01:640:103 Lecture 20: 10464-Topics_In_Mathematics-2019-04-18

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sangrialouse846

29 Apr 2019

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[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) >

Out of these, which is the false statement?

a) f(x) is concave-up on at least two disjoint open intervals

b) f'''(x) is defined everywhere

c) f''(x) is defined everywhere

d) f'(x) is defined everywhere

e) f''(x) < 0

f) f(x) is concave-up on at least two disjoint closed intervals

g) f(VI) = 0

h) There are at least 2 critical numbers

i) f'(0) < 0

j) There is at least 1 inflection point

01:640:103 Lecture 20: 10464-Topics_In_Mathematics-2019-04-18

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