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7. (a) -(A= B) is logically equivalent to (-A) = (-B). (b) -( V ES, P(x)) is logically equivalent to Va e S.-P()). (c) If a, b, c, d e Z, ab, bc, and cd, then ad. (d) If a, b, c e Z, and a be, then a b or a c. ( 1 cos 1 + i sin ) 1- Δ― 2 +2 (f) A polar form of the complex number z = -18i is 18(cos( + isin ). (g) god (29931005101 29831005102) = 29831005101

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Elin Hessel
Elin HesselLv2
17 Jun 2018
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5. (a) The negation of A B is AV-B). (b) The polynomial r3 - 3xΒ² + x + 2 has a rational root. (c) 75 is a valid n in the RSA encryption scheme. (d) If f(z) is a polynomial with complex coefficients and z = i is a root of f(z), then z =-i must also be a root of f(z). (e) A B is logically equivalent to (-B) = (-A). (f) 1 + i is an eighth root of unity.

[1] 1. (a) Solve 2 – 31 < 7. Solution: We have – 7<-3 < 7 which gives -4 0 on (-0, 1] U [2,). Next, we need to see where Vr2 – 3r +2 -1 > 0. If r2 - 3.c + 2 – 1 > 0, then V.22 – 3.0 +2 > . Observe that this is definitely true if r < 0. If r > 0, then squar- ing both sides we get 22 - 3.+2 > 1 2 > 3. V A VICO Thus, the domain of f is (-00, ). [2] (c) Find the exact value of tan(sec-14). Solution: Let y = sec-14. Then, sec y = 4. From the diagram we get that tan(sec-? 4) = tan y = 15 [2] (d) Find all r such that sin(2.x) = COS I. Solution: We have 2 sin r cos r = COS I, SO 0 = 2 sin Cos I -Cos I = cos .r (2 sin c - 1). Hence, sin(2x) = cost when cos x = 0, or when sin r = Thus, r= +ik, kez, I= +2nk, ke Z, or r= +2nk, kez [2] (e) State the precise mathematical definition of lim f(x) = 0. Solution: For every e > 0 there exists a 8 >0 such that if 0 < x-al <, then f(c) >

Problem 4 Circle your answer. a) (6 points) If f(x) = 2x + 3x2 - 361 + 5, then the tangent line to the graph y= f(x) is horizontal when i) 1 = -2,3 ii) I = 2,-3 iii) r = 0 iv) not listed b) [6 points] Suppose Given h(x) = f(g(T)), where g(2) = 3, g'(2) = 7, f'(2) = -1 and 1'(3) = -2. Then h'(2) is equal to i) - 1 ii) -2 iii) - 14 iv) not listed (x+3)(1 - 1)3 c) (6 points) The function f(1) == (x2 - 2x + 1)(x + 3) wa has vertical asymptotes at i) x = -3,0 ii) = -3,0,1 iii) x = 0,1 iv) not listed d) (6 points] The solution to the initial value problem v'(x) = 473 – cos(x) +e", y(0) = 3 is i) y(x) = x + sin(x) + e* + 2 ii) y(I) = x4 – sin(x) +et+2 iii) y(x) = 2* – sin(1) + et + 3 iv) not listed

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