MULTIPLE CHOICE QUESTIONS 1 Functions 1.1 Domain and range of a function s 9 Question 1 (L.O.1). Find the domain of function y = ln 3 + . x 9 9 x⩽− x⩽− 9 9 2 2 C x⩽− A − ⩽x<0 B D 2 2 x>0 x⩾0 E x>0 Question 2 (L.O.1). Find the range of the function 9x2 − 3, if x > 0 f (x) = 21x3 , if x ⩽ 0 A (0, +∞) B (−∞, +∞) D (−3, +∞) C (−∞, 0) E (−3, 0) Question 3 (L.O.2). A family plans to organize a party at a restaurant. The overall service cost for the event (venue, decoration, etc.) is 4 million Vietnamese dong. The restaurant requires a minimum of 2 tables, with each table accommodating 8 people. The cost per guest is 258 thousand Vietnamese dong. The family plans to spend a maximum of 36.6 million Vietnamese dong. If we denote x as the number of invited guests, C(x) (in million Vietnamese dong) represents the cost for the party. Find the domain D and the range R of C. A D = N ∩ [8; 142], R = [2.064; 36.6] B D = [16; 142], R = [8.128; 36.6] C D = N ∩ [16; 142], R = [8.128; 36.6] D D = N ∩ [0; 142], R = [0; 36.6] E D = [16; 36.6], R = [4.016; 142] 1.2 Description of a function Question 4 (L.O.2). A rectangular storage container with an open top has a volume of 14m3 . The length of its base is quadruple its width. Material for the base costs 11 dollars per square meter; material for the sides costs 8 dollars per square meter. Express the cost of materials as a function of the width w of the base. A 44w2 + 280 w B 44w2 + 280 C 32w2 + 1120 w D 32w2 + 385 w E 44w2 + 140 w Question 5 (L.O.2). To create an animated image appealing to players in a game, the game creator started by creating a rectangular image with dimensions of 29 × 71 pixels. Then, each side is increased at a rate of 15 pixels per second. Determine the area of the image after t seconds. A 2059 D 225t2 EXAM B 225t2 + 1500t + 2059 C 225t2 + 1065t E 225t2 + 435t Page 1/1 — Question code: 1234 Question 6 (L.O.2). The amount of adrenaline in the body changes rapidly. Let’s assume that initially the body has 18 mg of adrenaline, create a function describing the adrenaline level A = f (t) in the body after t (minutes), given that A increases 0.39 mg per minute. A 18t + 0.39 1.3 C 0.39 − 18t B 7.02t E 18 − 0.39t D 0.39t + 18 Inverse functions Question 7 (L.O.1). Let f (x) = x5 + 3x3 + 6x − 5. What is the value of f −1 (63). A 1 2 B 0 D 2 C 6 E 3 The limit of a function, continuous functions 2.1 The limit of a function e3.19x − cos(x) lim x→0 x Question 8 (L.O.1). Find m > 0 such that A 1.6481 2 .m + ln(1 + 6.04x) lim x→0 x C 1.2978 D 1.3338 6x 6x + m2 = e26.55m . Question 9 (L.O.1). Find m > 1 such that lim x→∞ 6x − 1 C 27.3814 A 26.5123 B 26.5475 D 26.8591 12 Question 10 (L.O.1). Find lim 1 + 2x2 e7x 9x . A e−7/9 B 1.2026 .m = 18.62. E 2.3985 E 26.4543 x→0 C −e2/9 B 1 E e2/9 D 0 Question 11 (L.O.1). Find S = 4a + 8b where a and b are the two real constants such that 2 lim x ea/x + + b = 15. x→+∞ x A 43 B 40 C 48 D 44 e4ax − 1 − 9x · x→0 x + 13x2 9 B 4a − 9, ∀a ̸= 4 E 4a − 9, ∀a ∈ R E 42 Question 12 (L.O.1). Evalulate I = lim 9 4 D 4a − 9, ∀a ̸= 0 A 4a, ∀a ̸= 2.2 C 4a, ∀a ∈ R Continuity Question 13 (L.O.1). Find real values of a such that f (x) = at x = 13. 2 A 13 B − 24 13 C − 37 13 ax − 2, √ arctan( x − 13), D − 50 13 3 5 arctan x − 31 x−3 Question 14 (L.O.1). Find real values of a such that f (x) = a, at x = 3. 5π 3π 5π A − B C 2 2 2 7π D a does not exist E 2 EXAM if x ⩽ 13 is continuous if x > 13 E − if x ̸= 3 11 13 is continuous if x = 3 Page 2/1 — Question code: 1234 2.3 Asymptotes Question 15 (L.O.1). Find all real numbers a so that the line of equation x = 0 is a vertical asymptote of ln(7x + a − 8) the curve y = · x+9 C 6 A 11 B 7 D 8 E 12 9x − 5. 1 − e−x C y = 9x − 5 Question 16 (L.O.1). Find all asymptotes of the graph of the function y = 4x + √ A x = 0, y = 13x − 5 B y = 13x − 5 D x = 0, y = 4x − 5 E y = 4x − 5 Question 17 (L.O.2). A plastic molding company states that if they use x% of the company’s machines, the total monthly cost for operating these machines is given by the function: C(x) = 19x2 − 57x − 60610 4x2 − 100x − 7656 (hundreds of dollars). The company employs an alternating maintenance mode to utilize close to 58% of the machines (optimal capacity).What is the total cost that the company has to incur for operating this number of machines if the machine usage reaches the optimal capacity? A 5.8984 3 B 5.9918 C 5.7198 D 6.6802 E 5.7032 Derivatives 3.1 Tangents Question 18 (L.O.1). A point M (a, b) is on the graph of the function f (x) = 1.55ex − x. Given that the tangent line to the graph at M is parallel to the line y = 2.47x + 2.82. Find the value of a. A 0.8059 C 1.7944 B 0.4466 D 1.4552 E 1.2394 Question 19 (L.O.1). Find ALL points on the graph of f (x) = 4x3 − 192x where the tangent line is parallel to the x−axis. A (4; 512) B (4; −512) and (−4; 512) D (4; 512) and (−4; −512) C (4; −512) E (−4; 512) Question 20 (L.O.1). Find ALL points on the graph of f (x) = 9 where the tangent line is perpendicular to x the line y = 16x − 4. A (−12; 3/4) C (12; −3/4) and (−12; 3/4) B (12; 3/4) D (−12; −3/4) E (12; 3/4) and (−12; −3/4) √ Question 21 (L.O.1). Find the real value of a such that the tangent line to the graph of the function y = 3 x 21 at x = a has y−intercept . 2 C 49 A 46 B 52 D 44 E 48 3.2 Tangent to paramtric curve Question 22 (L.O.1). Let the curve be given by paramtric equations x(t) = (3 − 4t) e3t y(t) = 3t2 − 5t − 4 e3t EXAM Page 3/1 — Question code: 1234 37 Find the set of all values of t such that the tangent at M (x(t), y(t)) has the slope of − ? 31 38 503 410 131 148 A − ;3 B − C − D − E ; −1 ;7 ;4 ;1 93 93 93 93 93 3.3 Differentiable functions x2 − 11 is not differentiable. x2 − 25 A x ̸= 5 B x = −5 D x = ±5 E x ̸= ±5 C x=5 p Question 24 (L.O.2). Determine ALL values of x such that f (x) = 3 (x2 − 25)2 is not differentiable. Question 23 (L.O.2). Determine ALL values of x such that f (x) = A x = ±5 B x ̸= ±5 C x ̸= 5 D x = −5 E x=5 Question 25 (L.O.2). Find S = a + b such that function f is differentiable at x = 0, if 2x2 + 7x + 5, x ⩽ 0 f (x) = ln(ax + b), x > 0 A 7e5 3.4 B 8 C 8e5 D 9 E 6e5 The chain rule Question 26 (L.O.1). Let f (x) and g(x) be functions satisfying the conditions: f (−4) = f ′ (−4) = 9 and g(9) = g ′ (9) = −4. Consider the function h(x) = (f ◦g◦f )(x) = f (g(f (x))). Find the value of h′ (−4). A −328 B −325 C −322 D −324 E −329 Question 27 (L.O.2). An importer of Rwandan coffee estimates that local consumers will buy approxi4346 pounds of the coffee per week when the price is p dollars per pound. It is estimated mately D(p) = p2 that t weeks from now, the price of Rwandan coffee will be p(t) = 0.02t2 + 0.08t + 9 dollars per pound. At what rate will the weekly demand for the coffee be changing with respect to time 6 weeks from now? A −3.1978 B −2.3627 C −3.4126 D −3.2408 E −2.621 Question 28 (L.O.2). When electric blenders are sold for p dollars apiece, local consumers will buy approx7928 imately D(p) = blenders per month. It is estimated that t months from now, the price of the blenders p√ will be p(t) = 0.03 t3 + 4 dollars. Compute the rate at which the monthly demand for the blenders will be changing with respect to time 16 months from now. A −40.9061 3.5 B −40.7186 C −40.9638 D −40.9057 Derivatives of inverse function Question 29 (L.O.1). If f (x) = 2x + 2 sinh (x − 2), find (f −1 )′ (4). 9 19 13 15 A B − C D − 4 4 4 4 3.6 E −41.3092 E 1 4 Undetermined Coefficients Question 30 (L.O.2). Let y = Ax3 + Bx + C, where A, B, C ∈ R, satisfy the equation y ′′′ + 2y ′′ − 5y ′ + 9y = 27x3 − 45x2 + 54x + 89. Find the value of S = A + B + C. A 11 EXAM B 9 C 16 D 15 E 14 Page 4/1 — Question code: 1234 Question 31 (L.O.2). Let y = Ax cos x + Bx sin x, where A, B ∈ R, satisfy the equation y ′′ + y = −4 sin x + 10 cos x. Find the value of S = A + B. A 6 B 9 C 8 D 5 E 7 Question 32 ([L.O.2). Let y = Axp + Bx + C, where A, B, C, p ∈ R, satisfy the equation y ′′ = y ′ (1) = −3.0, y(1) = 5.0. Find the value of S = (A + B + C).p. A 11 3.7 B 7 C 9 D 8 30 and x7 E 12 Rates of change Question 33 (L.O.2). The management fee for running services in an area is calculated by the formula f (x) = 0.0692x2 + 0.0902x (in dollars), where x is the number of homes being present in the area. Suppose that at the moment when 63 homes are present in this area, the number of homes is increasing at a rate of 5 homes per week. What is the rate of change (in dollars/week) in the management fee at this moment? A 44.047 B 44.5156 D 43.2984 C 43.0986 E 44.6353 25 · If the abscissa of x the particle increases by 0.3234 units per second, how does its ordinate vary as it passes through the point Question 34 (L.O.2). A particle is moving along a hyperbolic trajectory given by y = (5; 5)? A Increases 0.0971 units per second B Increases 0.8569 units per second C Decreases 0.4064 units per second D Decreases 0.1947 units per second E Decreases 0.3234 units per second 3.8 Related Rates Question 35 (L.O.2). Recall that volume of a right cylinder is V = π.r2 .h, where r is the radius of the base and h is the height. Given a right cylinder with the radius of the base of 9 cm and the height of 5 cm. The radius of the base is increasing at a rate of 0.8 cm/s, and its height is increasing at a rate of 0.21 cm/s. How fast is the volume of cylinder increasing? A 278.9661 B 280.2004 C 279.6332 D 279.1033 E 279.8283 Question 36 (L.O.2). Water pours into a conical tank of height 10(m) and radius 4(m) at a rate of 7(m3 /min). 4 r 10 h At what rate is the water level rising when the level is 5(m) high? A 0.5570 EXAM B −0.3594 C 0.1141 D 0.0975 E 0.7804 Page 5/1 — Question code: 1234 Question 37 (L.O.2). A leaky water tank is in the shape of an inverted right circular cone with depth of 11(m) and top radius 3(m). When the water in the tank is 3(m) deep, it is leaking out at a rate of 1/6(m3 /min). 3 r 11 h How fast is the water level in the tank dropping at that time? A 0.0792 B 0.3362 C −0.6292 D 0.1284 E −0.5812 Question 38 (L.O.2). Boyle’s law states that when a sample of gas is compressed at a constant temperature, the pressure P and volume V satisfy equation P V = C = const. Suppose that at a certain instant the volume is 647cm3 , the pressure is 184kP A, and the pressure is increasing at a rate of 28kP A/min. At what rate is the volume decreasing at this instant? A 98.1919 3.9 B 99.3825 C 98.4565 D 97.6455 E 98.362 D −299dx2 E −294dx2 Differentials Question 39 (L.O.1). Find d2 y(0) if y = cos6 (7x). A −292dx2 3.10 B −289dx2 C −298dx2 Applications of derivatives in physics Question 40 (L.O.1). A particle moving on the x−axis has position s(t) = t3 − 39t2 + 432t + 12 (m) after an elapsed time of t seconds. What is the total distance travelled by the particle during the first 31 seconds? A 6699 B 6707 C 6704 D 6709 E 6702 Question 41 (L.O.1). A particle moving on the x−axis has position s(t) = t3 − 45t2 + 600t + 9 (m) after an elapsed time of t seconds. What is the displacement travelled by the particle during the first 37 seconds? A 11243 3.11 B 11250 C 11248 D 11251 E 11253 Applications of derivatives in Economics Question 42 (L.O.2). For Luce Landscaping, the total revenue from the yard maintenance of x homes is given by R(x) = 1868x − 4x2 (dollars) and the total cost is given by C(x) = 2938 + 18x (dollars). Suppose that Luce is adding 19 homes per day at the moment when the 458th customer is signed. At that moment, what is the rate of change of total profit P in dollars per day, if the total profit P (x) = R(x) − C(x)? A −34469 EXAM B −34465 C −34471 D −34466 E −34463 Page 6/1 — Question code: 1234 3.12 Linear Approximations Question 43 (L.O.1). Use the linear approximation formula to find an approximate value of f (x) at x = 12.17, given that f (12) = 4 and f ′ (12) = 0.74. A 3.9803 B 4.4361 C 4.1258 D 3.8998 E 3.8271 Question 44 (L.O.2). A spherical balloon is inflated so that its radius increases from 20 cm to 20.44 cm in 4 seconds. By approximately how much has its volume increased (using linear approximation) (in cm3 /s)? A 552.2614 B 552.9203 C 553.2891 D 552.9905 E 552.0326 Question 45 (L.O.2). A company estimates that volume sales S(x) of a type of commodity depends on the amount of money x (in thousands dollars) spent on advertising, that is given by: S(x) = −0.0079x3 + 0.412x2 + 2.8147x + 338. Use the differential (also known as: the linear approximation formula) to estimate the change in the total sales of this commodity if the advertising expenditures are increased from 41 to 41.7583 (thousands dollars). A −2.4577 B −1.7600 C −2.4103 D −2.4215 E −1.9167 Question 46 (L.O.2). A ball of ice melts so that its radius decreases from 30 cm to 29.1647 cm. By approximately how much does the volume of the ball decrease? A −9447.0725 B −9446.1664 C −9446.7456 D −9447.01 E −9447.2685 Question 47 (L.O.2). A water reservoir has a base shaped like a rectangle combined with half of a circle, as illustrated in the picture. x 4x If we can measure x = 5 ± 0.0599(m). Approximate the error in the area of the reservoir’s base using differential formula. A 6.2392 B 5.6269 C 6.6628 D 6.6852 E 6.1596 Question 48 (L.O.1). Let f be an even function and f is differentiable in R. Given that f ′ (1) = 2.81, f (1) = −3. Use differential formula to approximate f (−0.22). A −6.0772 EXAM B −5.1918 C −5.673 D −6.0849 E −6.0731 Page 7/1 — Question code: 1234 3.13 Monotonicity ln(3x + 4)2 , then which statement is always true? 3x + 4 4 e 4 e A The function y is decreasing on − − , − + 3 3 3 3 4 4 e 4 e 4 and − , − + B The function y is increasing on − − , − 3 3 3 3 3 3 4 e 4 e C The function y is increasing on − − , − + 3 3 3 3 Question 49 (L.O.1). If y = D The function y is increasing on R 4 e 4 e E The function y is increasing on − − , 0 ∪ 0, − + 3 3 3 3 tan x − 3 Question 50 (L.O.1). Find all real numbers m such that y = is increasing on the interval tan x − m π . 