MATH 32 FALL 2013 FINAL EXAM SOLUTIONS. 1 cos( 2. is in the first quadrant, so its sine is positive. Finally, csc( π 8 ) = 2 2.

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1 MATH FALL 01 FINAL EXAM SOLUTIONS (1) (1 points) Evalute the following (a) tan(0) Solution: tan(0) = 0. (b) csc( π 8 ) Solution: csc( π 8 ) = 1 sin( π 8 ) To find sin( π 8 ), we ll use the half angle formula: π 8 = 1 ( π 4 ). sin( π 1 cos( π 8 ) = ± 4 ) 1 = ± = ± 4 = ± Now π 8 is in the first quadrant, so its sine is positive. Finally, csc( π 8 ) =. (c) ln((e )(e )) Solution: ln((e )(e )) = ln(e + ) = 5. (d) log (sin( π 4 )) log (cos( π )) Solution: log (sin( π 4 )) log (cos( π )) = log ( ) log ( 1 ) = log ( ) log () log ( 1 ) = 1 1 ( 1) = 1 1

2 MATH FALL 01 FINAL EXAM SOLUTIONS () In this problem, consider the following functions: g(x) = ln(x ) h(x) = 1 ln(x) (a) ( points) What is the domain of g? Solution: The domain of ln is (0, ). So for x to be in the domain of g, we must have x > 0. In fact, x 0 for all x, with x = 0 only when x = 0. So the only problem occurs when x = 0. The domain of g is all real numbers except 0, or, in interval notation, (, 0) (0, ). (b) ( points) What is the domain of h? Solution: For x to be in the domain of h, we must have x > 0 (so that ln(x) is defined) and 1 ln(x) 0 (so that 1 ln(x) is defined). Solving for x in the second inequality, 0 1 ln(x) ln(x) 1 e ln(x) e 1 x e Putting these conditions together, we find that the domain of h is (0, e]. (c) (6 points) Find a formula for the function g (h 1 ) and simplify. Solution: First we find a formula for h 1. Letting y = 1 ln(x) and solving for x, we have y = 1 ln(x) y 1 = ln(x) ln(x) = 1 y x = e 1 y So h 1 (x) = e 1 x. Now g(h 1 (x)) = g(e 1 x ) = ln((e 1 x ) ) = ln(e (1 x) ) = x. () (9 points) A triangle has a side of length and a side of length. All three angles of the triangle are acute. The area of the triangle is. How long is the third side of the triangle? Hint: First find the measure of one of the angles of the triangle. Partial credit if you draw a good picture and label this angle correctly. Solution: We will use the area formula Area = 1 ab sin(c) to find the measure of the angle C between the side a of length and the side b of length.

3 MATH FALL 01 FINAL EXAM SOLUTIONS = 1 ()() sin(c) = sin(c) sin(c) = Since C is acute, C is the angle in the first quadrant with sine, that is, C = π. Now we can use the law of cosines to find the length of the third side, c. So c = 7. (4) Consider the rational function a + b ab cos(c) = c + ()() cos( π ) = c f(x) = ( 1 ) = c 7 = c (x + 1)(x ) 4(x 1). (a) (6 points) Does f have any horizontal or vertical asympototes? If so, what are they? Solution: The numerator and denominator are of equal degree. The leading term of the numerator (after expanding) is 6x, and the leading term of the denominator (after expanding) is 4x. The ratio of these two terms is 6 4 =, so f has a horizontal asympotote at y =. The vertical asymptotes of f are at the zeros of the denominator. The denominator has one zero, at x = 1, so f has one vertical asympotote, at x = 1. (b) ( points) Does f have any zeros? If so, what are they? Solution: The zeros of f are the zeros of the numerator. (x + 1)(x ) = 0 when x + 1 = 0 or x = 0, that is, when x = 1 or x =.

