MATH 1301, Solutions to practice problems

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1 MATH 1301, Solutions to practice problems 1. (a) (C) and (D); x = 7. In 3 years, Ann is x + 3 years old and years ago, when was x years old. We get the equation x + 3 = (x ) which is (D); (C) is obtained by dividing by on both sides. (b) (A); x = 7. Five years ago, Ann was x 5 and a year ago, she was x 1. We get the equation x 5 = (x 1) 10 which is (A). (c) In 11 years, Ann is x + 11 and 3 years ago, she was x 3. The first is twice the second. (b) The area of the square is x and the area of the rectangle is 3(x + 1). The first is 3 units larger than the second. 3. (a) x = 4x 3 which gives the quadratic equation x 4x+3 = 0 which has solutions x = 1 and x = 3. (b) x = x + 3 which gives the quadratic equation x x 3 = 0 which has solutions x = 1 and x = 3. As x cannot be negative, the only possible solution is x = 1. (c) (x) = (x + 1) which gives 4x = x + x + 1 the quadratic equation 3x x 1 = 0 which has solutions x = 1/3 and x = 1. As x cannot be negative, the only possible solution is x = The solutions are given by x1± 1 a and the number of solutions therefore depends on the sign of 1 a. There are (a) solutions if a < 1 (b) 1 solution if a = 1 (c) no solutions if a > First consider Ann s strategy. Her fortune after time t is f(t) = at + b, and as she starts with fortune 5, we have f(0) = 5 which gives b = 5. Moreover, as her fortune doubles after a year, we have f(1) = 10 which gives f(1) = a + b = a + 5 = 10, that is, a = 5. Ann s linear function is f(t) = 5t + 5. Next, consider Bob s strategy. His fortune after time t is g(t) = at + bt + c, and as he starts with fortune, we have g(0) = which gives c =. Moreover, as his fortune doubles after a year, we have f(1) = 4, which gives

2 f(1) = a+b+c = a+b+ = 4, which gives a+b =. Finally, after years, his fortune is 10 and we have f() = 10, which gives f() = a = b +c = 4a + b + = 10, that is 4a + b = 8. Now note that 4a + b = (a + b) + a where we know that (a+b) = 4. Thus, we get 4+a = 8 which gives a =, and as a + b =, we get b = 0. Bob s quadratic function is g(t) = t +. Their are equal at the time t that has f(t) = g(t). This gives 5t + 5 = t + which is equivalent to the quadratic equation t 5t 3 = 0 which has solutions t = 1/ and t = 3. As t cannot be negative, the solution is t = 3 years. 6. In the twenties is the interval [0, 30) and the inequalities for F are which gives 68 F < F < (a) The inequality can be rewritten x 5x and the zeros of the LHS are x = and x = 3. A sign diagram shows that the inequality is satisfied when x 3, or in interval form [, 3] (three test values needed). (b) The single zero is x = 1 and a sign diagram shows that the inequality is satisfied when x 1 (two test values needed). (c) There are no zeros and only one test value is needed to show that the inequality is never satisfied; there are no solutions. 8. The inequality is x 5x + 6 and the zeros are x = 1 and x = 6. A sign diagram shows that the inequality is satisfied when x 1 or x 6. Because x cannot be negative, the solution is x 6. 9(a) 1. Move everything to the LHS so that RHS = 0: x 1 + x x 0. Make LHS into one fraction: x x 0

3 3. Find forbidden values (zeros of the denominator): x = 1 4. Find zeros (of the numerator): x = 1 5. Do a sign diagram with forbidden values, zeros and (test values): x ( ) 1 (0) 1 () LHS LHS 0? yes no no yes yes The inequality is satisfied when x < 1 or x 1 (the symbol denotes that LHS is undefined). (b) Rewrite the inequality on the equivalent form x 1 + x < 0 where the LHS has the forbidden value x = 1 and the zero x =. A sign diagram shows that the solution is 1 < x <. 10. With P 1 : ( 8, 5), P : (x, y ), M = ( 3, 3), the midpoint formula gives 3 = m x = x 1 + x = 8 + x which gives x = =. Similarly, we get y = 3 5 = 11. Thus, P has coordinates (, 1) and the taxi distance is td(p 1, P ) = x x 1 + y y 1 = ( 11) = = 6 11(a) Let M : (m x, m y ) be the midpoint between P 1 and P and let the coordinates of A be (q x, q y ). Note that Q is the midpoint between P 1 and M and apply the midpoint formula to P 1 and M to get q x = x 1 + m x q y = y 1 + m y