0; 4 A 1⩽m<3 B m>3 C m⩾0 D m ⩽ 0 or 1 ⩽ m < 3 3.14 E m⩽0 Local extreme values Question 51 (L.O.1). Studying the local extreme values of the function f (x) = arctan 3x − 9 x2 + 72 , which statement is always true? A f has local minimum when x = −6, and no local maximum B f has local minimum when x = −6, and local maximum when x = 12 C f has local maximum when x = −6, and local minimum when x = 12 D f has no local minimum, and local maximum when x = 12 E f has no local minimum, and no local maximum 3.15 Local extreme values of the parametric curve Question 52 (L.O.1). Let y = y(x) be the parametric curve defined by x(t) = −5t ln(t + 4) (t ⩾ 0). y(t) = 2t3 − 24t2 + 72t + 4 Which statement is always TRUE? A y attains local minimum value when x = −10 ln(6) B y attains local maximum value when x = −10 ln(6) C y attains local minimum value when x = 6 D y attains local maximum value when x = −30 ln(10) E y attains local maximum value when x = 2 3.16 Concavity and inflection point Question 53 (L.O.1). If y = x3 e(12/11)x , then the number of point(s) of inflection of the graph of y is A 4 EXAM B 1 C 0 D 2 E 3 Page 8/1 — Question code: 1234 3.17 Global maximum and minimum values Question 54 (L.O.2). Find the maximum value of e2−x + x2 + 9x + 5, if 0 ⩽ x ⩽ 2 f (x) = x2 − 24x + 72, if 2 < x ⩽ 12 A 23 B 31 C 25 D 30 E 28 Question 55 (L.O.2). A company specializing in producing athletic uniforms for students stated that the cost to produce x batches (0 < x ⩽ 264) is C(x) = 0.3x2 + 12x + 4915.2, (hundreds dollars) The average cost per batch when producing x batches is given by the formula Cave = C(x) · Find x such that x the average cost atains minimum value. A 125 3.18 B 131 C 124 D 123 E 128 Maclaurin-Taylor approximations Question 56 (L.O.1). Find the coefficient of x8 in the Maclaurin polynomial of degree 9 for the function f (x) = (2x − 5) sin (2x). 614 299 A B 315 315 C −16/315 D − 1591 315 E − 646 315 Question 57 (L.O.1). Write the Taylor polynomial of degree 3 for the function f (x) = 2x2 − 8x + 11 e5x where x is near x0 = 2. 79 (x − 2)2 e10 145 (x − 2)3 e10 + + O (x − 2)4 ; x → 2 2 2 79 (x − 2)2 e10 145 (x − 2)3 e10 10 10 3e − 2 + 15 (x − 2) e + + + x + O (x − 2)4 ; x → 2 2 2 2 10 3 10 79 (x − 2) e 145 (x − 2) e 3e10 + 15 (x − 2) e10 + + − (x − 2)3 + O (x − 2)4 ; x → 2 2 2 79 (x − 2)2 e10 145 (x − 2)3 e10 2 4 10 10 3e + 15 (x − 2) e + − (x − 2) + + O (x − 2) ; x → 2 2 2 2 10 3 10 79 (x − 2) e 145 (x − 2) e 3e10 + 15 (x − 2) e10 + + + O (x − 2)4 ; x → 2 2 2 A 3e10 + 1 + 15 (x − 2) e10 + B C D E 4 Mixed problems [L.O.2] Using the following information, answer the questions from 58 to 59. A family plans to organize a party at a restaurant. The overall service cost for the event (venue, decoration, etc.) is 8 million Vietnamese dong. The restaurant requires a minimum of 2 tables, with each table accommodating 11 people. The cost per guest is 250 thousand Vietnamese dong. The family plans to spend a maximum of 28.25 million Vietnamese dong. If we denote x as the number of invited guests, C(x) (in million Vietnamese dong) represents the cost for the party. Question 58. Find the domain of C(x) is A N ∩ [22; 81] EXAM B [0; 81] C [22; 28.25] D N ∩ [0; 22] E N ∩ [8; 28.25] Page 9/1 — Question code: 1234 Question 59. The range of C(x) is A [22; 81] B [13.5; 81] C [13.5; 28.25] D [8; 28.25] E [5.5; 28.25] [L.O.2] Using the following information, answer the questions from 60 to 61. In a certain country, income tax is assessed as follows. There is no tax on income up to $12340. Any income over $12340 is taxed at a rate of 10%, up to an income of $23446. Any income over $23446 is taxed at 13%. Question 60. Express the tax T as a function of income I. 0, I ⩽ 12340 A T = 0.1, 12340 < I ⩽ 23446 0.13, I > 23446 0, I ⩽ 12340 B T = C T = D T = 0.1I, 12340 < I ⩽ 23446 0.1I, I > 23446 0, I ⩽ 12340 0.13I, 12340 < I ⩽ 23446 0.13I, I > 23446 0, I ⩽ 12340 0.1I, 12340 < I ⩽ 23446 0.13I, I > 23446 0, I ⩽ 12340 E T = 0.13I, 12340 < I ⩽ 23446 0.1I, I > 23446 Question 61. How much tax (in dollars) is assessed on an income of $27148. A 3529.6 B 3529.5 C 3528.7 D 3529.7 E 3529.2 [L.O.2] Using the following information, answer the questions from 62 to 64. A water filter equipment manufacturer has a fixed monthly cost of 340 million Vietnamese dong, and the variable cost to produce x devices is C(x) = −0.005x2 + 6.0x, (0 ⩽ x ⩽ 800) (million Vietnamese dong). The selling price per device is p(x) = −0.002x + 8.6, (0 ⩽ x ⩽ 800) (million Vietnamese dong). Question 62. The company’s revenue (million Vietnamese dong) when selling 144 devices in a month is A 1197.9280 B 1193.9280 C 1199.9280 D 1196.9280 E 1192.9280 Question 63. The company’s profit (million Vietnamese dong) when selling 144 devices in a month is A 99.6080 B 94.6080 C 91.6080 D 96.6080 E 98.6080 Question 64. To achieve the profit of 2300 million Vietnamese dong in a month, how many devices does the company need to sell in that month? A 605 B 604 C 601 D 600 E 602 [L.O.2] Using the following information, answer the questions from 65 to 67. EXAM Page 10/1 — Question code: 1234 All pyramids with a square base and a height of h = 4 have a surface area given by S(v) = 0.75v + 2 p 0.28125v 2 + 12v, where v is the volume of the pyramid. If x is the length of each side of the base then we can consider the volume dv (7) = 18.67. of a pyramid as a function v(x) of x. In addition, suppose that v(7) = 65.33 and dx Question 65. Compute the slope of the slant asymptote of the graph of function S(v) with respect to v. A 1.1961 B 1.0757 C 2.3389 D 2.7592 E 1.8107 Question 66. Compute the rate of change of the surface area of the pyramid with respect to the length x of each side when x = 7. A 34.1516 B 34.2931 C 34.1693 D 34.4337 E 35.2992 Question 67. Approximate the change in surface area of a pyramid when the length of each side of the base increases from x = 7 to x = 7.75. A 25.8357 5 B 25.1149 C 25.2789 D 25.8253 E 26.8070 Integration 5.1 Indefinite integrals Question 68 (L.O.1). Find the antiderivative F (x) of the function f (x) = 8 sin x + 4 cos x which satisfies π the condition F = 12. 2 C −8 cos x + 4 sin x − 8 A −8 cos x + 4 sin x + 8 B −8 cos x − 4 sin x + 8 D 8 cos x − 4 sin x − 8 E 8 cos x + 4 sin x + 8 Question 69 (L.O.1). Find the antiderivative F (x) of the function f (x) = 8ex + 4x which satisfies the condition F (0) = 14. A 8ex + 2x2 − 6 B −8ex + x2 + 6 D 8ex + 2x2 + 6 C −8ex + 2x2 − 6 E ex + 2x2 + 6 Question 70 (L.O.1). Suppose that f ′′ (x) = −18x − 100 sin(5x) + 9e3x and f ′ (0) = 23, f (0) = −379. Evaluate f (2). A −2.2641 5.2 B −2.7473 C −2.0485 D −3.6915 E −3.5396 Definite integrals Z Question 71 (L.O.1). Let π 2 Z f (x) dx = 9.51. Evaluate I = 0 π 2 [6.49f (x) + 2.63 sin x] dx. 0 A 63.8129 B 64.3499 C 64.0289 D 64.5836 E 64.7637 Z 6.4 Z 6.4 Z 6.4 h Question 72 (L.O.1). Let f (x) dx = 6.88 and g(x) dx = −2.47. Evaluate I = 18.86x + −2.5 −2.5 −2.5 i 5.55f (x) − 7.76g(x) dx. A 384.0257 EXAM B 384.9296 C 385.2162 D 384.6665 E 385.0722 Page 11/1 — Question code: 1234 Question 73 (L.O.2). Let g(x) = f −1 (x) be the inverse function of function y = f (x) = ex/2 +5.34. Evaluate Z 11.01 g(x)dx. 6.34 A 9.6536 5.3 B 10.3370 C 10.9418 D 10.7950 E 9.4292 Riemann sum Question 74 (L.O.2). Using the midpoint rule with 5 equal interval on [3; 11] to estimate the following Z 11 f (x)dx where f (x) is given by the following table integral I = 3 x 3 3.8 4.6 5.4 6.2 7.0 7.8 8.6 9.4 10.2 f (x) 6.8 4.3 6.6 5.0 3.4 5.8 5.1 5.5 4.2 A 41.8 5.4 B 41.88 C 42.24 5.8 11 6.0 D 41.6 E 42.26 The fundamental theorem of calculus Z x Question 75 (L.O.2). Find the maximum and minimum values of f (x) = (t2 − 9t − 22)dt on the interval −9 [−9, 9]. A fmax ≈ 433.8311; fmin ≈ −0.1873 B fmax ≈ 433.4807; fmin ≈ 0.638 C fmax ≈ 432.9731; fmin ≈ 0.1885 D fmax ≈ 433.7281; fmin ≈ 0.0568 E fmax ≈ 432.8333; fmin ≈ 0 6 Applications of integration 6.1 The Area problems 1 ; y = 0; x = 0; Question 76 (L.O.2). Find the area of the region bounded by the graphs of y = √ 9 16 − x2 and x = 4. A 0.5049 B 0.3040 C 0.2790 D 0.2973 E 0.1745 Question 77 (L.O.2). Find the area of the region bounded by the graphs of y = e2.53x ; y = e4.75x and x = 3.03. A 374034.3816 6.2 B 374034.2241 C 374034.3701 D 374034.9554 E 374034.1488 The Volume of the solid of revolution Question 78 (L.O.2). Let the region D be bounded by the curve y = p 15x2 + 2, x−axis and the lines x = 0, x = 5. Evaluate the volume V of the solid, revolving the region D about x−axis. A 1995.0119 B 1995.0890 C 1994.9113 D 1994.7434 E 1993.9313 Question 79 (L.O.2). Find the volume of the solid obtained by rotating the region bounded by the curves x = 0, y = −16, y = x2 − 8x about the y−axis. A 134.0465 EXAM B 134.0413 C 133.9908 D 133.6074 E 134.2850 Page 12/1 — Question code: 1234 6.3 The arc length Question 80 (L.O.2). Find the length of the arc y = 3 ln x, where 6 ⩽ x ⩽ 9. A 3.2399 B 2.8595 C 3.9077 D 2.5529 E 3.1169 Question 81 (L.O.2). A steady wind blows a kite due west. The kite’s height above the ground from hori√ 3 zontal position x = 0 to x = a > 0 meters is given by y = 3 + 2x . Find the real number a such that the distance travelled by the kite is 20.66 meters. A 3.6074 B 3.2822 C 2.5416 D 2.8492 E 3.5988 Question 82 (L.O.2). Find the length of the arc y = P (x), where 3 ⩽ x ⩽ 9. Here P (x) is a polynomial of degree 2 such that P (3) = 62, P (4) = 96, P (9) = 386. C 323.3469 D 324.0597 E 324.9452 Z xp Question 83 (L.O.2). Find the length of the arc y = 5t2 + 7dt, where 2 ⩽ x ⩽ 8. A 323.5810 B 323.6091 A 69.4857 B 69.7190 2 6.4 C 69.7552 D 69.0281 E 68.6696 The Area of the surface of revolution √ Question 84 (L.O.2). Find the area of the surface of revolution obtained by rotating the curve y = 2 x + 6, where 1 ⩽ x ⩽ 8 about the x−axis. A 502.2993 B 502.5967 C 502.8761 D 502.4058 E 502.0702 p Question 85 (L.O.2). Rotating the curve x = − y 2 + 3, where 0 ⩽ y ⩽ 3 about the y−axis, evaluate the surface area of the obtained surface of revolution. A 53.5843 6.5 B 54.2794 C 54.0413 D 53.1674 E 54.2563 Applications of integration in Physics Question 86 (L.O.2). Evaluate the approximation of the displacement of one particle with velocity v(t) = 9 arcsin(t) (m/s) from t = 0s to t = 0.7s. A 3.0163 B 3.2834 C 2.5228 D 1.9853 E 2.3123 Question 87 (L.O.2). The velocity of a dragster t seconds after leaving the starting line is v(t) = 72te−0.7t ft/sec. Find the approximation of the distance traveled by the dragster during the first 9 seconds. A 145.0934 B 144.0773 C 143.9867 D 144.9122 E 144.9691 Question 88 (L.O.2). When a particle is located a distance x meter from the origin, a force of given F (x) = √ 1 (newton) acts on it. How much work is done in moving it from x = 5 to x = 7 3 − 2. 2 x + 4x + 53 A −0.0604 B −0.0687 C 0.0374 D 0.0634 E −0.4302 Question 89 (L.O.2). When a particle is located a distance x meter from the origin, a force of given F (x) = 1 √ (newton) acts on it. How much work is done in moving it from x = 17/2 to x = −16x2 + 128x + 1040 13. A 0.2618 B −0.7293 C 0.5967 D −0.3280 E −0.6007 Question 90 (L.O.2). When a particle is located a distance x meter from the origin, a force of given F (x) = ln(6 + 5x) (newton) acts on it. How much work is done in moving it from x = 10 to x = 17. EXAM Page 13/1 — Question code: 1234 A 30.1241 C 29.4318 B 30.4412 D 30.0137 E 29.2949 Question 91 (L.O.2). When a particle is located a distance x meter from the origin, a force of given F (x) = 60 √ (newton) acts on it. How much work is done in moving it from x = 1 to x = 6. 