Final examination semester 2 academic year 2019-2020 course name Calculus 1 giúp các bạn học sinh có thêm tài liệu ôn tập, luyện tập nhằm nắm vững được những kiến thức, kĩ năng cơ bản, đồng thời vận dụng kiến thức để giải các bài tập một cách thuận lợi.
HCMC UNIVERSITY OF TECHNOLOGY AND EDUCATION HIGH QUALITY TRAINING FACULTY GROUP OF MATHEMATICS - FINAL EXAM, SEMESTER 2, 2019-2020 Subject: Calculus Course code: MATH141601E Number of pages: 02 pages Duration: 90 minutes Date of exam: 31/07/2020 Materials are allowed during the exam Question (1 point) Given f ( x ) = x + 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 (1.5 points) a) Find the value of the constant m such that the following piecewise - defined function is continuous everywhere e3 x − , x≠0 f ( x) = x m , x=0 b) With m found in question a), find f ' ( x ) , ∀x Question (1 point) Let y be an implicit function of x satisfying the equation: x + xy − y = Find the tangent line to the graph of the equation at the point M (1; −1) Question (1 point) Find the relative extrema of g ( x ) = ( x − 2)e − 0.5 x Question (1 point) x+2 Let f ( x ) = Find the average value of f on the interval [ 0, 1] − x2 Question (1 point) Find the particular solution of the separable differential equation dy y ln y = 2x dx e satisfying the initial condition y ( ) = e Question (1.5 points) Assume that the position at time t of an object moving along a line is given by s ( t ) = 2t − 15t + 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: Question (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 cm? Question (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 [G 2.1]: Present mathematical information using words, statements, numbers, formulas, graphs and diagrams [G 1.1, 1.3, 5.2]: Students are able to find basic limits and test the continuity of a function Students are able to find derivative and differential [G 3.1, 5.4] : Apply important rules and theorems effectively, such 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 equations [G 4.1] Identify and analyze the information given in formulas, 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 relating to rates of change and optimization Questions 2, 3, 8,9 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 Because of g ( − x) = g ( x ) , g(x) is even b) ( f g )( x) = (cos(2 x))3 + 3;( g f )( x) = cos(2( x + 3)) (0.25pt) (0.25pt) (0.25pt+0,25pt) Q2:(1,5 points) D f = ℝ a) x ≠ : f ( x ) is continuous with all x ≠ f ( x) is continuous with all x ∈ ℝ ⇔ f ( x ) is continuous with at x = (0.25pt) e3 x − (0.25pt+0,25pt) ⇔ lim f ( x ) = f (0) ⇔ lim = m ⇔ m = x →0 x →0 x 3e3 x x − e3 x + (0.25pt) b) x ≠ : f ′( x) = x2 e3 x − − x (0.25pt+0,25pt) = f ′(0) = lim x→0 x2 Q3: (1 point) Let F ( x, y ) = x + xy − y − = (*) Differentiate both sides of the equation (*) with respect to x, we get 2x + y (0.25pt+0,25pt) x + y + xy '− y y ' = ⇔ y ' = 3y − x At M(1; -1): y '(1) = (0.25pt) The quation of the tangent line to the graph of the equation at the point M is : 1 (0.25pt) (d ) : y = ( x − 1) − ⇔ y = x − 2 Q4 (1 point) Dg = ℝ We have g '( x ) = e −0.5 x (2 − 0.5 x ) g '( x ) = ⇔ x = g(x) is increasing on ( −∞, 4) and is decreasing on (4, +∞ ) So, g(x) has a relative maximum at x = 4, f(4) = 2e-2 (0.25pt) (0.25pt) (0.25pt) (0.25pt) Q5 (1 point) The average value of f on [0, 1] is AV = ∫ No.: BM1/QT-PĐBCL-RĐTV x+2 − x2 dx (0.25pt) Page: AV = ∫ I1 = ∫ 4− x x − x2 I = 2∫ x dx + ∫ − x2 dx = I1 + I dx = − − x = − π x dx = 2sin = 20 22 − x So AV = − + (0.25pt) −1 π (0.25pt) (0.25pt) Q6 (1 point): dy dx dy dx We get = 2x ⇒ ∫ = ∫ 2x y ln y e y ln y e −2 x e Then we have ln | ln y |= + C −2 e −2 x 1 Because of y (0) = e , C = Then ln | ln y |= + −2 Q7(1.5 points): v(t ) = s '(t ) = 6t − 30t + 36 ⇒ v(1) = 12 a) a(t ) = v '(t ) = 12t − 30 ⇒ a(1) = −18 b) The object is advancing ⇔ v(t ) > ⇔ t ∈ [1, 2) or t ∈ (3, 4] The object is retreating ⇔ v (t ) < ⇔ t ∈ (2,3) c) The object is accelerating ⇔ a (t ) > ⇔ t ∈ (2.5, 4] The object is decelerating ⇔ a (t ) < ⇔ t ∈ [1, 2.5) (0.25pt) (0.25pt+0.25pt) (0.25pt) (0.25pt) (0.25pt) (0.25pt) (0.25pt) (0.25pt) (0.25pt) Q8 ( point) V = π R3 dV dV dR dR = = 4π R dt dR dt dt dR R = ⇒ 10 = 4π 32 dt dR ⇒ = (cm / s ) dt 18π So the radius is increasing at the rate of No.: BM1/QT-PĐBCL-RĐTV (0.25pt) (0.25pt) (0.25pt) (cm / s ) 18π (0.25pt) Page: Q9 (1 point) V = π r h = 24π ⇒ h = 24 r2 (0.25pt) 120π ( r > 0) (0.25pt) r 120π 60 TC '( r ) = 14π r − ( r > 0); TC '( r ) = ⇔ r = (0.25pt) r 240π TC '( r ) = 14π + > ∀ r > r The total cost is TC ( r ) = 7π r + 5π rh = 7π r + 60 ⇒ MinTC = 7π + 120π No.: BM1/QT-PĐBCL-RĐTV 60 when r = , h = 24 60 60 (0.25pt) Page: ... rate of No.: BM1/QT-PĐBCL-RĐTV (0 .25 pt) (0 .25 pt) (0 .25 pt) (cm / s ) 18 π (0 .25 pt) Page: Q9 (1 point) V = π r h = 24 π ⇒ h = 24 r2 (0 .25 pt) 12 0 π ( r > 0) (0 .25 pt) r 12 0 π 60 TC '( r ) = 14 π r − ( r... f(4) = 2e -2 (0 .25 pt) (0 .25 pt) (0 .25 pt) (0 .25 pt) Q5 (1 point) The average value of f on [0, 1] is AV = ∫ No.: BM1/QT-PĐBCL-RĐTV x +2 − x2 dx (0 .25 pt) Page: AV = ∫ I1 = ∫ 4− x x − x2 I = 2? ?? x dx... + ∫ − x2 dx = I1 + I dx = − − x = − π x dx = 2sin = ? ?2? ??0 22 − x So AV = − + (0 .25 pt) ? ?1 π (0 .25 pt) (0 .25 pt) Q6 (1 point): dy dx dy dx We get = 2x ⇒ ∫ = ∫ 2x y ln y e y ln y e ? ?2 x e Then