4 Because M is the midpoint of P 1 and P, the midpoint formula applied to these two points gives m x = x 1 + x m y = y 1 + y which substituted into the expressions for q x and q y gives q x = x 1 + x 1+x = 3x 1 + x 4 q y = y 1 + y 1+y = 3y 1 + y 4 (b) q x = 1 = (3 0 + x )/4 gives x = 4 and q y = 1 = y )/4 gives y = 11. Thus, P has coordinates (4, 11). 1(c) The given line has slope m 1 =, and any perpendicular line must have m = 1/m = (e) The equation for a parabola with vertex (0, 3) is ax + 3 and only (e) satisfies this equation with a = m = (y y 1 )/(x x 1 ) = ( ( 3))/( ( 1)) = (b) odd, (e) odd, (g) even, (h) even, (i) odd, (l) odd, (o) even. Note that (m) is not the graph of any function, and that the rest have sharp edges whereas graphs of polynomials are always smooth. 16(a) degree 3, leading term x 3, leading coefficient 1 (b) degree 7, leading term x 7, leading coefficient 17(a) degree, leading coefficient 1, f(x) as x ± (b) degree 1, leading coefficient 1, f(x) as x and f(x) as x (c) degree 8, leading coefficient, f(x) as x ± (d) degree 5, leading coefficient 1, f(x) as x and f(x) as x.

5 18. (a), (c), (e), (f), (h) 19. (a) and (c) (too many turning points and the degree in (c) is even). 0(a) f(x) = (x 1)(x 5x + 6) (b) f(x) = (x + 1)(x x + ) (c) f(x) = (x + 1)(x 1) 1(a) b 3ac = 3 3 ( ) = 1 > 0: two turning points (b) b 3ac = = 0: one inflexion point (c) b 3ac = = 3 < 0: no critical points. (a) Zero: x = 1, b 3ac = ( 4) = < 0 and there are no critical points. The symmetry point has x-coordinate b/3a = ( 4)/3 = 4/3 which is to the right of the zero x = 1. Graph:

6 (b) Zeros: x = 1, x =, two turning points. Graph: (c) Zeros: x =, x = 1, x = 4, two turning points. Graph: 3. Set y = f(x) to get the equation y = x 1 which has the solution x = 5 5y + 1 = f 1 (y). Replacing y by x gives the inverse f 1 (x) = 5x + 1, x R. 4. (c) and (f) are graphs of one-to-one functions. (a), (d), and (e) fail the horizontal line test and (b) is not the graph of any function (fails the vertical line test). 5. Any domain of the type x a where a 0 works, as does any domain of the type x a where a 0. The largest possible domains are either x 0 or x 0. 6(a) Not possible; a rational function cannot have two different horizontal

7 asymptotes. (b) Domain: (0, ) (or 0 < x < ); range: (, ) (or x R). (c) f 1 (x) has vertical asymptotes at x = 0 and x = and looks something like this: 7. Solve the equation y = f(x): y = x 1 x + 1 xy + y = x 1 xy x = 1 y x(y 1) = 1 y x = 1 y y 1 = 1 + y 1 y = f 1 (y) Replace y by x to get the inverse f 1 (x) = 1 + x. To find the domain of 1 x f 1, recall that this equals the range of f. As n = k = 1 and a = A = 1, f has a horizontal asymptote at y = 1. Also, f(0) = (0 1)/(0+1) = 1, and