196 − x2 A 22.2853 B 22.7360 C 23.0572 D 21.5142 E 23.0559 6.6 Applications of integration in Economics Question 92 (L.O.1). The accumulated, or total, future value after T years of an income stream of R(t) thousand dollars per year, earning interest at the rate of r per year compounded continuously, is given by Z T R(t)e−rt dt. A company recently bought an automatic car-washing machine that is expected to A = erT 0 generate 3 thousand dollars in revenue per year, t years from now, for the next 7 years. If the income is reinvested in a business earning interest at the rate of r = 11% per year compounded continuously, find the total accumulated value of this income stream at the end of 7 years. A 30.6382 6.7 C 32.4296 B 31.6300 D 32.2997 E 31.7604 The average value of a funtion Question 93 (L.O.2). The interest rates charged by Madison Finance on auto loans for used cars over a certain 12−month period in 2022 are approximated by the function r(t) = − 1 3 1 2 t + t − 2t + 27, 21 3 (0 ⩽ t ⩽ 12) where t is measured in months and r(t) is the annual percentage rate. What is the average rate on auto loans extended by Madison over the 12−month period? A 11.1063 6.8 B 11.2527 C 10.2379 D 10.4286 E 10.6659 D 1.5139 E 1.3924 D 0.7021 E −0.0613 Improper integrals +∞ Z 3x2 dx Question 94 (L.O.2). Evaluate the integral 7 + x4 1 A 1.8614 B 2.4366 C 1.9137 Z Question 95 (L.O.2). Evaluate the integral A 1.1263 B 0.3787 +∞ dx x2 + 625 671 x · C −0.6059 √ √ 4 Question 96 (L.O.2). The total profit of a firm in dollars, from producing x units of an item, is P (x). The firm is able to determine that its marginal profit is given by P ′ (x) = 190e−1.05x . Suppose that it were possible for the firm to make infinitely many units of this item. What would its total profit be? A 180.8926 B 181.1377 C 180.4869 D 181.9074 E 180.9524 D 1.1703 E 1.7271 +∞ Z 8dx √ Question 97 (L.O.2). Evaluate the integral . x2 + x4 5 A 0.9952 EXAM B 0.9963 C 1.5895 Page 14/1 — Question code: 1234 Question 98 (L.O.2). Let a be the real number such that lim p x→+∞ x2 + 8x + 2 − x − a = 0. Evaluate +∞ Z dx · (x + a)(x − 2)2 a+ 6 A 4.5249 D 3.4233 E 4.8000 Z 9 Z 5 2 mdx x dx √ √ Question 99 (L.O.2). Find the approximation of m which satisfies = 2 x−1 81 − x 0 1 C 17.6462 A 18.2816 B 17.4858 D 18.3127 E 18.3730 7 B 4.0162 C 3.4425 Ordinary differential equations (ODE) 7.1 The first order differential equations Question 100 (L.O.2). Which function is a solution of the following differential equation y ′ + where C is any constant? 1 A y = x2 + Cx−7 3 3 D y = x2 + Cx7 8 7.2 7 y = 3x, x 3 1 B y = x2 + Cx−7 C y = x2 + Cx7 8 3 3 2 −7 E y = x + Cx 10 The second order differential equations Question 101 (L.O.2). Let y(x) be a function satisfying y ′′ − 4y ′ − 32y = 0, y(0) = 6, y ′ (0) = 24. Evaluate y(1.88). A 13609714.8215 B 13609714.2745 D 13609713.4611 C 13609713.4938 E 13609714.0102 Question 102 (L.O.2). Let y(x) be a function satisfying y ′′ − 5y ′ = 2x2 + 5, y(0) = 0, y ′ (0) = 0. Evaluate y(2). A 4543.1199 B 4542.7416 C 4542.6055 D 4542.9902 E 4541.7701 Question 103 (L.O.2). Let y(x) = axe4x be a function satisfying y ′′ − 10y ′ + 24y = −6.3894e4x . Find the value of a. A 3.5403 B 3.6631 C 3.1947 D 3.2759 E 3.0536 ESSAY QUESTIONS 1 1.1 Differentiation Related rates Question 104. EXAM Page 15/1 — Question code: 1234 [L.O.2] Two crates, A and B, are on the floor of a warehouse. The crates are joined by a rope 15m long, each crate being hooked at floor level to an end of the rope. The rope is stretched tight and pulled over a pulley P that is attached to a rafter P Q = 4m above a point Q on the floor directly between the two crates. If crate A is 3m from Q and is being pulled directly away from Q at a rate of 0.5 m/s, how fast is crate B moving toward Q? Question 105. [L.O.2] A major network is televising the launching of the rocket. At a distance of x = 12000 feet from the launch site, a spectator is observing a rocket being launched vertically. A camera tracking the liftoff of the rocket is located at point A, where ϕ denotes the angle of elevation of the camera at A. When the rocket is z = 12434 ft from the camera and this distance is increasing at the rate of 480 ft/sec, how fast is ϕ changing? 2 Applications of integration Question 106 (L.O.2). Let V1 and V2 be the volumes of the solids that result when the plane region bounded 4 1 1 by y = , y = 0, x = , and x = c, where c > , is revolved about the x−axis and the y−axis, respectively. x 5 5 Find the value of c for which V1 = V2 . Question 107 (L.O.2). Find the area of the surface obtained by rotating the arc y = 2x3/2 , (0 ⩽ x ⩽ 4), about the x−axis. 3 3.1 Ordinary differential equations (ODE) The first order differential equations Question 108 (L.O.2). A tank initially contains 80 grams of salt dissolved in 100 liters of water. Pure water flows into the tank at the rate of 12 liters per minute, and the well stirred mixture flows out of the tank at the same rate. How much salt does the tank contain after 2 minutes? Question 109 (L.O.2). A tank with a capacity of 130 liters initially contains 80 grams of salt dissolved in 100 liters of water. Pure water flows into the tank at the rate of 12 liters per minute, and the well stirred mixture flows out of the tank at the rate of 2 liters per minute. How much salt does the tank contain when the tank is full? Question 110 (L.O.2). Recall that a simple electric circuit with a resistor of R ohms, an inductor of L henries and a battery with a voltage of E volts will produce a current I(t) amperes (t is measured in seconds) satisfying the following differential equation LI ′ (t) + RI = E. Let study one such circuit with R = 2 ohms, L = 3 henries and E = 12 volts. Given initial condition I(0) = 3 amperes. Evaluate current’s value at t = 3 seconds. EXAM Page 16/1 — Question code: 1234 3.2 The second order differential equations Question 111 (L.O.2). Solve the differential equation y ′′ − 8y ′ + 12y = 120e12x Question 112 (L.O.2). Solve the differential equation y ′′ − 14y ′ + 74y = 10e7x cos(5x) Question 113 (L.O.2). Let yp be the particular solution of the differential equation y ′′ −4y ′ −12y = −144x− 72. Evaluate yp (2). Question 114 (L.O.2). Recall that a simple electric circuit with a resistor of R ohms, an inductor of L henries, a capacitor of C farads and a battery with a voltage E(t) = 100 sin(ωt) (volts) will produce a current 1 I(t) amperes (t is measured in seconds) satisfying the following differential equation LI ′′ (t) + RI ′ + I = C E ′ (t) = 100ω cos(ωt). Let study one such circuit with R = 8 ohms, L = 5 henries, C = 0.2 farads and ω = 11 rad/seconds. Given initial condition I(0) = 0, I ′ (0) = 0, evaluate current’s value at t = 6.2 seconds. Question 115 (L.O.2). Recall that a spring with mass m, spring constant k relaxing after being stretched will be in motion, where its position x(t) satisfies the following differential equation mx′′ (t) + kx = 0. Let study one spring with m = 2 and k = 32. Given initial condition x(0) = 6, x′ (0) = 4, evaluate numerically its position at t = 2. 3.3 System of the first order ODEs Question 116 (L.O.2). Consider the system of ODEs: x′ (t) = 4x − y − t y ′ (t) = 5x + 10y + (t − 1) Find a solution of the system such that x (0) = 3, y (0) = 1. ———— END ———— EXAM Page 17/1 — Question code: 1234 ANSWER SHEET FOR MULTIPLE CHOICE QUESTIONS 7-D 3-C 4-A 5-B 6-D 8-A 9-A 10 - E 18 - A 19 - B 20 - E 27 - E 28 - B 29 - E 30 - E 36 - A 37 - A 38 - C 39 - E 40 - C 45 - A 46 - D 47 - E 48 - B 49 - B 50 - D 54 - E 55 - E 56 - C 57 - E 58 - A 59 - C 60 - D 63 - D 64 - D 65 - E 66 - D 67 - D 68 - A 69 - D 70 - B 72 - D 73 - B 74 - C 75 - E 76 - E 77 - A 78 - C 79 - B 80 - A 81 - D 82 - D 83 - A 84 - A 85 - C 86 - E 87 - E 88 - C 89 - A 90 - D 91 - A 92 - B 93 - D 94 - C 95 - B 96 - E 97 - C 98 - B 99 - B 100 - A 101 - E 102 - C 103 - C 1-B 2-B 11 - D 12 - E 13 - A 14 - D 15 - D 16 - B 17 - A 21 - C 22 - A 23 - D 24 - A 25 - C 26 - D 31 - E 32 - B 33 - A 34 - E 35 - C 41 - C 42 - D 43 - C 44 - B 51 - B 52 - B 53 - E 61 - E 62 - D 71 - B QUESTIONS AND THE ANSWER Page 1/1 — Question code: 1234 MULTIPLE CHOICE QUESTIONS 1 Functions 1.1 Domain and range of a function s 9 . Question 1 (L.O.1). Find the domain of function y = ln 3 + x 9 9 x ⩽ − x⩽− 9 9 2 2 A − ⩽x<0 B C x⩽− D 2 2 x>0 x⩾0 E x>0 Solution 9 x⩽− 9 9 2x + 9 2 ⩾0⇒3+ ⩾1⇒ The function is defined when ln 3 + ⩾0⇒ x x x x>0 The correct answer is B □ Question 2 (L.O.1). Find the range of the function 9x2 − 3, if x > 0 f (x) = 21x3 , if x ⩽ 0 A (0, +∞) B (−∞, +∞) C (−∞, 0) D (−3, +∞) E (−3, 0) Solution When x > 0, we have f (x) = 9x2 − 3 > −3. On the another hand, when x ⩽ 0, we have f (x) = 21x3 ⩽ 0.Therefore, the range of the function f (x) is (−∞, +∞). The correct answer is B □ Question 3 (L.O.2). A family plans to organize a party at a restaurant. The overall service cost for the event (venue, decoration, etc.) is 4 million Vietnamese dong. The restaurant requires a minimum of 2 tables, with each table accommodating 8 people. The cost per guest is 258 thousand Vietnamese dong. The family plans to spend a maximum of 36.6 million Vietnamese dong. If we denote x as the number of invited guests, C(x) (in million Vietnamese dong) represents the cost for the party. Find the domain D and the range R of C. A D = N ∩ [8; 142], R = [2.064; 36.6] B D = [16; 142], R = [8.128; 36.6] C D = N ∩ [16; 142], R = [8.128; 36.6] D D = N ∩ [0; 142], R = [0; 36.6] E D = [16; 36.6], R = [4.016; 142] Solution QUESTIONS AND THE ANSWER Page 2/1 — Question code: 1234 The restaurant requires a minimum of 2 tables, each table accommodating 8 people, so x ⩾ 2 × 8 = 16. Furthermore, the cost per guest is 258 thousand Vietnamese dong, so C(x) ⩾ 4 + 0.258 × 16 = 8.128 and the family plans to spend a maximum of 36.6 million, hence the maximum number of invited guests is 36.6 = 142. Since the number of invited guests is a natural number, the domain for the cost function of the 0.258 party is D = N ∩ [16; 142] and the range for the cost function is R = [8.128; 36.6]. The correct answer is C 1.2 □ Description of a function Question 4 (L.O.2). A rectangular storage container with an open top has a volume of 14m3 . The length of its base is quadruple its width. Material for the base costs 11 dollars per square meter; material for the sides costs 8 dollars per square meter. Express the cost of materials as a function of the width w of the base. A 44w2 + 280 w B 44w2 + 280 C 32w2 + 1120 w D 32w2 + 385 w E 44w2 + 140 w Solution Let w and 4w be the width and length of the base, respectively, and h be the height. The total cost is C = 11 × w × 4w + 8 × 2(w + 4w)h. Using the fact that the volume is 14, we have w × 4w × h = 14 ⇒ h = C(w) = 11 × w × 4w + 8 × 2(w + 4w) × 14 . Therefore, 4w2 14 280 = 44w2 + 2 4w w The correct answer is A □ Question 5 (L.O.2). To create an animated image appealing to players in a game, the game creator started by creating a rectangular image with dimensions of 29 × 71 pixels. Then, each side is increased at a rate of 15 pixels per second. Determine the area of the image after t seconds. B 225t2 + 1500t + 2059 A 2059 D 225t2 C 225t2 + 1065t E 225t2 + 435t Solution The area of the image after t seconds is (29 + 15t) × (71 + 15t) = 225t2 + 1500t + 2059 The correct answer is B □ Question 6 (L.O.2). The amount of adrenaline in the body changes rapidly. Let’s assume that initially the body has 18 mg of adrenaline, create a function describing the adrenaline level A = f (t) in the body after t (minutes), given that A increases 0.39 mg per minute. A 18t + 0.39 QUESTIONS AND THE ANSWER B 7.02t C 0.39 − 18t D 0.39t + 18 E 18 − 0.39t Page 3/1 — Question code: 1234 Solution The function describing the adrenaline level in the body after t (minutes) is A = f (t) = 0.39t + 18. The correct answer is D 1.3 □ Inverse functions Question 7 (L.O.1). Let f (x) = x5 + 3x3 + 6x − 5. What is the value of f −1 (63). A 1 C 6 B 0 D 2 E 3 Solution We have f ′ (x) = 5x4 + 9x2 + 6 > 0, ∀x ∈ R. Thus, f is one-to-one function and has inverse function. Let f −1 (63) = x0 ⇒ f (x0 ) = 63 ⇒ x50 + 3x30 + 6x0 − 5 = 63 ⇒ x0 = 2. The correct answer is D 2 □ The limit of a function, continuous functions 2.1 The limit of a function Question 8 (L.O.1). Find m > 0 such that e3.19x − cos(x) lim x→0 x 2 .m + ln(1 + 6.04x) lim x→0 x .m = 18.62. A 1.6481 B 1.2026 C 1.2978 D 1.3338 E 2.3985 Solution Using the L’Hospital rule, we have e3.19x − cos(x) 3.19e3.19x + sin(x) = lim = 3.19, x→0 x→0 x 1 lim and ln(1 + 6.04x) = lim x→0 x→0 x lim 6.04 1+6.04x 1 = 6.04. Therefore, we receive 3.19m2 + 6.04m = 18.62. Choosing m > 0, we get m = 1.64814172658632 ≈ 1.6481. The correct answer is A □ 6x + m2 x→∞ 6x − 1 C 27.3814 Question 9 (L.O.1). Find m > 1 such that lim A 26.5123 B 26.5475 6x = e26.55m . D 26.8591 E 26.4543 Solution We have lim x→∞ 6x + m2 6x − 1 6x QUESTIONS AND THE ANSWER = lim x→∞ m2 + 1 1+ 6x − 1 6x(m2 +1) 6x−1 · 6x−1 2 m +1 2 +1 = em = e26.55m ⇒ m2 + 1 = 26.55m. Page 4/1 — Question code: 1234 Choosing m > 1, we receive m = 26.5122816318155 ≈ 26.5123. The correct answer is A □ 2 7x Question 10 (L.O.1). Find lim 1 + 2x e A e 1 9x2 x→0 −7/9 . C −e2/9 B 1 E e2/9 D 0 Solution 12 ln(1 + 2x2 e7x ) Let y = 1 + 2x2 e7x 9x ⇒ ln y = · Then 9x2 ln(1 + 2x2 e7x ) 2x2 e7x 2 ln(1 + 2x2 e7x ) 2 = lim × lim =1× = 2 2 7x 2 x→0 x→0 9x x→0 9x 2x e 9 9 lim ln y = lim x→0 Therefore, lim y = lim eln y = e2/9 x→0 x→0 The correct answer is E □ Question 11 (L.O.1). Find S = 4a + 8b where a and b are the two real constants such that 2 lim x ea/x + + b = 15. x→+∞ x A 43 B 40 C 48 D 44 E 42 Solution Let t = 1 eat + 2 × t + b then I = lim = 15. The constant b must satisfy the equation x t t→0+ ea×0 + 2 × 0 + b = 0 ⇒ b = −1. Otherwise, I = ∞. When b = −1, using L’ Hospital rule, we obtain I = lim t→0+ aeat + 2 = a + 2 = 15 ⇒ a = 15 − 2 = 13. 1 Therefore, S = 4 × 13 − 1 × 8 = 44 The correct answer is D □ e4ax − 1 − 9x · x→0 x + 13x2 9 C 4a, ∀a ∈ R B 4a − 9, ∀a ̸= 4 E 4a − 9, ∀a ∈ R Question 12 (L.O.1). Evalulate I = lim 9 4 D 4a − 9, ∀a ̸= 0 A 4a, ∀a ̸= Solution Using the L’Hospital rule, we have I = lim x→0 (4a − 9)x = 4a − 9. x The correct answer is E □ QUESTIONS AND THE ANSWER Page 5/1 — Question code: 1234 2.2 Continuity Question 13 (L.O.1). Find real values of a such that f (x) = uous at x = 13. 