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18 Jan 2018

14. The solution to the separable differential equation 222, subject to y(-2)=1, y> 0 is ) = /431 c) v= f(x+3* +4 ) >= 3=42! D) Y= {+3+2 E) none of these

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Sixta Kovacek
Sixta KovacekLv2
21 Jan 2018
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subject to the initial 13. The solution to the separable differential equation given by 2xy + y'= 1, with x>- condition y(0)=5 is C) y = 2ln(2x+1)+5 A) y = 2ln(2x+1) B) y= ln(2x+1)+5 D) y=--In(2x+1)+5 E) none of these

Problem 14. Find the minimum value of the objective function 2.r + y subject to the constraints: > 0 y > 0 +y > 10 A) 40; B) 20; C) 22; D) 10; E) there is no minimum value; F) none of these.

[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) >

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