8 the range of f is thus [ 1, 1). We get the inverse f 1 (x) = 1 + x, 1 x < 1. 1 x 8. As h(x) = 0 has no solutions, the degree of h(x) must be even. 9. The zeros of h(x) are 0 and ; the zeros of g(x) are 1 and 3. Sign diagram of the relevant values: x 1 (3/) (5/) 3 f(x) f(x)? We have f(x) as x and f(x) as x +. 30(a) n =, k = 1, no horizontal asymptote (b) n = 4, k = 5, horizontal asymptote at y = 0 (c) n = 11, k = 11, a = 6, A =, horizontal asymptote at y = To get exactly one crossing of the x-axis, g(x) must have exactly one zero and must therefore be of odd degree (but not every odd-degree polynomial will work of course). To get no vertical asymptotes, h(x) must have no zeros and must therefore be of even degree (again, not every even-degree polynomial will work). To get no horizontal asymptotes, the degree of g(x) must be higher than the degree of h(x). One example is f(x) = x3 x Solve the equation y = 3x 1 to get x = y + 1 = f 1 (y). As the range 3 of f(x) is R, the domain of f 1 is also R and we get the inverse function f 1 (x) = x + 1, x R Solve the equation y = x y to get x = x + 1 y = f 1 (y). As f(0) = 1 and f has a horizontal asymptote at y = (n = k = 1, a =, A = 1) the range of f is [ 1, 1) which is therefore the domain of f 1. Replace y be x to

9 get the inverse function f 1 (x) = 1 + x, 1 x <. x 34(b) Remember that the graphs of f and f 1 are symmetric around the line y = x. 35(a) q(0) = 1000 (b) q() = 1000e (c) Find t such that 1500 = q(t) = 1000e 0.1t which gives the equation e 0.1t ln 1.5 = 1.5. Take logarithms: 0.1t = ln1.5 which gives t = years. 36. ln x 3 ln = ln x 1 y 3 ( ) ( ) 1 1 ln y = ln(x ) ln ln(y ) y y 3 ln(y ) = ln(x y 3 ) ln(y ) ( x y 3 ) = ln = ln(x y) 37. ln ( x y z 3 y ) = ln(x y) ln(z 3 ) = ln(x )+lny ln(z 3 ) = lnx+ln y 3 lnz. 38. Because 3 = 9, we have log 3 9 = and get log 3 (3 log 3 9 ) = log 3 (3 ) =. 39(a) π (b) π/ (c) 4π (d) 3π 40. If θ is an acute angle (an angle between 0 and π/), a right triangle with sides opp = 6 and hyp = 10 has sin θ6/10 = 0.6. The remaining side is adj = 10 6 = 64 = 8 (by Pythagoras Theorem) and we get cosθ = adj/hyp = 8/10 = 0.8. There are however many angles with the same sin-value. In general, the identity sin θ + cos θ = 1 solved for cosθ gives cos θ = 1 sin θ which gives cosθ = ± 1 sin θ which in our case equals ± = ± 0.64 = ±0.8. Thus, an angle with sin θ = 0.6 either has cos θ = 0.8 or cos θ = A right triangle with adj = 6 and hyp = 10 has cosθ = 0.6. The remaining side is opp = 8 (as above) and we get tanθ = 8/6 = 4/3.

10 As above, there are many possible angles that have cosθ = 0.6. As tanθ = sin θ/ cosθ and sin θ = ± 1 cos θ = ±0.8 we get tan θ = ±0.8/0.6 = ±4/3. Thus, an angle with cosθ = 0.6 either has tan θ = 4/3 or tan θ = 4/3. 4. The x-coordinate is x = 3 and the y-coordinate is y = 4. The distance from the origin to the point is z = = 5 and we get sin θ = y/z = 0.8, cosθ = x/z = 0.6 and tan θ = y/x = 4/ (d) is the graph of f(x) = 0.5+cosx, the graph of cos x shifted 0.5 units up. The other graphs are of (a) f(x) = 0.5+sin x (b) f(x) = 0.5+cos x (c) f(x) = sin x. 44. Draw the 6-by-6 diagram with 36 dots, mark those that are in the respective events, count them and divide the number by 36. (a) 11/36 (b) 5/36 (c) 13/36 (d) 10/36. For example, the event in (b) (sum equal to 8) is shown below ( ) = Probability to win = 1 and expected gain 37 1 EV = a.7 cent loss for each dollar wagered. + ( 1) =

11 47. The probability to win is 5/38. Denote the payout by x to get the expected gain x 33 EV = x + ( 1) = To find x, solve the equation EV = 38 : 5x 33 = which gives 5x 33 = which has the solution x = 31/5 = 6. so the payout should be 6.. (b) The expected gain is 5 EV = ( 1)33 38 = 3 38 = 7.9 a 7.9 cent loss for each dollar wagered.