2 A 13 B − 24 13 C − ax − 2, √ arctan( x − 13), 37 13 D − 50 13 if x ⩽ 13 is contin- if x > 13 E − 11 13 Solution The function f is continuous at x = 13 if lim f (x) = 0 = lim f (x) = f (13) = 13 × a − 2 ⇒ a = x→13− x→13+ 2 13 The correct answer is A □ 3 5 arctan x − 31 x−3 Question 14 (L.O.1). Find real values of a such that f (x) = a, ous at x = 3. 5π 3π 5π B C A − 2 2 2 7π D a does not exist E 2 if x ̸= 3 is continu- if x = 3 Solution The function f is continuous at x = 3 if lim f (x) = lim f (x) = f (3) x→3− x→3+ We have lim f (x) = − x→3+ 5π 5π ̸= lim f (x) = − 2 2 x→3 Therefore, there is no exist a such that f is continuous at x = 3. The correct answer is D 2.3 □ Asymptotes Question 15 (L.O.1). Find all real numbers a so that the line of equation x = 0 is a vertical asymptote ln(7x + a − 8) of the curve y = · x+9 C 6 A 11 B 7 D 8 E 12 Solution The line of equation x = 0 is a vertical asymptote of the curve y = lim x→0 QUESTIONS AND THE ANSWER ln(7x + a − 8) when x+9 ln(7x + a − 8) ln(7 × 0 + a − 8) ln(a − 8) = = = ∞. x+9 0+9 9 Page 6/1 — Question code: 1234 Therefore, a = 8. The correct answer is D □ 9x − 5. 1 − e−x C y = 9x − 5 Question 16 (L.O.1). Find all asymptotes of the graph of the function y = 4x + √ A x = 0, y = 13x − 5 B y = 13x − 5 D x = 0, y = 4x − 5 E y = 4x − 5 Solution Domain: x > 0. We have lim f (x) = lim 4x + √ x→0+ x→0+ 9x 9x − 5 = −5 + lim √ = −5. −x + x→0 1−e 1 − e−x The line x = 0 is not the vertical asymptote of the curve y = f (x). On the other hand, lim 4x + √ x→+∞ 9x − 5 = +∞. 1 − e−x Thus, the function does not have horizontal asymptote. Consider the asymptote of the form y = mx + b where 4x + m = lim √ 9x 1−e−x −5 = lim 4 − x x→+∞ x→+∞ 5 9 = 13 +√ x 1 − e−x and b = lim 4x + √ x→+∞ 9x 9x 9x − 5 − mx = lim 4x + √ − 5 − 13x = lim −5 + √ − 9x = −5. −x −x x→+∞ x→+∞ 1−e 1−e 1 − e−x Therefore, the function has only 01 slant asymptote y = 13x − 5. The correct answer is B □ Question 17 (L.O.2). A plastic molding company states that if they use x% of the company’s machines, the total monthly cost for operating these machines is given by the function: C(x) = 19x2 − 57x − 60610 4x2 − 100x − 7656 (hundreds of dollars). The company employs an alternating maintenance mode to utilize close to 58% of the machines (optimal capacity).What is the total cost that the company has to incur for operating this number of machines if the machine usage reaches the optimal capacity? A 5.8984 B 5.9918 C 5.7198 D 6.6802 E 5.7032 Solution According to the requirement of our problem, we need to find: lim C(x) = lim x→58 x→58 (x + 55) (19x − 1102) 19 (x + 55) = lim = 5.8984 x→58 4 (x + 33) (x + 33) (4x − 232) The correct answer is A □ QUESTIONS AND THE ANSWER Page 7/1 — Question code: 1234 3 Derivatives 3.1 Tangents Question 18 (L.O.1). A point M (a, b) is on the graph of the function f (x) = 1.55ex − x. Given that the tangent line to the graph at M is parallel to the line y = 2.47x + 2.82. Find the value of a. A 0.8059 B 0.4466 C 1.7944 D 1.4552 E 1.2394 Solution Since the tangent line to the graph at M is parallel to the line y = 2.47x + 2.82, then 2.47 + 1 ′ a f (a) = 2.47 ⇒ 1.55e − 1 = 2.47 ⇒ a = ln = 0.805899663027613 ≈ 0.8059. 1.55 The correct answer is A □ Question 19 (L.O.1). Find ALL points on the graph of f (x) = 4x3 − 192x where the tangent line is parallel to the x−axis. A (4; 512) B (4; −512) and (−4; 512) D (4; 512) and (−4; −512) C (4; −512) E (−4; 512) Solution The tangent line is parallel to the x−axis when its slope is equal to 0. It means that f ′ (x) = 12x2 − 192 = 0 ⇔ x = ±4. Therefore, all points on the graph of f (x) where the tangent line is parallel to the x−axis are (4; −512) and (−4; 512). The correct answer is B □ Question 20 (L.O.1). Find ALL points on the graph of f (x) = 9 where the tangent line is perpendicular x to the line y = 16x − 4. A (−12; 3/4) C (12; −3/4) and (−12; 3/4) B (12; 3/4) D (−12; −3/4) E (12; 3/4) and (−12; −3/4) Solution The tangent line is perpendicular to the line y = 16x − 4 when its slope is equal to − f ′ (x) = − 1 · It means that 16 9 1 = − ⇔ x = ±12. 2 x 16 Therefore, all points on the graph of f (x) where the tangent line is perpendicular to the line y = 16x − 4, are (12; 3/4) and (−12; −3/4) . The correct answer is E □ QUESTIONS AND THE ANSWER Page 8/1 — Question code: 1234 Question 21 (L.O.1). Find the real value of a such that the tangent line to the graph of the function √ 21 y = 3 x at x = a has y−intercept . 2 A 46 B 52 C 49 D 44 E 48 Solution √ 3 The tangent line at x = a has an equation of the form y = 3 a + √ (x − a). This tangent line has 2 a √ 21 3 a = ⇒ a = 49. y−intercept 21/2 when 2 2 The correct answer is C □ 3.2 Tangent to paramtric curve Question 22 (L.O.1). Let the curve be given by paramtric equations x(t) = (3 − 4t) e3t y(t) = 3t2 − 5t − 4 e3t 37 Find the set of all values of t such that the tangent at M (x(t), y(t)) has the slope of − ? 31 503 410 131 148 38 C − ; −1 ;7 ;4 ;1 D − E A − ;3 B − 93 93 93 93 93 Solution (6t − 5) e3t + 3 · 3t2 − 5t − 4 e3t y ′ (t) The slope of the tangent at M (x(t), y(t)) is y (x) = ′ = · Therefore, x (t) 3 · (3 − 4t) e3t − 4e3t 38 2 t = − −9t + 9t + 17 37 37 ′ 93 =− ⇔ y (x) = − ⇔ 31 12t − 5 31 t = 3 ′ The correct answer is A 3.3 □ Differentiable functions Question 23 (L.O.2). Determine ALL values of x such that f (x) = x2 − 11 is not differenx2 − 25 tiable. A x ̸= 5 B x = −5 D x = ±5 C x=5 E x ̸= ±5 Solution The domain of f (x) = x2 − 11 is D = R\{±5}. x2 − 25 ⇒ y′ = −28x − 25)2 (x2 So the function f is not differentiable when x = ±5. The correct answer is D □ QUESTIONS AND THE ANSWER Page 9/1 — Question code: 1234 Question 24 (L.O.2). Determine ALL values of x such that f (x) = p 3 (x2 − 25)2 is not differen- tiable. A x = ±5 B x ̸= ±5 C x ̸= 5 D x = −5 E x=5 Solution The domain of f (x) = p 3 (x2 − 25)2 is D = R. ⇒ y′ = 2 4x × (x2 − 25)−1/3 × 2x = p 3 3 3 (x + 5)(x − 5) So the function f is not differentiable when x = ±5. The correct answer is A □ Question 25 (L.O.2). Find S = a + b such that function f is differentiable at x = 0, if 2x2 + 7x + 5, x ⩽ 0 f (x) = ln(ax + b), x > 0 A 7e5 C 8e5 B 8 D 9 E 6e5 Solution In order to be differentiable, function has to be continuous lim f (x) = lim f (x) ⇒ ln b = 5 ⇒ b = e5 . x→0+ x→0− Function f is differentiable at x = 0 if and only if f+′ (0) = f−′ (0) ⇒ a = 7 ⇒ a = 7b = 7e5 . b Therefore, S = a + b = 8e5 The correct answer is C 3.4 □ The chain rule Question 26 (L.O.1). Let f (x) and g(x) be functions satisfying the conditions: f (−4) = f ′ (−4) = 9 and g(9) = g ′ (9) = −4. Consider the function h(x) = (f ◦ g ◦ f )(x) = f (g(f (x))). Find the value of h′ (−4). A −328 B −325 C −322 D −324 E −329 Solution Using the chain rule, we have h′ (x) = f ′ (g(f (x))).g ′ (f (x)).f ′ (x) ⇒ h′ (−4) = f ′ (g(f (−4))).g ′ (f (−4)).f ′ (−4) = = f ′ (−4).g ′ (9).f ′ (−4) = 9 × (−4) × 9 = −324. The correct answer is D □ QUESTIONS AND THE ANSWER Page 10/1 — Question code: 1234 Question 27 (L.O.2). An importer of Rwandan coffee estimates that local consumers will buy approxi4346 mately D(p) = 2 pounds of the coffee per week when the price is p dollars per pound. It is estimated p that t weeks from now, the price of Rwandan coffee will be p(t) = 0.02t2 + 0.08t + 9 dollars per pound. At what rate will the weekly demand for the coffee be changing with respect to time 6 weeks from now? A −3.1978 B −2.3627 C −3.4126 D −3.2408 E −2.621 Solution 2 Using the chain rule, we have D(t) = D(p(t)) ⇒ D (t) = D (p).p (t) = 4346 × − 3 × (0.04t + 0.08) p 2 When t = 6 ⇒ p(6) = 0.02 × 6 + 0.08 × 6 + 9 = 10.2 and 2 ′ × (0.04 × 6 + 0.08) = −2.62101303420253 ≈ −2.621. D (6) = 4346 × − 10.23 ′ ′ ′ The correct answer is E □ Question 28 (L.O.2). When electric blenders are sold for p dollars apiece, local consumers will buy 7928 approximately D(p) = blenders per month. It is estimated that t months from now, the price of p √ the blenders will be p(t) = 0.03 t3 + 4 dollars. Compute the rate at which the monthly demand for the blenders will be changing with respect to time 16 months from now. A −40.9061 B −40.7186 C −40.9638 D −40.9057 E −41.3092 Solution √ 1 Using the chain rule, we have D(t) = D(p(t)) ⇒ D′ (t) = D′ (p).p′ (t) = 7928 × − 2 × (0.045 t) p √ 3 When t = 16 ⇒ p(16) = 0.03 × 16 + 4 ≈ 5.92 and √ 1 ′ D (16) ≈ 7928 × − × (0.045 × 16) = −40.7185902118335 ≈ −40.7186. 5.922 The correct answer is B 3.5 □ Derivatives of inverse function Question 29 (L.O.1). If f (x) = 2x + 2 sinh (x − 2), find (f −1 )′ (4). 9 19 13 15 A B − C D − 4 4 4 4 E 1 4 Solution Let y = f (x) = 2x + 2 sinh (x − 2) ⇒ x′ (y) = 1 1 = . When y = 4 we have x = 2. y ′ (x) 2 cosh (x − 2) + 2 Therefore, (f −1 )′ (4) = x′ (4) = 1 y ′ (2) = 1 4 The correct answer is E □ QUESTIONS AND THE ANSWER Page 11/1 — Question code: 1234 3.6 Undetermined Coefficients Question 30 (L.O.2). Let y = Ax3 +Bx+C, where A, B, C ∈ R, satisfy the equation y ′′′ +2y ′′ −5y ′ +9y = 27x3 − 45x2 + 54x + 89. Find the value of S = A + B + C. A 11 B 9 C 16 D 15 E 14 Solution We have y = Ax3 + Bx + C ⇒ ′ 2 y = 3Ax + B y ′′ = 6Ax y ′′′ = 6A Substituting these derivatives into the equation y ′′′ + 2y ′′ − 5y ′ + 9y = 27x3 − 45x2 + 54x + 89, we receive 6A + 2 × 6Ax − 5 × (3Ax2 + B) + 9 × (Ax3 + Bx + C) = 27x3 − 45x2 + 54x + 89 ⇔ 9Ax3 − 15Ax2 + (12A + 9B)x + (6A − 5B + 9C) = 27x3 − 45x2 + 54x + 89 9A = 27 A=3 −15A = −45 ⇔ ⇔ B=2 12A + 9B = 54 C=9 6A − 5B + 9C = 89 Therefore, S = A + B + C = 14 The correct answer is E □ Question 31 (L.O.2). Let y = Ax cos x + Bx sin x, where A, B ∈ R, satisfy the equation y ′′ + y = −4 sin x + 10 cos x. Find the value of S = A + B. A 6 B 9 C 8 D 5 E 7 Solution We have y = Ax cos x + Bx sin x y ′ = A cos x − Ax sin x + B sin x + Bx cos x ⇒ y ′′ = −A sin x − A sin x − Ax cos x + B cos x + B cos x − Bx sin x Substituting these derivatives into the equation y ′′ + y = −4 sin x + 10 cos x, we receive −2A × sin x + 2B × cos x = −4 sin x + 10 cos x −2A = −4 A=2 ⇔ ⇔ 2B = 10 B=5 Therefore, S = A + B = 7 The correct answer is E □ Question 32 ([L.O.2). Let y = Axp + Bx + C, where A, B, C, p ∈ R, satisfy the equation y ′′ = y ′ (1) = −3.0, y(1) = 5.0. Find the value of S = (A + B + C).p. A 11 QUESTIONS AND THE ANSWER B 7 C 9 D 8 30 and x7 E 12 Page 12/1 — Question code: 1234 Solution If y = Axp + Bx + C y ′ = Apxp−1 + B ⇒ y ′′ = Ap(p − 1)xp−2 Substituting these derivatives into the equation y ′′ = 30 , we receive x7 Ap(p − 1)xp−2 = 30x−7 Ap(p − 1) = 30 A=1 ⇒ ⇒ p = −5 p − 2 = −7 Also, B=2 A × 1p + 1B + C = 5.0 ⇒ C=2 Ap × 1p−1 + B = −3.0 Therefore, S = (A + B + C).p = 7 The correct answer is B 3.7 □ Rates of change Question 33 (L.O.2). The management fee for running services in an area is calculated by the formula f (x) = 0.0692x2 + 0.0902x (in dollars), where x is the number of homes being present in the area. Suppose that at the moment when 63 homes are present in this area, the number of homes is increasing at a rate of 5 homes per week. What is the rate of change (in dollars/week) in the management fee at this moment? A 44.047 B 44.5156 C 43.0986 D 43.2984 E 44.6353 Solution The rate of change (in dollars/week) in the management fee at the moment when x = 63 homes are present in the area and x′ (t) = 5 homes per week is f ′ (t) = f ′ (x).x′ (t) = f ′ (63) × 5 = 44.047 ≈ 44.047. The correct answer is A □ 25 · If the abscissa x of the particle increases by 0.3234 units per second, how does its ordinate vary as it passes through the Question 34 (L.O.2). A particle is moving along a hyperbolic trajectory given by y = point (5; 5)? A Increases 0.0971 units per second B Increases 0.8569 units per second C Decreases 0.4064 units per second D Decreases 0.1947 units per second E Decreases 0.3234 units per second QUESTIONS AND THE ANSWER Page 13/1 — Question code: 1234 Solution The rate of change of the ordinate (y-coordinate) as the particle passes through the point (5; 5) is y ′ (t) = y ′ (x).x′ (t) = y ′ (5) × 0.3234 = −0.3234 ≈ −0.3234. The correct answer is E 3.8 □ Related Rates Question 35 (L.O.2). Recall that volume of a right cylinder is V = π.r2 .h, where r is the radius of the base and h is the height. Given a right cylinder with the radius of the base of 9 cm and the height of 5 cm. The radius of the base is increasing at a rate of 0.8 cm/s, and its height is increasing at a rate of 0.21 cm/s. How fast is the volume of cylinder increasing? A 278.9661 B 280.2004 C 279.6332 D 279.1033 E 279.8283 Solution The formula for the volume of a right cylinder is V = π.r2 .h, (1) where r is the radius of the base and h is the height. In this problem, V, r and h are functions of the time t in seconds. Taking the derivative of both sides of equation (1) with respect to time yields dh dr dV = π r2 + h.2r dt dt dt (2) Since the radius of the base is increasing at a rate of 0.8 cm/s and the height is increasing at a rate of 0.21 cm/s, dr = 0.8, dt dh = 0.21. dt Substituting these, as well as r = 9 and h = 5, into equations (2) yields dV = π 92 × 0.21 + 5 × 2 × 9 × 0.8 = 89.01π ≈ 279.6332 dt Because the sign of dV is positive, the volume of the right cylinder is increasing at a rate of 279.6332 cm3 /s. dt The correct answer is C □ Question 36 (L.O.2). Water pours into a conical tank of height 10(m) and radius 4(m) at a rate of 7(m3 /min). QUESTIONS AND THE ANSWER Page 14/1 — Question code: 1234 4 r 10 h At what rate is the water level rising when the level is 5(m) high? A 0.5570 B −0.3594 D 0.0975 C 0.1141 E 0.7804 Solution dh Let V and h be the volume and height of the water in the tank at time t. Our problem is: Compute at dt dV 1 h = 5 given that = 7m3 /min. When the water level is h, the volume of water in the cone is V = πhr2 , dt 3 where r is the radius of the cone at height h, but we cannot use this relation unless we eliminate the variable r. Using similar triangles, we see that dV r 4 1 dV dh dh dt = ⇒ r = 2/5 × h ⇒ V = πh(2/5 × h)2 ⇒ = 4/25πh2 ⇒ = h 10 3 dt dt dt 4/25πh2 When h = 5, the level is rising at a rate of 7 7 dh = = ≈ 0.5570(m/min). 