 point) Find the relative extrema of g ( x ) = ( x − 2)e − 0.5 x . Question 5 (1 point) x+2 Let f ( x ) = . Find the average value of f on the interval [ 0, 1] . 4 − x2 Question 6 (1 point) Find the particular solution of the separable differential equation dy y ln y = 2x dx e satisfying the initial condition y ( 0 ) = e. Question 7 (1.5 points) Assume that the position at time t of an object moving along a line is given by s ( t ) = 2t 3 − 15t 2 + 36t for t on [1, 4]. a) Find the initial velocity and acceleration for the object. b) When is the object advancing and retreating? c) When is the object accelerating and decelerating? No.: BM1/QT-PĐBCL-RĐTV Page: 1
Question 8 (1 point) The volume of a spherical balloon is increasing at a constant rate of 10 cm3 / s . At what rate is the radius of the balloon increasing when the radius is 3 cm? Question 9 (1 point) A cylinder box (Figure 1) is constructed with the volume V = 24π cm3 . The cost of the material used for the bottom is $2/cm2, the cost of the material used for the lateral side is $2.5/cm2 and the cost of the material used for the top is $5/cm2. Find the dimension of the box r and h that minimizes the total cost. Notice: Invigilators should not explain the questions on the exam papers. Expected Learning Outcomes Questions [G 2.1]: Present mathematical information using words, statements, 1 numbers, formulas, graphs and diagrams [G 1.1, 1.3, 5.2]: Students are able to find basic limits and test the 2, 3, 4 continuity of a function. Students are able to find derivative and differential. [G 3.1, 5.4] : Apply important rules and theorems effectively, such 5 as the mean value. Students are able to apply theory to evaluate indefinite and definite integrals. [ELO 1.4, 5.4]: Students are able to solve basic differential 6 equations. [G 4.1] Identify and analyze the information given in formulas, 7 graphs and tables relating to (a) rectilinear motion; (b) linear; (c) optimization and applications in physics; (d) Riemann sum and integration [G 5.3]: Students are able to use derivative to solve problems 8,9 relating to rates of change and optimization July 24th, 2020 Head of group of mathematics
Đáp án Q1: (1 point) a) Because of f ( − x ) ≠ ± f ( x ) , f(x) is neither odd nor even. (0.25pt) Because of g ( − x) = g ( x ) , g(x) is even. (0.25pt) b) ( f g )( x) = (cos(2 x))3 + 3;( g f )( x) = cos(2( x 3 + 3)). (0.25pt+0,25pt) Q2:(1,5 points) D f = ℝ . a) x ≠ 0 : f ( x ) is continuous with all x ≠ 0 . f ( x) is continuous with all x ∈ ℝ ⇔ f ( x ) is continuous with at x = 0 .(0.25pt) e3 x − 1 ⇔ lim f ( x ) = f (0) ⇔ lim = m ⇔ m = 3. (0.25pt+0,25pt) x →0 x →0 x 3e3 x .x − e3 x + 1 b) x ≠ 0 : f ′( x) = . (0.25pt) x2 e3 x − 1 − 3 x 9 f ′(0) = lim = . (0.25pt+0,25pt) x→0 x2 2 Q3: (1 point) Let F ( x, y ) = x 2 + xy − y 3 − 1 = 0 (*) Differentiate both sides of the equation (*) with respect to x, we get 2x + y 2 x + y + xy '− 3 y 2 y ' = 0 ⇔ y ' = 2 (0.25pt+0,25pt) 3y − x 1 At M(1; -1): y '(1) = . (0.25pt) 2 The quation of the tangent line to the graph of the equation at the point M is : 1 1 3 (d ) : y = ( x − 1) − 1 ⇔ y = x − . (0.25pt) 2 2 2 Q4 (1 point) Dg = ℝ . We have g '( x ) = e −0.5 x (2 − 0.5 x ). (0.25pt) g '( x ) = 0 ⇔ x = 4. (0.25pt) g(x) is increasing on ( −∞, 4) and is decreasing on (4, +∞ ) (0.25pt) So, g(x) has a relative maximum at x = 4, f(4) = 2e-2. (0.25pt) Q5 (1 point) 1 x+2 The average value of f on [0, 1] is AV = ∫ dx . (0.25pt) 0 4 − x2 No.: BM1/QT-PĐBCL-RĐTV Page: 3
1 1 x 2 AV = ∫ 0 4− x 2 dx + ∫ 0 4 − x2 dx = I1 + I 2 1 3 x I1 = ∫ dx = − 4 − x 2 = 2 − 3. (0.25pt) 0 4 − x2 2 1 1 1 x π I 2 = 2∫ −1 dx = 2sin   = . (0.25pt) 0 22 − x 2 20 3 π So AV = 2 − 3 + . (0.25pt) 3 Q6 (1 point): dy dx dy dx We get = 2x ⇒ ∫ = ∫ 2x (0.25pt) y ln y e y ln y e −2 x e Then we have ln | ln y |= + C. (0.25pt+0.25pt) −2 1 e −2 x 1 Because of y (0) = e , C = . Then ln | ln y |= + . (0.25pt) 2 −2 2 Q7(1.5 points): v(t ) = s '(t ) = 6t 2 − 30t + 36 ⇒ v(1) = 12 (0.25pt) a) a(t ) = v '(t ) = 12t − 30 ⇒ a(1) = −18 (0.25pt) b) The object is advancing ⇔ v(t ) > 0 ⇔ t ∈ [1, 2) or t ∈ (3, 4] (0.25pt) The object is retreating ⇔ v (t ) < 0 ⇔ t ∈ (2,3) (0.25pt) c) The object is accelerating ⇔ a (t ) > 0 ⇔ t ∈ (2.5, 4] (0.25pt) The object is decelerating ⇔ a (t ) < 0 ⇔ t ∈ [1, 2.5) (0.25pt) Q8 ( 1 point) 4 V = π R3 (0.25pt) 3 dV dV dR dR = . = 4π R 2 . (0.25pt) dt dR dt dt dR R = 3 ⇒ 10 = 4π .32. (0.25pt) dt dR 5 ⇒ = (cm / s ) dt 18π 5 So the radius is increasing at the rate of (cm / s ). (0.25pt) 18π No.: BM1/QT-PĐBCL-RĐTV Page: 4
Q9 (1 point) 24 V = π r 2 h = 24π ⇒ h = (0.25pt) r2 120π The total cost is TC ( r ) = 7π r 2 + 5π rh = 7π r 2 + ( r > 0) (0.25pt) r 120π 60 TC '( r ) = 14π r − 2 ( r > 0); TC '( r ) = 0 ⇔ r = 3 (0.25pt) r 7 240π TC '( r ) = 14π + 3 > 0 ∀ r > 0 r 2 2 (0.25pt)  60  7 60  7  ⇒ MinTC = 7π  3  + 120π 3 when r = 3 , h = 24  3   7  60 7  60  No.: BM1/QT-PĐBCL-RĐTV Page: 5