MATH 3 FALL 0 FINAL EXAM - PRACTICE EXAM SOLUTIONS () You cut a slice from a circular pizza (centered at the origin) with radius 6 along radii at angles 4 and 3 with the positive horizontal axis. (a) (3 points) What is the area of your slice? The interior angle of the slice is So the area of the slice is 3 4 = 4 3 =. θr = 6 = 36 4 = 3. (b) (3 points) What is the arc length of the outer portion of crust on your slice? rθ = 6 =. () (6 points) Find all values of θ in the interval [0, ] satisfying sin (θ) + cos(θ) =. Rewrite sin (θ) = cos (θ). Then, noticing that we have a quadratic in cos(θ), write x = cos(θ). ( cos (θ)) + cos(θ) = cos (θ) + cos(θ) = 0 x + x = 0 x(x ) = 0 So x = 0 or, and so cos(θ) = 0 or. For each of these two cosine values, there are two corresponding points on the unit circle. Since we only want angle measures between 0 and, each of these points corresponds to only one angle, so there should be four solutions. They are θ = 3,, 3, 5 3.
MATH 3 FALL 0 FINAL EXAM - PRACTICE EXAM SOLUTIONS (3) (3 points) Find an equation for the line perpendicular to y = 3x + 7 through the point (8, 6). This line should have slope 3. In point-slope form, y 6 = 3(x 8). This is an acceptable solution. You may also rewrite the answer as y = 3x + 50. (4) (3 points) What is 0 radians in degrees? 0 80 = 80 0 = 8. (5) In the triangle below, let A = 6, B = 8, and a = 5. (a) (3 points) Find sin(b). This is just asking us to find sin 8, which we can do using the half-angle formula. sin 8 = cos 4 = = =. 4 (b) (3 points) Find b.
MATH 3 FALL 0 FINAL EXAM - PRACTICE EXAM SOLUTIONS 3 By the law of sines, sin A a (6) Consider the rational function sin 6 5 = sin B b = sin 8 b 5 = sin 8 b 0 b = sin 8 b = 0 b = 5. f(x) = x 7x + 3x (a) (3 points) Does f have a horizontal asymptote? If so, what is it? Yes, y = 3. (b) (6 points) Solve the inequality f(x) 0. Factoring the numerator, we have (x 3)(x 4) f(x) = 3x. f could switch sign at x = 0, x = 3, or x = 4. We ll do some sign analysis. (, 0) (0, 3) (3, 4) (4, ) (x 3) + + (x 4) + 3x + + + + f(x) + + + So f(x) < 0 when 3 < x < 4, and f(x) = 0 when x = 3 or x = 4, so the solutions are 3 x 4. (7) (6 points) Simplify the following expression: e ln(x+3) ln(x+). e ln(x+3) ln(x+) = e ln((x+3) ) e ln((x+) ) = x + 3 = x + 3 (x + ) (x + )
4 MATH 3 FALL 0 FINAL EXAM - PRACTICE EXAM SOLUTIONS (8) (6 points) Show that for all θ, sin(3θ) = 3 sin(θ) 4 sin 3 (θ). We ll apply the angle sum formula and the double angle formulas. sin(3θ) = sin(θ + θ) = sin(θ) cos(θ) + cos(θ) sin(θ) = ( sin(θ) cos(θ)) cos(θ) + ( sin (θ)) sin(θ) = sin(θ) cos (θ) + sin(θ) sin 3 (θ) = sin(θ)( sin (θ)) + sin(θ) sin 3 (θ) = sin(θ) + sin(θ) sin 3 (θ) sin 3 (θ) = 3 sin(θ) 4 sin 3 (θ) Note that since the end goal was an expression just involving sin, we chose the cosine double angle formula involving sine: cos(θ) = sin (θ). We also used the Pythagorean Identity in the form cos (θ) = sin (θ) to transform the remaining cosine terms to sines. (9) (a) (3 points) Find an equation for a circle with center (, 3) and radius 5. (x ) + (y + 3) = 5. (b) (3 points) What is the circumference of this circle? r = 5 = 0. (c) (3 points) What is its area? r = 5 = 5. (0) Consider the function f(x) = cos(x) +. (a) (6 points) Sketch a graph of this function. Clearly label the y-intercept and several x-intercepts.
MATH 3 FALL 0 FINAL EXAM - PRACTICE EXAM SOLUTIONS 5 (b) (3 points) What is the amplitude of this function? The amplitude is. (c) (3 points) What is the period of this function? The period is =. () (6 points) Sketch a graph of y = x + x +. Hint: Write this as a piecewise function with three cases. (x ) + (x + ), if x < y = (x ) + (x + ) if x < (x ) + (x + ) if < x x, if x < = if x < x if < x () You put $50 in a bank account with 8% interest compounded 4 times per year. (a) (3 points) Write down an expression for the amount of money you will have after t years. ( A = 50 +.08 ) 4t 4 (b) (3 points) After how many years will you have $80? We ll solve the following for t:
6 MATH 3 FALL 0 FINAL EXAM - PRACTICE EXAM SOLUTIONS ( 80 = 50 +.08 4 80 50 = (.0)4t ( ) 8 log.0 = 4t 5 t = 4 log.0(.6) ) 4t (3) Evaluate the following: (a) (3 points) cos(cos (.8)).8 (b) (3 points) sin (sin( 3 6 )) 3 6 is in the second quadrant, so it is not a possible output of sin. The angle in the first quadrant with the same sine value is 3 6 = 3 6. (c) (3 points) cos(tan ( 7 5 )) Draw a right triangle and label one angle θ = tan ( 7 5 ). Label the opposite side 7 and the adjacent side 5. Then the hypoteneuse c satisfies c = 5 + 7, so c = 74. Then cos(tan ( 7 5 5 )) = cos(θ) = 74. (4) (3 points) Find log 6 (3). Change of base formula: log 6 (3) = log (3) log (6) = 5 4.