2 dt 4/25π5 4π The correct answer is A □ Question 37 (L.O.2). A leaky water tank is in the shape of an inverted right circular cone with depth of 11(m) and top radius 3(m). When the water in the tank is 3(m) deep, it is leaking out at a rate of 1/6(m3 /min). 3 r 11 h How fast is the water level in the tank dropping at that time? A 0.0792 B 0.3362 C −0.6292 D 0.1284 E −0.5812 Solution dh Let V and h be the volume and depth of the water in the tank at time t. Our problem is: Compute at h = 3 dt dV 1 given that = 1/6m3 /min. When the water level is h, the volume of water in the cone is V = πhr2 , dt 3 where r is the radius of the cone at height h, but we cannot use this relation unless we eliminate the variable r. Using similar triangles, we see that dV r 3 1 dV dh dh dt = ⇒ r = 3/11 × h ⇒ V = πh(3/11 × h)2 ⇒ = 9/121πh2 ⇒ = h 11 3 dt dt dt 9/121πh2 QUESTIONS AND THE ANSWER Page 15/1 — Question code: 1234 121 dV dh −1/6 dh = − = −1/6 ⇒ = ≈ −0.0792(m/min). Because the sign of is dt dt 9/121π32 486π dt negative, the water level in the tank is dropping a rate of 0.0792(m/min). When h = 3; The correct answer is A □ Question 38 (L.O.2). Boyle’s law states that when a sample of gas is compressed at a constant temperature, the pressure P and volume V satisfy equation P V = C = const. Suppose that at a certain instant the volume is 647cm3 , the pressure is 184kP A, and the pressure is increasing at a rate of 28kP A/min. At what rate is the volume decreasing at this instant? A 98.1919 B 99.3825 D 97.6455 C 98.4565 E 98.362 Solution Since the pressure P and volume V satisfy equation (3) P V = C = const, where P and V are functions of the time t in minutes, so we can differentiate both sides of equation (3) with respect to time and receive dP dV ·V +P · = 0. dt dt Since the pressure P is increasing at a rate of 28 kPA/min, so (4) dP = 28 dt Substituting these, as well as V = 647cm3 and P = 184kP A, into equations (4) yields 28 × 647 + 184 · ⇒ dV =0 dt dV 28 × 647 =− = 98.4565217391304 ≈ −98.4565. dt 184 dV is negative, the volume is decreasing at a rate of 98.4565 (cm3 /min). dt The correct answer is C Because the sign of 3.9 □ Differentials Question 39 (L.O.1). Find d2 y(0) if y = cos6 (7x). A −292dx2 B −289dx2 C −298dx2 D −299dx2 E −294dx2 Solution We have y ′ = 6 cos5 (7x)(− sin(7x)) × 7 = −42 cos5 (7x) sin(7x) ⇒ y ′′ = −42 × 5 cos4 (7x)(− sin(7x)) × 7 × sin(7x) − 42 × 7 cos5 (7x) cos(7x). Thus y ′′ (0) = −42 × 7 = −294 ⇒ d2 y(0) = −294dx2 . The correct answer is E □ QUESTIONS AND THE ANSWER Page 16/1 — Question code: 1234 3.10 Applications of derivatives in physics Question 40 (L.O.1). A particle moving on the x−axis has position s(t) = t3 − 39t2 + 432t + 12 (m) after an elapsed time of t seconds. What is the total distance travelled by the particle during the first 31 seconds? A 6699 B 6707 C 6704 D 6709 E 6702 Solution We have s(t) = t3 − 39t2 + 432t + 12 ⇒ v(t) = s′ (t) = 3t2 − 78t + 432. 0<t<8 v(t) > 0 ⇔ and v(t) < 0 ⇔ 8 < t < 18 18 < t < 31 So the total distance travelled by the particle during the first 31 seconds is |s(8) − s(0)| + |s(18) − s(8)| + |s(31) − s(18)| = 6704. The correct answer is C □ Question 41 (L.O.1). A particle moving on the x−axis has position s(t) = t3 − 45t2 + 600t + 9 (m) after an elapsed time of t seconds. What is the displacement travelled by the particle during the first 37 seconds? A 11243 B 11250 C 11248 D 11251 E 11253 Solution We have s(t) = t3 − 45t2 + 600t + 9. So the displacement travelled by the particle during the first 37 seconds is s(37) − s(0) = 11248. The correct answer is C 3.11 □ Applications of derivatives in Economics Question 42 (L.O.2). For Luce Landscaping, the total revenue from the yard maintenance of x homes is given by R(x) = 1868x − 4x2 (dollars) and the total cost is given by C(x) = 2938 + 18x (dollars). Suppose that Luce is adding 19 homes per day at the moment when the 458th customer is signed. At that moment, what is the rate of change of total profit P in dollars per day, if the total profit P (x) = R(x) − C(x)? A −34469 B −34465 C −34471 D −34466 E −34463 Solution QUESTIONS AND THE ANSWER Page 17/1 — Question code: 1234 We have h i P (x(t)) = R(x(t)) − C(x(t)) ⇒ P ′ (t) = P ′ (x).x′ (t) = R′ (x) − C ′ (x) x′ (t). From the given information, we have x = 458 and x′ (t) = 19. Therefore, h i h i P ′ (t) = 1868 − 2 × 4x − 18 x′ (t) = 1868 − 2 × 4 × 458 − 18 × 19 = −34466. The correct answer is D 3.12 □ Linear Approximations Question 43 (L.O.1). Use the linear approximation formula to find an approximate value of f (x) at x = 12.17, given that f (12) = 4 and f ′ (12) = 0.74. A 3.9803 B 4.4361 C 4.1258 D 3.8998 E 3.8271 Solution The approximate value of f (x) at x = 12.17, is f (12.17) ≈ f (12) + f ′ (12)∆x = f (12) + f ′ (12) × (12.17 − 12) = 4.1258 ≈ 4.1258. The correct answer is C □ Question 44 (L.O.2). A spherical balloon is inflated so that its radius increases from 20 cm to 20.44 cm in 4 seconds. By approximately how much has its volume increased (using linear approximation) (in cm3 /s)? A 552.2614 B 552.9203 C 553.2891 D 552.9905 E 552.0326 Solution 4 The volume of the spherical balloon is V = πR3 . So the change of volume in 1 second is 3 20.44 − 20 ∆V ≈ V ′ (R).R′ (t) = 4πR2 .R′ (t) ≈ 4π × 202 × = 176.000000000001π ≈ 552.9203. 4 The correct answer is B □ Question 45 (L.O.2). A company estimates that volume sales S(x) of a type of commodity depends on the amount of money x (in thousands dollars) spent on advertising, that is given by: S(x) = −0.0079x3 + 0.412x2 + 2.8147x + 338. Use the differential (also known as: the linear approximation formula) to estimate the change in the total sales of this commodity if the advertising expenditures are increased from 41 to 41.7583 (thousands dollars). A −2.4577 QUESTIONS AND THE ANSWER B −1.7600 C −2.4103 D −2.4215 E −1.9167 Page 18/1 — Question code: 1234 Solution The change in the total sales of this commodity if the advertising expenditures are increased from 41 to 41.7583 (thousands dollars) is ∆S ≈ S ′ (41)∆x = (3 × (−0.0079) × 412 + 2 × 0.412 × 41 + 2.8147) × (41.7583 − 41) = = −2.45765283203125 ≈ −2.4577. The correct answer is A □ Question 46 (L.O.2). A ball of ice melts so that its radius decreases from 30 cm to 29.1647 cm. By approximately how much does the volume of the ball decrease? A −9447.0725 B −9446.1664 C −9446.7456 D −9447.01 E −9447.2685 Solution 4 The volume of the ball of ice is V = πR3 3 ⇒ ∆V ≈ dV = V ′ (R)dR = V ′ (R)∆R = 4πR2 ∆R Since ∆R = 29.1647 − 30 = −0.8353 cm and R = 30 cm, so ∆V ≈ 4π × 302 × (−0.8353) = −3007.08π ≈ −9447.01 cm3 . The correct answer is D □ Question 47 (L.O.2). A water reservoir has a base shaped like a rectangle combined with half of a circle, as illustrated in the picture. x 4x If we can measure x = 5 ± 0.0599(m). Approximate the error in the area of the reservoir’s base using differential formula. A 6.2392 QUESTIONS AND THE ANSWER B 5.6269 C 6.6628 D 6.6852 E 6.1596 Page 19/1 — Question code: 1234 Solution 1 The area of the reservoir’s base is S(x) = x × 4x + × π × 2 4x 2 2 ⇒ S ′ (x) = 8x + 4.0πx. Therefore, ∆S ≈ S ′ (5)∆x = (40 + 20.0π) × (±0.0599) = 2.396 + 1.198π ≈ 6.1596. The correct answer is E □ Question 48 (L.O.1). Let f be an even function and f is differentiable in R. Given that f ′ (1) = 2.81, f (1) = −3. Use differential formula to approximate f (−0.22). A −6.0772 B −5.1918 C −5.673 D −6.0849 E −6.0731 Solution Using property of even function and differential formula, we have f (−0.22) = f (0.22) ≈ f (1) + f ′ (1)(0.22 − 1) = −5.1918 ≈ −5.1918 The correct answer is B 3.13 □ Monotonicity ln(3x + 4)2 , then which statement is always true? 3x + 4 4 e 4 e A The function y is decreasing on − − , − + 3 3 3 3 4 e 4 4 4 e B The function y is increasing on − − , − and − , − + 3 3 3 3 3 3 4 e 4 e C The function y is increasing on − − , − + 3 3 3 3 Question 49 (L.O.1). If y = D The function y is increasing on R 4 e 4 e E The function y is increasing on − − , 0 ∪ 0, − + 3 3 3 3 Solution We have y= 4)2 ln(3x + 3x + 4 Thus QUESTIONS AND THE ANSWER = 2 ln |3x + 4| 2(3 − 3 ln |3x + 4|) ⇒ y′ = = 3x + 4 (3x + 4)2 3 · 2 − log (3x + 4)2 (3x + 4)2 −4 − e 4 e x= =− − 3 3 3 y ′ = 0 ⇔ |3x + 4| = e ⇔ −4 + e 4 e x= =− + 3 3 3 Page 20/1 — Question code: 1234 x −∞ f ′ (x) 4 e − − 3 3 − + 0 −4/3 +∞ 0 4 e − + 3 3 + − 0 4 e f − + 3 3 +∞ f (x) f 4 e − − 3 3 −∞ 0 4 e 4 e Therefore, y is increasing on − − , −4/3 and −4/3, − + . 3 3 3 3 The correct answer is B □ tan x − 3 Question 50 (L.O.1). Find all real numbers m such that y = is increasing on the interval tan x − m π 0; . 4 A 1⩽m<3 B m>3 C m⩾0 D m ⩽ 0 or 1 ⩽ m < 3 E m⩽0 Solution 1 Let t = tan x ⇒ t ∈ (0; 1). Then t′ (x) = > 0, ∀x ∈ R and y ′ (x) = y ′ (t).t′ (x). So if y ′ (x) > 0, ∀x ∈ cos2 x π ⇒ y ′ (t) > 0, ∀t ∈ (0; 1). We have 0; 4 y(t) = t−3 3−m ⇒ y ′ (t) = t−m (t − m)2 Thus y is increasing on (0; 1) when y ′ (t) > 0 1⩽m<3 m<3 ⇔ ⇔ m∈ m∈ / (0; 1) / (0; 1) m⩽0 The correct answer is D 3.14 □ Local extreme values Question 51 (L.O.1). Studying the local extreme values of the function f (x) = arctan 3x − 9 x2 + 72 , which statement is always true? A f has local minimum when x = −6, and no local maximum B f has local minimum when x = −6, and local maximum when x = 12 C f has local maximum when x = −6, and local minimum when x = 12 D f has no local minimum, and local maximum when x = 12 E f has no local minimum, and no local maximum Solution QUESTIONS AND THE ANSWER Page 21/1 — Question code: 1234 Since f (x) = arctan 3x − 9 x2 + 72 −3x2 + 18x + 216 3 −x2 + 6x + 72 −3x2 + 18x + 216 (x2 + 72)2 ′ ⇒ f (x) = = 4 = 2 (x + 72)2 + (3x − 9)2 x + 153x2 − 54x + 5265 3x − 9 2 1+ x2 + 72 ⇒ f ′ (x) = 0 ⇔ x = −6 ∨ x = 12 x −∞ f ′ (x) −6 − + 0 +∞ 12 0 0 − f (12) f (x) f (−6) 0 The correct answer is B 3.15 □ Local extreme values of the parametric curve Question 52 (L.O.1). Let y = y(x) be the parametric curve defined by x(t) = −5t ln(t + 4) (t ⩾ 0). y(t) = 2t3 − 24t2 + 72t + 4 Which statement is always TRUE? A y attains local minimum value when x = −10 ln(6) B y attains local maximum value when x = −10 ln(6) C y attains local minimum value when x = 6 D y attains local maximum value when x = −30 ln(10) E y attains local maximum value when x = 2 Solution Since x(t) = −t ln(t + 1) ⇒ x′ (t) = −5 ∗ t/(t + 4) − 5 ∗ log(t + 4) < 0, ∀t ⩾ 0. On another hand, t = 2 y(t) = 2t3 − 24t2 + 72t + 4 ⇒ y ′ (t) = 6t2 − 48t + 72 = 0 ⇔ t = 6 Also we have y ′′ (x) = y ′′ (t)x′ (t) − y ′ (t)x′′ (t) [x′ (t)]3 If y ′′ (x0 ) > 0 then y attains local minimum value at x0 ; if y ′′ (x0 ) < 0 then y attains local maximum value at x0 . If t = 2 then x = −10 log (6) and y ′′ (−10 log (6)) = QUESTIONS AND THE ANSWER 40 + 120 log (6) 3 ≈ −0.2126 −5 log (6) − 53 Page 22/1 — Question code: 1234 If t = 6 then x = −30 log (10) and y ′′ (−30 log (10)) = −120 log (10) − 72 ≈ 0.1139 (−5 log (10) − 3)3 The correct answer is B 3.16 □ Concavity and inflection point Question 53 (L.O.1). If y = x3 e(12/11)x , then the number of point(s) of inflection of the graph of y is A 4 C 0 B 1 D 2 E 3 Solution Domain: D = R y ′ = 3x2 e12/11x + 12 × x3 e12/11x . 11 2 12 2 12/11x 12 12 x e + 3 × x2 e12/11x + x3 e12/11x = 11 11 11 h 12 2 i 12 = xe12/11x x2 + 6 × x + 6 . 11 11 ⇒ y ′′ = 6xe12/11x + 3 × y ′′ = 0 ⇔ x = 0 ∨ x = −3/2 ∨ x = −4. −∞ x −3/2 −4 +∞ 0 f ′′ (x) − 0 + 0 − 0 + f (x) CD PI CU PI CD PI CU Therefore, the graph of y has 3 points of inflection. The correct answer is E 3.17 □ Global maximum and minimum values Question 54 (L.O.2). Find the maximum value of e2−x + x2 + 9x + 5, if 0 ⩽ x ⩽ 2 f (x) = x2 − 24x + 72, if 2 < x ⩽ 12 A 23 B 31 C 25 D 30 E 28 Solution If 0 ⩽ x ⩽ 2 then f (x) = e2−x + x2 + 9x + 5 ⇒ f ′ (x) = 2x − e2−x + 9 and f ′′ (x) = e2−x + 2 > 0, ∀x ∈ (0, 2). Therefore, f ′ (x) ⩾ f ′ (0) = −e2 + 9 > 0, ∀x ∈ (0, 2) ⇒ fmax = f (2) = 28, x ∈ [0, 2]. QUESTIONS AND THE ANSWER Page 23/1 — Question code: 1234 If 2 ⩽ x ⩽ 12 then f (x) = x2 − 24x + 72 ⇒ f ′ (x) = 2x − 24 < 0, ∀x ∈ (2, 12) ⇒ fmax = f (2) = 28, ∀x ∈ [2, 12]. The function f is continuous at x = 2. Thus, fmax = f (2) = 28, ∀x ∈ [0, 12]. The correct answer is E □ Question 55 (L.O.2). A company specializing in producing athletic uniforms for students stated that the cost to produce x batches (0 < x ⩽ 264) is C(x) = 0.3x2 + 12x + 4915.2, (hundreds dollars) The average cost per batch when producing x batches is given by the formula Cave = C(x) · Find x such x that the average cost atains minimum value. A 125 B 131 C 124 D 123 E 128 Solution We have Cave = C(x) = x 0.3x2 x = −128 (rejected) + 12x + 4915.2 4915.2 ′ ⇒ Cave = 0.3 − =0⇔ 2 x x x = 128 (accepted) Comparing the values of Cave when x = 128 and when x = 264 we obtain min Cave = min{88.8000; 109.8182} = 88.8000. Therefore, the average cost atains minimum value when x = 128. The correct answer is E 3.18 □ Maclaurin-Taylor approximations Question 56 (L.O.1). Find the coefficient of x8 in the Maclaurin polynomial of degree 9 for the function f (x) = (2x − 5) sin (2x). 