4 4 MATH FALL 01 FINAL EXAM SOLUTIONS (5) (6 points) Find the length of the highlighted portion of the circle of radius 4 pictured below. Solution: First, let s convert 15 to radians. 15 = = π + π 4 = π 4. Now the arc length of a portion of the circle with interior angle θ is rθ, where r is the radius of the circle. So the length of the highlighted curve is 4( π 4 ) = π. (6) (6 points) Is f(x) = sin(x) an even function, an odd function, or neither? Explain. Solution: It is an even function. f( x) = sin( x) = sin(x) = sin(x) = f(x). (7) (6 points) Find a solution to the equation sin (θ) = sin(θ) + 1 in the interval [π, π ]. Solution: This equation is a quadratic in sin(θ). Let x = sin(θ). We have x = x + 1 x x 1 = 0 (x 1 x 1 ) = 0 (x 1)(x + 1 ) = 0 So x = 1 or x = 1. Now we want an angle θ such that sin(θ) = 1 or sin(θ) = 1. There are many such angles, but in particular we want a solution in the interval [π, π ]. This is the third quadrant, and no angle in the third quadrant has sin(θ) = 1. The unique angle in the third quadrant with sin(θ) = 1 is θ = 7π 6. (8) (6 points) Give an example of a periodic function with period π and amplitude 1. Write down a formula for this function and sketch a graph. Solution: There are many examples. Here s one: f(x) = 1 cos(4x)

5 MATH FALL 01 FINAL EXAM SOLUTIONS 5 (9) A sequence is defined recursively by a 1 = 1 a n+1 = a n 1, for n 1 (a) ( points) List the first 4 terms of the sequence. Solution: 1,, 5, 14 (b) ( points) Is this sequence arithmetic, geometric, or neither? Explain. Solution: Neither. In an arithmetic sequence, each term is obtained from the previous by adding a constant value. But 1 and differ by 1, and 5 differ by, 5 and 14 differ by 9, etc. In a geometric sequence, each term is obtained from the previous by multiplying by a constant value. But the ratio of 1 and is, the ratio of and 5 is 5, etc. (10) (a) ( points) Evaluate ( ( )) 1 cos sin 1. Solution: Let s draw a triangle. Here A = sin 1 ( 1 opposite ). Note that A fits into this triangle because sin(a) = adjacent = 1. By the Pythagorean Theorem, we have 1 + b =, so b = 9 1 = 8, and b = 8. So cos(sin 1 ( 1 )) = cos(a) = 8 =.

6 6 MATH FALL 01 FINAL EXAM SOLUTIONS (b) ( points) Evaluate ( ( )) 1 cos sin 1. Solution: We ll use the double angle formula for cosine (choosing the one that depends only on cosine, since we know the value of cosine from part (a)). cos( sin 1 ( 1 )) = (cos(sin 1 ( 1 ))) 1 8 = ( ) 1 = = 7 9 (11) (6 points) An anonymous benefactor (Mr. X) deposited a secret amount of money into a bank account for you on the day of your birth. This account accrues interest at a rate of 4%, compounded quarterly (4 times per year). On your 5th birthday, you gain access to the account and discover that it contains exactly one million dollars! How much was Mr. X s original deposit? (No need to give a decimal approximation - just give an expression which evaluates to the correct answer.) Solution: The formula for compound interest is A(t) = P (1 + r n )nt. Here we have t = 5 years, A(t) = $ (the amount you have after 5 years), r =.04 (the interest rate), and n = 4 (the number of times per year the interest is compounded). We want to solve for P, the principal, or initial investment = P ( )4(5) = P (1.01) 100 P = If you re curious, this works out to P $ (1) (6 points) Use polynomial long division to divide x 4 6x + x + by x + x + 1. Label the quotient and the remainder. Solution:

7 MATH FALL 01 FINAL EXAM SOLUTIONS 7 ) x x 6 x +x +1 x 4 6x +x + (x 4 +x +x ) x 9x +x + ( x x x) 6x +4x + ( 6x 6x 6) 10x +9 The quotient is x x 6 and the remainder is 10x + 9.