614 299 A B 315 315 C −16/315 D − 1591 315 E − 646 315 Solution The Maclaurin polynomial for sin(2x) is 2x − 4x3 4x5 8x7 + − + O x9 3 15 315 Therefore, the Maclaurin polynomial for the function f (x) = (2x − 5) sin (2x) is −10x + 4x2 + 20x3 8x4 4x5 8x6 8x7 16x8 − − + + − + O x9 3 3 3 15 63 315 Thus, the coefficient of x8 in the Maclaurin polynomial of degree 9 for the function f (x) is − 16 315 The correct answer is C □ QUESTIONS AND THE ANSWER Page 24/1 — Question code: 1234 Question 57 (L.O.1). Write the Taylor polynomial of degree 3 for the function f (x) 2x2 − 8x + 11 e5x where x is near x0 = 2. 79 (x − 2)2 e10 145 (x − 2)3 e10 A 3e10 + 1 + 15 (x − 2) e10 + + + O (x − 2)4 ; x → 2 2 2 79 (x − 2)2 e10 145 (x − 2)3 e10 10 10 B 3e − 2 + 15 (x − 2) e + + + x + O (x − 2)4 ; x → 2 2 2 2 10 3 10 79 (x − 2) e 145 (x − 2) e C 3e10 + 15 (x − 2) e10 + + − (x − 2)3 + O (x − 2)4 ; x → 2 2 2 79 (x − 2)2 e10 145 (x − 2)3 e10 2 10 10 − (x − 2) + + O (x − 2)4 ; x → 2 D 3e + 15 (x − 2) e + 2 2 2 10 3 10 79 (x − 2) e 145 (x − 2) e E 3e10 + 15 (x − 2) e10 + + + O (x − 2)4 ; x → 2 2 2 = Solution The Taylor polynomial for e5x is 25 (x − 2)2 e10 125 (x − 2)3 e10 + + O (x − 2)4 ; x → 2 2 6 Therefore, the Taylor polynomial for the function f (x) = 2x2 − 8x + 11 e5x is e10 + 5 (x − 2) e10 + 3e10 + 15 (x − 2) e10 + 79 (x − 2)2 e10 145 (x − 2)3 e10 + + O (x − 2)4 ; x → 2 2 2 The correct answer is E 4 □ Mixed problems [L.O.2] Using the following information, answer the questions from 58 to 59. A family plans to organize a party at a restaurant. The overall service cost for the event (venue, decoration, etc.) is 8 million Vietnamese dong. The restaurant requires a minimum of 2 tables, with each table accommodating 11 people. The cost per guest is 250 thousand Vietnamese dong. The family plans to spend a maximum of 28.25 million Vietnamese dong. If we denote x as the number of invited guests, C(x) (in million Vietnamese dong) represents the cost for the party. Question 58. Find the domain of C(x) is A N ∩ [22; 81] D N ∩ [0; 22] B [0; 81] C [22; 28.25] E N ∩ [8; 28.25] Solution 250 × x. The restaurant requires a minimum of 2 tables, each table accommodating 1000 11 people, so x ⩾ 22. On the another hand, the family plans to spend a maximum of 28.25 million, hence 250 C(x) ⩽ 28.25 ⇔ 8 + × x ⩽ 28.25 ⇔ x ⩽ 81. 1000 The correct answer is A □ We have C(x) = 8 + QUESTIONS AND THE ANSWER Page 25/1 — Question code: 1234 Question 59. The range of C(x) is A [22; 81] C [13.5; 28.25] B [13.5; 81] D [8; 28.25] E [5.5; 28.25] Solution We have C(x) = 8 + 250 250 ×x⩾8+ × 22 = 13.5. 1000 1000 Furthermore, C(x) ⩽ 28.25. The correct answer is C □ [L.O.2] Using the following information, answer the questions from 60 to 61. In a certain country, income tax is assessed as follows. There is no tax on income up to $12340. Any income over $12340 is taxed at a rate of 10%, up to an income of $23446. Any income over $23446 is taxed at 13%. Question 60. Express the tax T as a function of income I. 0, I ⩽ 12340 A T = 0.1, 12340 < I ⩽ 23446 0.13, I > 23446 0, I ⩽ 12340 B T = 0.1I, 12340 < I ⩽ 23446 0.1I, I > 23446 0, I ⩽ 12340 C T = D T = E T = 0.13I, 12340 < I ⩽ 23446 0.13I, I > 23446 0, I ⩽ 12340 0.1I, 12340 < I ⩽ 23446 0.13I, I > 23446 0, I ⩽ 12340 0.13I, 12340 < I ⩽ 23446 0.1I, I > 23446 Solution The tax function is T = 0, I ⩽ 12340 0.1I, 12340 < I ⩽ 23446 0.13I, I > 23446 The correct answer is D □ Question 61. How much tax (in dollars) is assessed on an income of $27148. A 3529.6 QUESTIONS AND THE ANSWER B 3529.5 C 3528.7 D 3529.7 E 3529.2 Page 26/1 — Question code: 1234 Solution Using the formula for tax function, we have T (27148) = 3529.2 The correct answer is E □ [L.O.2] Using the following information, answer the questions from 62 to 64. A water filter equipment manufacturer has a fixed monthly cost of 340 million Vietnamese dong, and the variable cost to produce x devices is C(x) = −0.005x2 + 6.0x, (0 ⩽ x ⩽ 800) (million Vietnamese dong). The selling price per device is p(x) = −0.002x + 8.6, (0 ⩽ x ⩽ 800) (million Vietnamese dong). Question 62. The company’s revenue (million Vietnamese dong) when selling 144 devices in a month is A 1197.9280 B 1193.9280 C 1199.9280 D 1196.9280 E 1192.9280 Solution The company’s revenue is R(x) = x.p(x) = x(−0.002x + 8.6). Therefore, the company’s revenue (million Vietnamese dong) when selling 144 devices in a month is R(144) = 144(−0.002 × 144 + 8.6) = 1196.9280 The correct answer is D □ Question 63. The company’s profit (million Vietnamese dong) when selling 144 devices in a month is A 99.6080 B 94.6080 C 91.6080 D 96.6080 E 98.6080 Solution The company’s profit is P (x) = R(x) − C(x) − 340 = x(−0.002x + 8.6) − (−0.005x2 + 6.0x) − 340. Therefore, The company’s profit (million Vietnamese dong) when selling 144 devices in a month is P (144) = 144(−0.002 × 144 + 8.6) − (−0.005 × (144)2 + 6.0 × 144) − 340 = 96.6080 The correct answer is D □ Question 64. To achieve the profit of 2300 million Vietnamese dong in a month, how many devices does the company need to sell in that month? A 605 B 604 C 601 D 600 E 602 Solution The company’s profit is P (x) = R(x) − C(x) − 340 = x(−0.002x + 8.6) − (−0.005x2 + 6.0x) − 340. QUESTIONS AND THE ANSWER Page 27/1 — Question code: 1234 Therefore, To achieve the profit of 2300 million Vietnamese dong in a month, the company need to sell x devices in that month, it means that x(−0.002x + 8.6) − (−0.005x2 + 6.0x) − 340 = 2300 ⇒ x = 600 The correct answer is D □ [L.O.2] Using the following information, answer the questions from 65 to 67. All pyramids with a square base and a height of h = 4 have a surface area given by S(v) = 0.75v + 2 p 0.28125v 2 + 12v, where v is the volume of the pyramid. If x is the length of each side of the base then we can consider the volume dv of a pyramid as a function v(x) of x. In addition, suppose that v(7) = 65.33 and (7) = 18.67. dx Question 65. Compute the slope of the slant asymptote of the graph of function S(v) with respect to v. A 1.1961 B 1.0757 C 2.3389 D 2.7592 E 1.8107 Solution The slope of the slant asymptote of the graph of function S(v) with respect to v is r √ √ S(v) 0.75v + 2 0.28125v 2 + 12v 3 3 2 12 m = lim = lim = lim 0.75 + 2 0.28125 + = + ≈ 1.8107 v→+∞ v v→+∞ v→+∞ v v 4 4 The correct answer is E □ Question 66. Compute the rate of change of the surface area of the pyramid with respect to the length x of each side when x = 7. A 34.1516 B 34.2931 C 34.1693 D 34.4337 E 35.2992 Solution The rate of change of the surface area of the pyramid with respect to the length x of each side is 2 × 0.28125 × v + 12 ′ ′ ′ √ S (x) = S (v).v (x) = 0.75 + × v ′ (x). 2 0.28125v + 12v When x = 7, we have S (7) = 0.75 + √ ′ 2 × 0.75 × 65.33 + 12 0.28125 × 65.332 + 12 × 65.33 × 18.67 ≈ 34.4337 The correct answer is D □ Question 67. Approximate the change in surface area of a pyramid when the length of each side of the base increases from x = 7 to x = 7.75. A 25.8357 B 25.1149 C 25.2789 D 25.8253 E 26.8070 Solution QUESTIONS AND THE ANSWER Page 28/1 — Question code: 1234 The change in surface area of a pyramid when the length of each side of the base increases from x = 7 to x = 7.75 is ∆S ≈ S ′ (7).(7.75 − 7) ≈ 25.8253 The correct answer is D 5 □ Integration 5.1 Indefinite integrals Question 68 (L.O.1). Find the antiderivative F (x) of the function f (x) = 8 sin x+4 cos x which satisfies π = 12. the condition F 2 A −8 cos x + 4 sin x + 8 B −8 cos x − 4 sin x + 8 C −8 cos x + 4 sin x − 8 D 8 cos x − 4 sin x − 8 E 8 cos x + 4 sin x + 8 Solution Z f (x) dx = 4 sin (x) − 8 cos (x) + C. Moreover, We have F (x) = F π 2 = 4 + C = 12 ⇒ C = 8. The correct answer is A □ Question 69 (L.O.1). Find the antiderivative F (x) of the function f (x) = 8ex + 4x which satisfies the condition F (0) = 14. A 8ex + 2x2 − 6 B −8ex + x2 + 6 D 8ex + 2x2 + 6 C −8ex + 2x2 − 6 E ex + 2x2 + 6 Solution Z We have F (x) = f (x) dx = 8ex + 2x2 + C. Moreover, F (0) = 8e0 + 2 × 02 + C = 14 ⇒ C = 6 The correct answer is D □ Question 70 (L.O.1). Suppose that f ′′ (x) = −18x − 100 sin(5x) + 9e3x and f ′ (0) = 23, f (0) = −379. Evaluate f (2). A −2.2641 B −2.7473 C −2.0485 D −3.6915 E −3.5396 Solution QUESTIONS AND THE ANSWER Page 29/1 — Question code: 1234 We have f ′′ (x) = −18x − 100 sin(5x) + 9e3x ⇒ f ′ (x) = −9x2 + 20 cos(5x) + 3e3x + C1 . Since f ′ (0) = 23, so C1 = 0. Thus, f (x) = −3x3 + 4 sin(5x) + e3x + 0x + C2 . Since f (0) = −379, nên C2 = −380. Therefore, f (2) ≈ −2.7473. The correct answer is B 5.2 □ Definite integrals π 2 Z Question 71 (L.O.1). Let Z f (x) dx = 9.51. Evaluate I = 0 A 63.8129 π 2 [6.49f (x) + 2.63 sin x] dx. 0 C 64.0289 B 64.3499 D 64.5836 E 64.7637 Solution We have Z I= π 2 π 2 Z [6.49f (x) + 2.63 sin x] dx = 6.49 0 Z f (x) dx + 2.63 0 π 2 sin x dx = 0 π 2 = 61.7199 − 2.63 cos x = 61.7199 + 2.63 = 64.3499(≈ 64.3499). 0 The correct answer is B □ Z Question 72 (L.O.1). Let i 5.55f (x) − 7.76g(x) dx. 6.4 Z −2.5 Z 6.4 g(x) dx = −2.47. Evaluate I = f (x) dx = 6.88 and −2.5 B 384.9296 A 384.0257 6.4 C 385.2162 h 18.86x + −2.5 D 384.6665 E 385.0722 Solution We have Z 6.4 Z 6.4 [18.86x + 5.55f (x) − 7.76g(x)] dx = 18.86 −2.5 Z x2 2 Z 6.4 f (x) dx − 7.76 x dx + 5.55 −2.5 = 18.86 6.4 −2.5 g(x) dx = −2.5 6.4 + 5.55 × 6.88 − 7.76 × (−2.47) = 384.6665(≈ 384.6665). −2.5 The correct answer is D □ Question 73 (L.O.2). Let g(x) = f −1 (x) be the inverse function of function y = f (x) = ex/2 + 5.34. Z 11.01 Evaluate g(x)dx. 6.34 A 9.6536 B 10.3370 C 10.9418 D 10.7950 E 9.4292 Solution x Since y = f (x) = e 2 + 5.34 ⇒ x = 2 ln(y − 5.34). Thus g(x) = f −1 (x) = 2 ln(x − 5.34). Therefore, Z 11.01 g(x)dx ≈ 10.3370. 6.34 QUESTIONS AND THE ANSWER Page 30/1 — Question code: 1234 The correct answer is B 5.3 □ Riemann sum Question 74 (L.O.2). Using the midpoint rule with 5 equal interval on [3; 11] to estimate the following Z 11 f (x)dx where f (x) is given by the following table integral I = 3 x 3 3.8 4.6 5.4 6.2 7.0 7.8 8.6 9.4 10.2 f (x) 6.8 4.3 6.6 5.0 3.4 5.8 5.1 5.5 4.2 A 41.8 B 41.88 C 42.24 11 5.8 6.0 D 41.6 E 42.26 Solution 11 − 3 = 1.6. Therefore, the subintervals consist of [3, 4.6]; [4.6, 6.2]; 5 [6.2, 7.8]; [7.8, 9.4]; [9.4, 11.0]. The midpoints of these subintervals are 3.8; 5.4; 7.0; 8.6; 10.2. Thus, the approxEach subinterval has length of ∆x = imation of the integral using midpoint rule is h i I ≈ ∆x × 4.3 + 5.0 + 5.8 + 5.5 + 5.8 ≈ 42.24 The correct answer is C 5.4 □ The fundamental theorem of calculus Z x (t2 − 9t − 22)dt on the Question 75 (L.O.2). Find the maximum and minimum values of f (x) = −9 interval [−9, 9]. A fmax ≈ 433.8311; fmin ≈ −0.1873 B fmax ≈ 433.4807; fmin ≈ 0.638 C fmax ≈ 432.9731; fmin ≈ 0.1885 D fmax ≈ 433.7281; fmin ≈ 0.0568 E fmax ≈ 432.8333; fmin ≈ 0 Solution We have Z x f (x) = −9 (t2 − 9t − 22)dt ⇒ f ′ (x) = x2 − 9x − 22 = 0 ⇔ x = −2 x = 11 −9 Z (t2 − 9t − 22)dt = 0; f (−9) = −9 Z −2 (t2 − 9t − 22)dt ≈ 432.8333; f (−2) = −9 Z 9 f (9) = (t2 − 9t − 22)dt ≈ 90. −9 QUESTIONS AND THE ANSWER Page 31/1 — Question code: 1234 The correct answer is E 6 □ Applications of integration 6.1 The Area problems Question 76 (L.O.2). Find the area of the region bounded by the graphs of y = x = 0; and x = 4. A 0.5049 C 0.2790 B 0.3040 D 0.2973 1 √ ; y = 0; 9 16 − x2 E 0.1745 Solution The required area is Z 4 dx 1h b π x i4 1 √ A= = lim arcsin − arcsin 0 = ≈ 0.1745 arcsin = − 2 9 4 9 4 18 b→4 0 0 9 16 − x The correct answer is E □ Question 77 (L.O.2). Find the area of the region bounded by the graphs of y = e2.53x ; y = e4.75x and x = 3.03. A 374034.3816 B 374034.2241 C 374034.3701 D 374034.9554 E 374034.1488 Solution The required area is Z 3.03 A= 0 e2.53x i3.03 = 4.75 2.53 0 = 374034.381624145 ≈ 374034.3816 (e4.75x − e2.53x )dx = h e4.75x − The correct answer is A 6.2 □ The Volume of the solid of revolution Question 78 (L.O.2). Let the region D be bounded by the curve y = p 15x2 + 2, x−axis and the lines x = 0, x = 5. Evaluate the volume V of the solid, revolving the region D about x−axis. A 1995.0119 B 1995.0890 C 1994.9113 D 1994.7434 E 1993.9313 Solution We have Z V =π 5 p 15x2 0 +2 2 Z dx = π 0 5 5 15x2 + 2 dx = π 5x3 + 2x 0 = 635π ≈ 1994.9113 The correct answer is C □ QUESTIONS AND THE ANSWER Page 32/1 — Question code: 1234 Question 79 (L.O.2). Find the volume of the solid obtained by rotating the region bounded by the curves x = 0, y = −16, y = x2 − 8x about the y−axis. A 134.0465 B 134.0413 C 133.9908 D 133.6074 E 134.2850 Solution We have VOy = 2π Z 4 0 Z 4 128π ≈ 134.0413 |x(−16)| − |x(x2 − 8x)| dx = 2π x(16 − 8x + x2 )dx = 3 0 The correct answer is B 6.3 □ The arc length Question 80 (L.O.2). Find the length of the arc y = 3 ln x, where 6 ⩽ x ⩽ 9. A 3.2399 B 2.8595 C 3.9077 D 2.5529 E 3.1169 Solution 3 · The arc length is x s 2 Z 9 Z 9p √ √ 3 1 1 1 + [y ′ (x)]2 dx = 1+ L= dx = −3 5 − 3 asinh + 3 asinh + 3 10 ≈ 3.2399 x 3 2 6 6 We have y = 3 ln x ⇒ y ′ = The correct answer is A □ Question 81 (L.O.2). A steady wind blows a kite due west. The kite’s height above the ground from 3 √ 3 + 2x . Find the real number a such horizontal position x = 0 to x = a > 0 meters is given by y = that the distance travelled by the kite is 20.66 meters. A 3.6074 B 3.2822 C 2.5416 D 2.8492 E 3.5988 Solution We have y = √ 3 + 2x 3 √ ⇒ y ′ = 3 2x + 3· The distance travelled by the kite is √ 3 Z ap Z ar h √ i2 (18a + 28) 2 56 7 ′ 2 L= 1 + [y (x)] dx = 1 + 3 2x + 3 dx = − = 20.66 27 27 0 0 ⇒ a ≈ 2.8492. The correct answer is D □ Question 82 (L.O.2). Find the length of the arc y = P (x), where 3 ⩽ x ⩽ 9. Here P (x) is a polynomial of degree 2 such that P (3) = 62, P (4) = 96, P (9) = 386. A 323.5810 QUESTIONS AND THE ANSWER B 323.6091 C 323.3469 D 324.0597 E 324.9452 Page 33/1 — Question code: 1234 Solution Let P (x) = ax2 + bx + c. Since P (3) = 62, P (4) = 96, P (9) = 386, so we have 9a + 3b + c = 62 16a + 4b + c = 96 ⇒ a = 4; b = 6; c = 8. 81a + 9b + c = 386 Thus, P (x) = 4x2 + 6x + 8 ⇒ P ′ (x) = 8x + 6. The arc length is Z 9r Z 9p h i2 1 + [y ′ (x)]2 dx = 1 + 8x + 6 dx = L= 3 3 √ √ 15 901 asinh (30) asinh (78) 39 6085 =− − + + ≈ 324.0597. 8 16 16 8 The correct answer is D □ Z xp Question 83 (L.O.2). Find the length of the arc y = 5t2 + 7dt, where 2 ⩽ x ⩽ 8. 2 A 69.4857 B 69.7190 C 69.7552 D 69.0281 E 68.6696 Solution We have y ′ (x) = p 5x2 + 7. The arc length is Z 8r Z 8p hp i2 ′ 2 1 + [y (x)] dx = 1+ 5x2 + 7 dx = L= 2 2 √ = −2 7 − √ √ 4 5 asinh 210 5 √ √ √ 4 5 asinh 2 10 + + 8 82 ≈ 69.4857 5 The correct answer is A 6.4 □ The Area of the surface of revolution Question 84 (L.O.2). Find the area of the surface of revolution obtained by rotating the curve y = √ 2 x + 6, where 1 ⩽ x ⩽ 8 about the x−axis. A 502.2993 B 502.5967 C 502.8761 D 502.4058 E 502.0702 Solution √ 1 We have y = 2 x + 6 ⇒ y ′ = √ · The surface area is x s Z 8 p Z 8 √ 1 2 ′ 2 A = 2π |y| 1 + [y (x)] dx = 2π (2 x + 6) 1 + √ dx = x 1 1 √ ! √ √ 26 2 √ = −2π 6 log 1 + 2 + + 2π 6 asinh 2 2 + 36 + 36 2 ≈ 502.2993. 3 The correct answer is A □ QUESTIONS AND THE ANSWER Page 34/1 — Question code: 1234 p Question 85 (L.O.2). Rotating the curve x = − y 2 + 3, where 0 ⩽ y ⩽ 3 about the y−axis, evaluate the surface area of the obtained surface of revolution. A 53.5843 C 54.0413 B 54.2794 D 53.1674 E 54.2563 Solution p y We have x = − y 2 + 3 ⇒ x′ = − p · The surface area is 2 y +3 3 Z A = 2π s Z 3 p p |x| 1 + [x′ (y)]2 dy = 2π | − y 2 + 3| 1 + 0 Z s 3 2π = 2π 0 0 2y 2 + 3 p 2 y + 3dy = 2π y2 + 3 y2 dy = y2 + 3 √ √ √ ! 3 2 asinh 6 3 21 ≈ 54.0413. + 4 2 The correct answer is C 6.5 □ Applications of integration in Physics Question 86 (L.O.2). Evaluate the approximation of the displacement of one particle with velocity v(t) = 9 arcsin(t) (m/s) from t = 0s to t = 0.7s. A 3.0163 B 3.2834 C 2.5228 D 1.9853 E 2.3123 Solution The displacement of one particle is Z D= 0.7 Z 0.7 v(t)dt = 0 9 arcsin(t)dt. 0 u = arcsin t du = √ dt 1 − t2 Let ⇒ dv = dt v = t Therefore, Z 0.7 Z 0.7 0.7 tdt 9 0.7 √ −9 + D = 9t arcsin t = 9t arcsin t (1 − t2 )−1/2 d(1 − t2 ) = 2 0 0 0 1 − t2 0 p 0.7 0.7 = 9t arcsin t + 9 1 − t2 = 2.31228981433631 ≈ 2.3123. 0 0 The correct answer is E □ Question 87 (L.O.2). The velocity of a dragster t seconds after leaving the starting line is v(t) = 72te−0.7t ft/sec. Find the approximation of the distance traveled by the dragster during the first 9 seconds. A 145.0934 B 144.0773 C 143.9867 D 144.9122 E 144.9691 Solution QUESTIONS AND THE ANSWER Page 35/1 — Question code: 1234 The distance is 9 Z Z 9 |v(t)|dt = D= 72te−0.7t dt = 0 0 Z 9 72te−0.7t dt = 144.969057569783 ≈ 144.9691 0 The correct answer is E □ Question 88 (L.O.2). When a particle is located a distance x meter from the origin, a force of given 1 (newton) acts on it. How much work is done in moving it from x = 5 to x = F (x) = 2 x + 4x + 53 √ 7 3 − 2. A −0.0604 B −0.0687 C 0.0374 D 0.0634 E −0.4302 Solution We have Z √ 7 3−2 W = 5 √ 7 3−2 Z 7√3−2 1 d(x + 2) 1 (x + 2) F (x)dx = = = arctan 2 2 x + 4x + 53 (x + 2) + 49 7 7 5 5 √ 1 1 π π π = (arctan 3 − arctan 1) = − = ≈ 0.0374. 7 7 3 4 84 Z √ 7 3−2 5 = The correct answer is C □ Question 89 (L.O.2). When a particle is located a distance x meter from the origin, a force of given 1 F (x) = √ (newton) acts on it. How much work is done in moving it from x = 17/2 2 −16x + 128x + 1040 to x = 13. B −0.7293 A 0.2618 C 0.5967 D −0.3280 E −0.6007 Solution We have Z 1 d(x − 4) 1 13 p W = F (x)dx = dx = = 2 4 81 − (x − 4)2 17/2 17/2 −16x + 128x + 1040 17/2 1 (x − 4) 13 1 1 1 π π π = arcsin = arcsin 1 − arcsin = − = ≈ 0.2618. 4 9 4 2 4 2 6 12 17/2 Z 13 Z 13 √ The correct answer is A □ Question 90 (L.O.2). When a particle is located a distance x meter from the origin, a force of given F (x) = ln(6 + 5x) (newton) acts on it. How much work is done in moving it from x = 10 to x = 17. A 30.1241 B 30.4412 C 29.4318 D 30.0137 E 29.2949 Solution Z 17 We have W = Z 17 ln(6 + 5x)dx = − F (x)dx = 10 10 91 log (91) 56 log (56) −7+ ≈ 30.0137. 5 5 The correct answer is D □ QUESTIONS AND THE ANSWER Page 36/1 — Question code: 1234 Question 91 (L.O.2). When a particle is located a distance x meter from the origin, a force of given 60 (newton) acts on it. How much work is done in moving it from x = 1 to x = F (x) = √ 196 − x2 6. A 22.2853 B 22.7360 C 23.0572 D 21.5142 E 23.0559 Solution Z We have W = 6 Z F (x)dx = 1 1 6 60 √ dx = −60 asin 196 − x2 1 14 3 + 60 asin ≈ 22.2853. 7 The correct answer is A 6.6 □ Applications of integration in Economics Question 92 (L.O.1). The accumulated, or total, future value after T years of an income stream of R(t) thousand dollars per year, earning interest at the rate of r per year compounded continuously, is given by Z T R(t)e−rt dt. A company recently bought an automatic car-washing machine that is expected A = erT 0 to generate 3 thousand dollars in revenue per year, t years from now, for the next 7 years. If the income is reinvested in a business earning interest at the rate of r = 11% per year compounded continuously, find the total accumulated value of this income stream at the end of 7 years. A 30.6382 C 32.4296 B 31.6300 D 32.2997 E 31.7604 Solution We have r = 11%, R(t) = 3, T = 7. Therefore, the total accumulated value of this income stream at the end of 7 years is rT Z A=e T R(t)e−rt dt = 31.6299887395886 ≈ 31.6300 0 The correct answer is B 6.7 □ The average value of a funtion Question 93 (L.O.2). The interest rates charged by Madison Finance on auto loans for used cars over a certain 12−month period in 2022 are approximated by the function r(t) = − 1 3 1 2 t + t − 2t + 27, 21 3 (0 ⩽ t ⩽ 12) where t is measured in months and r(t) is the annual percentage rate. What is the average rate on auto loans extended by Madison over the 12−month period? A 11.1063 B 11.2527 C 10.2379 D 10.4286 E 10.6659 Solution The average rate on auto loans extended by Madison over the 12−month period is Z 12 Z 1 1 12 1 73 1 rave = r(t)dt = − t3 + t2 − 2t + 27 dt = ≈ 10.4286. 12 − 0 0 12 0 21 3 7 QUESTIONS AND THE ANSWER Page 37/1 — Question code: 1234 The correct answer is D 6.8 □ Improper integrals +∞ Z 3x2 dx Question 94 (L.O.2). Evaluate the integral 7 + x4 1 A 1.8614 B 2.4366 C 1.9137 D 1.5139 E 1.3924 Solution We have +∞ +∞ Z Z 3x2 dx 3/x2 dx I= = 7 + x4 1 + 7/x4 1 Let t = 1 dx ⇒ dt = − 2 · x x Z0 I= 1 −3dt = 1 + 7t4 1 Z1 3dt ≈ 1.9137 1 + 7t4 0 The correct answer is C □ Z Question 95 (L.O.2). Evaluate the integral A 1.1263 B 0.3787 +∞ √ 671 x √ 4 · C −0.6059 dx x2 + 625 D 0.7021 E −0.0613 Solution Let t = p 4 x2 + 625 ⇒ t4 = x2 + 625 ⇒ 2t3 dt = xdx. Then, Z +∞ Z +∞ Z +∞ 2t3 dt dx xdx √ √ I= √ = = = √ 4 4 2 t(t4 − 625) x2 + 1 x2 + 625 671 x · 671 x · 6 Z +∞ Z +∞ 2 Z +∞ Z +∞ 2t2 dt (t + 25) + (t2 − 25)dt dt dt = = = + = 4 2 2 2 2 t − 625 (t − 25)(t + 25) t − 25 t + 25 6 6 6 6 = 1 t−5 ln 10 t+5 π 1 6 1 1 − arctan − ln = 10 5 5 10 11 6 6 π = −0.175211610119639 + 0.1 log (11) + ≈ 0.3787 10 +∞ + 1 t arctan 5 5 +∞ = The correct answer is B □ Question 96 (L.O.2). The total profit of a firm in dollars, from producing x units of an item, is P (x). The firm is able to determine that its marginal profit is given by P ′ (x) = 190e−1.05x . Suppose that it were possible for the firm to make infinitely many units of this item. What would its total profit be? A 180.8926 B 181.1377 C 180.4869 D 181.9074 E 180.9524 Solution QUESTIONS AND THE ANSWER Page 38/1 — Question code: 1234 Z The total profit is P (x) = ∞ ′ ∞ Z −1.05x 190e P (x)dx = 0 0 190 −1.05x dx = e −1.05 ∞ = 0 190 ≈ 180.9524. 1.05 The correct answer is E □ +∞ Z 8dx √ . Question 97 (L.O.2). Evaluate the integral x2 + x4 5 A 0.9952 B 0.9963 C 1.5895 D 1.1703 E 1.7271 Solution We have +∞ +∞ Z Z 8dx 8/x2 dx √ p I= = x2 + x4 1 + 1/x2 5 Let t = 5 dx 1 ⇒ dt = − 2 · x x Z0 I= −8dt √ = 1 + t2 1/5 Z1/5 8dt √ 1 + t2 0 = 1.58952088279393 ≈ 1.5895 The correct answer is C □ Question 98 (L.O.2). Let a be the real number such that lim x→+∞ p x2 + 8x + 2 − x − a = 0. Evaluate +∞ Z dx · (x + a)(x − 2)2 a+ 6 A 4.5249 B 4.0162 C 3.4425 D 3.4233 E 4.8000 Solution We have a = lim x→+∞ p x2 + 8x + 2 − x = lim √ x→+∞ 8x + 2 = 4. Then, x2 + 8x + 2 + x +∞ Z dx log (10) log (4) 1 + + ≈ 4.0162 =4+ − (x + a)(x − 2)2 36 36 24 a+ 6 The correct answer is B □ Z 5 2 x dx mdx √ = x−1 81 − x2 1 0 D 18.3127 E 18.3730 Z Question 99 (L.O.2). Find the approximation of m which satisfies A 18.2816 B 17.4858 C 17.6462 9 √ Solution Z Let A = 0 9 √ h dx x i9 π = arcsin = · 9 0 2 81 − x2 QUESTIONS AND THE ANSWER Page 39/1 — Question code: 1234 Z 5 Let B = 1 √ x2 dx √ · Using substitution method, let t = x − 1 ⇒ t2 = x − 1 ⇒ 2tdt = dx, we have x−1 √ √ 5 √ 4 Z √4 3 √ 2( 4)5 4( 4)3 2t 4t 2 2 B= 2(t + 1) dt = + + 2t = + + 2 4. 5 3 5 3 0 0 B 824 = ≈ 17.4858 A 15π The correct answer is B Therefore, m = 7 □ Ordinary differential equations (ODE) 7.1 The first order differential equations 7 Question 100 (L.O.2). Which function is a solution of the following differential equation y ′ + y = 3x, x where C is any constant? 1 3 1 A y = x2 + Cx−7 B y = x2 + Cx−7 C y = x2 + Cx7 3 8 3 3 2 3 2 7 −7 D y = x + Cx E y = x + Cx 8 10 Solution This is the first order linear differential equation with p(x) = Z − Z − y=e 7 dx hZ x 3xe Z p(x)dx hZ y=e Z 7 and q(x) = 3x. Therefore, x p(x)dx q(x)e dx + C i 7 h 9 i i dx x dx + C = x−7 3 x + C = 1 x2 + Cx−7 9 3 The correct answer is A 7.2 □ The second order differential equations Question 101 (L.O.2). Let y(x) be a function satisfying y ′′ −4y ′ −32y = 0, y(0) = 6, y ′ (0) = 24. Evaluate y(1.88). A 13609714.8215 B 13609714.2745 D 13609713.4611 C 13609713.4938 E 13609714.0102 Solution 1. Step 1. Solve the homogeneous equation y ′′ − 4y ′ − 32y = 0. The characteristic equation k 2 − 4k − 32 = 0 has 2 real different roots k1 = 8, k2 = −4 QUESTIONS AND THE ANSWER Page 40/1 — Question code: 1234 2. Step 2. The homogeneous solution is yh = C1 e8x + C2 e−4x 3. Step 3. Thus, yh′ = 8C1 e8x + (−4)C2 e−4x . Since y(0) = 6 ⇒ C1 + C2 = 6 and y ′ (0) = 8C1 + (−4)C2 = 24 ⇒ C1 = 4, C2 = 2. Therefore, y(1.88) ≈ 13609714.0102 The correct answer is E □ Question 102 (L.O.2). Let y(x) be a function satisfying y ′′ − 5y ′ = 2x2 + 5, y(0) = 0, y ′ (0) = 0. Evaluate y(2). A 4543.1199 B 4542.7416 C 4542.6055 D 4542.9902 E 4541.7701 Solution 1. Step 1. Solve the homogeneous equation y ′′ − 5y ′ = 0. The characteristic equation k 2 − 5k = 0 has 2 real different roots k1 = 5, k2 = 0 2. Step 2. The homogeneous solution is yh = C1 e5x + C2 e0x = C1 e5x + C2 3. Step 3. Find a particular solution of nonhomogeneous equation y ′′ − 5y ′ = 2x2 + 5. The particular solution has a form yp = xs .e0x .(Ax2 + Bx + C). Since α = 0 is a root of characteristic equation then s = 1 and yp = x(Ax2 + Bx + C) = Ax3 + Bx2 + Cx. 0× yp = Ax3 + Bx2 + Cx −5× yp′ = 3Ax2 + 2Bx + C 1× yp′′ = 6Ax + 2B yp′′ − 5yp′ = −15Ax2 + (−10B + 6A)x + 2B − 5C = 2x2 + 5 ⇒A=− QUESTIONS AND THE ANSWER 2 2 129 ;B = − ;C = − 15 25 125 Page 41/1 — Question code: 1234 4. Step 4. The general solution is 2 2 129 3 2 ygen = yh + yp = C1 e + C2 + − x + − x + − x 15 25 125 2 4 129 ′ 5x 2 ⇒ ygen = 5C1 e + − x + − x+ − . 5 25 125 129 129 Since y(0) = 0 ⇒ C1 + C2 = 0 và y ′ (0) = 0 ⇒ C1 = , C2 = −C1 = − . Therefore, 625 625 6857 129e10 y(2) = − + ≈ 4542.6055 1875 625 5x The correct answer is C □ Question 103 (L.O.2). Let y(x) = axe4x be a function satisfying y ′′ − 10y ′ + 24y = −6.3894e4x . Find the value of a. A 3.5403 B 3.6631 C 3.1947 D 3.2759 E 3.0536 Solution Since y(x) = axe4x is the solution of y ′′ − 10y ′ + 24y = −6.3894e4x . Then, 24× yp = axe4x −10× yp′ = ae4x (1 + 4x) 1× yp′′ = ae4x (8 + 16x) y ′′ − 10y ′ + 24y = −2ae4x = −6.3894e4x ⇒ a = 3.1947. The correct answer is C □ ESSAY QUESTIONS 1 Differentiation 1.1 Related rates Question 104. [L.O.2] Two crates, A and B, are on the floor of a warehouse. The crates are joined by a rope 15m long, each crate being hooked at floor level to an end of the rope. The rope is stretched tight and pulled over a pulley P that is attached to a rafter P Q = 4m above a point Q on the floor directly between the two crates. If crate A is 3m from Q and is being pulled directly away from Q at a rate of 0.5 m/s, how fast is crate B moving toward Q? Solution QUESTIONS AND THE ANSWER Page 42/1 — Question code: 1234 1. The Pythagorean theorem can give us AP + BP = 15 ⇔ p p x2 + 42 + y 2 + 42 = 15 2. Differentiating both sides of the equation, we receive dx y dy dy x x √ · +p · = 0 ⇒ y ′ (t) = = −√ · x′ (t) · 2 2 2 dt x + 9 dt x +9 y + 9 dt p y2 + 9 y 3. When x = 3 ⇒ y ≈ 9.1652 and x′ (t) = 0.5, then √ 3 84 + 9 y (t) = − √ × 0.5 × ≈ −0.3720 (m/s) 2 9.1652 3 +9 ′ □ Question 105. [L.O.2] A major network is televising the launching of the rocket. At a distance of x = 12000 feet from the launch site, a spectator is observing a rocket being launched vertically. A camera tracking the liftoff of the rocket is located at point A, where ϕ denotes the angle of elevation of the camera at A. When the rocket is z = 12434 ft from the camera and this distance is increasing at the rate of 480 ft/sec, how fast is ϕ changing? Solution 1. We have cos ϕ = x 12000 12000 = ⇒ ϕ = arccos z z z 2. Differentiating both sides of the equation, we receive ′ 1 ⇒ ϕ (t) = − q 1− 120002 z2 12000 × − 2 .z ′ (t) z 3. When z(t0 ) = 12434; z ′ (t0 ) = 480, at the moment t0 , we have 1 12000 ′ × − × 480 ≈ 0.1423(rad/s) ϕ (t0 ) = − q 2 124342 1 − 12000 2 12434 □ 2 Applications of integration Question 106 (L.O.2). Let V1 and V2 be the volumes of the solids that result when the plane region 4 1 1 bounded by y = , y = 0, x = , and x = c, where c > , is revolved about the x−axis and the y−axis, x 5 5 respectively. Find the value of c for which V1 = V2 . Solution QUESTIONS AND THE ANSWER Page 43/1 — Question code: 1234 Z c 16dx 16π = 80π − 2 x c 1/5 Z c 4dx x· 2. We also have V2 = 2π = 2π × 4(c − 1/5) x 1/5 1. We have V1 = π 3. V1 = V2 ⇒ c = 10.0 □ Question 107 (L.O.2). Find the area of the surface obtained by rotating the arc y = 2x3/2 , (0 ⩽ x ⩽ 4), about the x−axis. Solution 3 1. We have y ′ = 2 × x1/2 . 2 2. Using the area of surface of revolution formula, we get Zb |f (x)| S = 2π r Z4 h i2 36 3/2 1 + f ′ (x) dx = 2π 2x × 1 + xdx ≈ 820.7836 4 r a 0 □ 3 Ordinary differential equations (ODE) 3.1 The first order differential equations Question 108 (L.O.2). A tank initially contains 80 grams of salt dissolved in 100 liters of water. Pure water flows into the tank at the rate of 12 liters per minute, and the well stirred mixture flows out of the tank at the same rate. How much salt does the tank contain after 2 minutes? Solution 1. Step 1. The rate of change of salt in the tank, dQ is equal to the rate at which salt is flowing in minus dt the rate at which it is flowing out. dQ = rate in − rate out dt 2. Step 2. The rate at which salt enters the tank is the concentration 0 gr/l times the flow rate 12 l/min or 0 gr/min. 3. Step 3. To find the rate at which salt leaves the tank, we need to multiply the concentration of salt in the tank by the rate of outflow, 12 l/min. The volume of water in the tank remains constant at 100 liters, and since the mixture is well-stirred, the concentration throughout the tank is the same, namely, Q 12Q gr/l. Therefore, the rate at which salt leaves the tank is gr/min. 100 100 QUESTIONS AND THE ANSWER Page 44/1 — Question code: 1234 4. Step 4. Solve the differential equation using separable method Z Z dQ 12Q dQ 12 dQ 12 =0− ⇒ =− dt ⇒ = − dt dt 100 Q 100 Q 100 ⇒ ln |Q| = − 12 t + ln C ⇒ Q = Ce−12t/100 100 5. Step 5. The initial condition is Q(0) = 80, so 80 = Ce0 = C. Thus Q = 80e−12t/100 ⇒ Q(2) = 80e−12×2/100 = 62.9302. □ Question 109 (L.O.2). A tank with a capacity of 130 liters initially contains 80 grams of salt dissolved in 100 liters of water. Pure water flows into the tank at the rate of 12 liters per minute, and the well stirred mixture flows out of the tank at the rate of 2 liters per minute. How much salt does the tank contain when the tank is full? Solution 1. Step 1. The rate of change of salt in the tank, dQ is equal to the rate at which salt is flowing in minus dt the rate at which it is flowing out. dQ = rate in − rate out dt 2. Step 2. The rate at which salt enters the tank is the concentration 0 gr/l times the flow rate 12 l/min or 0 gr/min. 3. Step 3. To find the rate at which salt leaves the tank, we need to multiply the concentration of salt in the tank by the rate of outflow, 2 l/min. The volume of water in the tank after t minutes is 100 + (12 − 2)t = 100+10t liters, and since the mixture is well-stirred, the concentration throughout the tank is the Q 2Q same, namely, gr/l. Therefore, the rate at which salt leaves the tank is gr/min. 100 + 10t 100 + 10t 4. Step 4. Solve the differential equation using separable method Z Z dQ 2Q dQ 2 dQ 2 =0− ⇒ =− dt ⇒ = − dt dt 100 + 10t Q 100 + 10t Q 100 + 10t ⇒ ln |Q| = − 2 2 ln |100 + 10t| + ln C ⇒ Q = C(100 + 10t)− 10 10 2 2 5. Step 5. The initial condition is Q(0) = 80, so 80 = C × 100− 10 ⇒ C = 80 × 100 10 . Thus 2 2 Q = 80 × 100 10 (100 + 10t)− 10 . The tank is full after 130 − 100 = 3 minutes. 12 − 2 2 2 ⇒ Q(3) = 80 × 100 10 (100 + 10 × 3)− 10 = 75.9104. □ QUESTIONS AND THE ANSWER Page 45/1 — Question code: 1234 Question 110 (L.O.2). Recall that a simple electric circuit with a resistor of R ohms, an inductor of L henries and a battery with a voltage of E volts will produce a current I(t) amperes (t is measured in seconds) satisfying the following differential equation LI ′ (t) + RI = E. Let study one such circuit with R = 2 ohms, L = 3 henries and E = 12 volts. Given initial condition I(0) = 3 amperes. Evaluate current’s value at t = 3 seconds. Solution 1. Step 1. Let consider the differential equation LI ′ (t) + RI = E in our case, we have 2 12 3I ′ (t) + 2I = 12 ⇔ I ′ (t) + I = 3 3 2. Step 2. Solving this linear differential equation with p(t) = Z − I(t) = e 2 12 and q(t) = , we obtain 3 3 Z p(t)dt hZ p(t)dt q(t)e i dt + C . Therefore, Z Z 2 2 Z h i − dt dt 12 3 3 I(t) = e e dt + C 3 Z h i 2 12 2 t I(t) = e− 3 t e 3 dt + C 3 − 23 t 12 23 t I(t) = e e +C 2 2 12 12 3. Step 3. The initial condition is I(0) = + C = 3, so C = 3 − . Thus I(t) = e− 3 t 2 2 2 12 2 ×3 ⇒ I(3) = e− 3 ×3 e 3 + C = 5.594. 2 12 2 t e3 + C 2 □ 3.2 The second order differential equations Question 111 (L.O.2). Solve the differential equation y ′′ − 8y ′ + 12y = 120e12x Solution 1. Step 1. Solve the homogeneous equation y ′′ − 8y ′ + 12y = 0. The characteristic equation k 2 − 8k + 12 = 0 has 2 real different roots k1 = 6, k2 = 2 QUESTIONS AND THE ANSWER Page 46/1 — Question code: 1234 2. Step 2. The homogeneous solution is yh = C1 e6x + C2 e2x 3. Step 3. Find a particular solution of nonhomogeneous equation y ′′ − 8y ′ + 12y = 120e12x . The particular solution has a form yp = xs .e12x .A. Since α = 12 is not the root of characteristic equation then s = 0 and yp = A.e12x . 12× yp = Ae12x −8× yp′ = 12Ae12x 1× yp′′ = 144Ae12x yp′′ − 8yp′ + 12yp = 60Ae12x = 120e12x ⇒A=2 4. Step 4. The general solution is ygen = yh + yp = C1 e6x + C2 e2x + 2e12x . □ Question 112 (L.O.2). Solve the differential equation y ′′ − 14y ′ + 74y = 10e7x cos(5x) Solution 1. Step 1. Solve the homogeneous equation y ′′ − 14y ′ + 74y = 0. The characteristic equation k 2 − 14k + 74 = 0 has 2 complex conjugate roots k1 = 7 + 5i, k2 = 7 − 5i 2. Step 2. The homogeneous solution is h i yh = e7x C1 cos(5x) + C2 sin(5x) 3. Step 3. Find a particular solution of nonhomogeneous equation y ′′ − 14y ′ + 74y = 10e7x cos(5x). h i The particular solution has a form yp = xs .e7x . A cos(5x) + B sin(5x) . Since 7 + 5i is one of two h i complex roots of the characteristic equation, so s = 1 and yp = x.e7x . A cos(5x) + B sin(5x) . Differentiating yp we obtain h i yp′ = e7x . (7A + 5B)x cos(5x) + (7B − 5A)x sin(5x) + A cos(5x) + B sin(5x) h i yp′′ = e7x . (24A + 70B)x cos(5x) + (24B − 70A)x sin(5x) + (14A + 10B) cos(5x) + (14B − 10A) sin(5x) h i y ′′ − 14y ′ + 74y = e7x . 10B cos(5x) − 10A sin(5x) = 10e7x cos(5x) ⇒ A = 0; B = 1 QUESTIONS AND THE ANSWER Page 47/1 — Question code: 1234 4. Step 4. The general solution is h i ygen = yh + yp = e7x C1 cos(5x) + C2 sin(5x) + x.e7x . sin(5x). □ Question 113 (L.O.2). Let yp be the particular solution of the differential equation y ′′ − 4y ′ − 12y = −144x − 72. Evaluate yp (2). Solution 1. Step 1. Solve the homogeneous equation y ′′ − 4y ′ − 12y = 0. The characteristic equation k 2 − 4k − 12 = 0 has 2 real different roots k1 = 6, k2 = −2 2. Step 2. Find a particular solution of nonhomogeneous equation y ′′ − 4y ′ − 12y = −144x − 72. The particular solution has a form yp = xs .e0x (Ax + B). Since α = 0 is not the root of characteristic equation then s = 0 and yp = Ax + B. −12× yp = Ax + B −4× yp′ = A 1× y”p = 0 y”p − 4yp′ − 12yp = −12Ax − (12B + 4A) = −144x − 72 A = 12 −12A = −144 ⇒ ⇔ −12B − 4A = −72 B=2 3. Step 3. The particular solution is yp = 12x + 2 ⇒ yp (2) = 12 × 2 + 2 = 26. □ Question 114 (L.O.2). Recall that a simple electric circuit with a resistor of R ohms, an inductor of L henries, a capacitor of C farads and a battery with a voltage E(t) = 100 sin(ωt) (volts) will produce a current I(t) amperes (t is measured in seconds) satisfying the following differential equation LI ′′ (t) + 1 RI ′ + I = E ′ (t) = 100ω cos(ωt). Let study one such circuit with R = 8 ohms, L = 5 henries, C = 0.2 C farads and ω = 11 rad/seconds. Given initial condition I(0) = 0, I ′ (0) = 0, evaluate current’s value at t = 6.2 seconds. Solution QUESTIONS AND THE ANSWER Page 48/1 — Question code: 1234 1. Step 1. Solve the homogeneous equation LI ′′ (t) + RI ′ + 1 I = 0. C 1 = 0 ⇒ 5k 2 + 8k + 5 = 0 has 2 complex conjugate roots C 4 3 4 3 k1 = − + i; k2 = − − i 5 5 5 5 The characteristic equation Lk 2 + Rk + 2. Step 2. The homogeneous solution is Ih = e − 54 t C1 cos 3 3 t + C2 sin t 5 5 3. Step 3. Find a particular solution of nonhomogeneous equation LI ′′ (t) + RI ′ + 1 I = 100ω cos(ωt). C h i The particular solution has a form Ip = ts .e0t A cos(ωt) + B sin(ωt) . Since 0 + ωi is not the root of characteristic equation then s = 0 and Ip = A cos(ωt) + B sin(ωt). 5× Ip = A cos(ωt) + B sin(ωt) 8× Ip′ = −Aω sin(ωt) + Bω cos(ωt) 5× I”p = −Aω 2 cos(ωt) − Bω 2 sin(ωt) 5I”p + 8Ip′ + 5Ip = 100ω cos(ωt) ω(5 − 5ω 2 ) × 100 (5 − 5ω 2 )A + 8ωB = 100ω A= (5 − 5ω 2 )2 + (8ω)2 ⇒ ⇔ 8ω 2 × 100 −8ωA + (5 − 5ω 2 )B = 0 B= (5 − 5ω 2 )2 + (8ω)2 4. Step 4. The general solution is 3 3 Igen = Ih + Ip = e t + C2 sin t + A cos(ωt) + B sin(ωt). C1 cos 5 5 4 4 3 3 4 3 3 ′ We have Igen = e− 5 t − C1 + C2 cos t + − C2 − C1 sin t −Aω sin(ωt)+Bω cos(ωt) 5 5 5 5 5 5 From the initial condition I(0) = 0, I ′ (0) = 0, we obtain C1 = −A C2 = − 5 Bω − 4A 3 3 3 3 − 54 ×6.2 C1 cos × 6.2 + C2 sin × 6.2 +A cos(ω×6.2)+B sin(ω× 5. Step 5. Evaluate Igen (6.2) = e 5 5 6.2) ≈ −1.3042. − 45 t □ Question 115 (L.O.2). Recall that a spring with mass m, spring constant k relaxing after being stretched will be in motion, where its position x(t) satisfies the following differential equation mx′′ (t) + kx = 0. Let study one spring with m = 2 and k = 32. Given initial condition x(0) = 6, x′ (0) = 4, evaluate numerically its position at t = 2. Solution QUESTIONS AND THE ANSWER Page 49/1 — Question code: 1234 1. Step 1. Solve the homogeneous equation mx′′ (t) + kx = 0. The characteristic equation mr2 + k = 0 has 2 complex conjugate roots r r r r k 32 k 32 r1 = i= i; r2 = − i=− i m 2 m 2 2. Step 2. The homogeneous solution is r x = C1 cos ! ! r 32 32 t + C2 sin t 2 2 ! ! r r 32 32 32 t + C2 cos t . From the initial condition, we have 3. Step 3. x = −C1 2 2 2 ! ! r r 32 32 × 0 + C2 sin × 0 = C1 = 6 x(0) = C1 cos 2 2 ! ! r r r r r 32 32 32 32 32 ′ × 0 + C2 cos × 0 = C2 =4 x (0) = −C1 2 sin 2 2 2 2 C1 = 6 r ⇔ 2 C2 = 4 × 32 ! ! r r 32 32 4. Step 4. Evaluate x(2) = C1 cos × 2 + C2 sin × 2 ≈ 0.1164. 2 2 r ′ 32 sin 2 r □ 3.3 System of the first order ODEs Question 116 (L.O.2). Consider the system of ODEs: x′ (t) = 4x − y − t y ′ (t) = 5x + 10y + (t − 1) Find a solution of the system such that x (0) = 3, y (0) = 1. Solution 1. Step 1. The first equation gives y in terms of x and x′ y= x′ − 4x + t −1 Differentiating this will give us y ′ in terms of x′ and x′′ y′ = x′′ − 4x′ + 1 −1 and we can now substitute these into the second equation, we have x′′ − 4x′ + 1 x′ − 4x + t = 5x + 10 × + (t − 1) −1 −1 ⇔ x′′ − 14x′ + 45x = 9t QUESTIONS AND THE ANSWER Page 50/1 — Question code: 1234 2. Step 2. Solve the homogeneous equation x′′ − 14x′ + 45x = 0. The characteristic equation k 2 − 14k + 45 = 0 has 2 real different roots k1 = 5, k2 = 9 3. Step 3. The homogeneous solution is xtn = C1 e5t + C2 e9t 4. Step 4. Find a particular solution of nonhomogeneous equation x′′ − 14x′ + 45x = 9t. The particular solution has a form xr = ts .e0t .(At + B). Since α = 0 is NOT a root of the characteristic equation, so s = 0 and xr = At + B. 45× xr = At + B −14× x′r = A 1× x′′r = 0 ′ = −14A + 45(At + B) = 9t ′′ x − 14x + 45x 1 14 ⇒ A = ;B = 5 225 5. Step 5. The general solution is xtq = xtn + xr = C1 e5t + C2 e9t + 1 14 t+ 5 225 1 ⇒ x′tq = 5C1 e5t + 9C2 e9t + . 5 Then x′tq − 4xtq + t t 11 = −C1 e5t − 5C2 e9t − + −1 5 225 14 11 Since x(0) = 3 ⇒ C1 + C2 + = 3 và y(0) = 1 ⇒ −C1 − 5C2 + = 1. Solve the system 225 225 14 C1 + C2 + =3 391 35 225 ⇒ C1 = ; C2 = − . 11 100 36 −C1 − 5C2 + =1 225 ytq = Therefore, the solution which satisfies the conditions of the system is t 35e9t 391e5t 14 x= − + + 5 369t 1005t 225 y = − t + 175e − 391e + 11 5 36 100 225 □ QUESTIONS AND THE ANSWER Page 51/1 — Question code: 1234