SOA/CAS MAY 2003 COURSE 1 EXAM SOLUTIONS

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1 SOA/CAS MAY 2003 COURSE 1 EXAM SOLUTIONS Prepared by S. Broverman 2brove@rogers.om website 1. We identify the following events:. - wathed gymnastis, ) - wathed baseball, : - wathed soer. We wish to find 7. q) q: µ. By DeMorgan's rules we have 7. q) q:µ~7.r)r:µ. We use the relationship 7.r)r:µ~7.µb7 )µb7 :µ 47.q)µb7.q:µb7 )q:µ 5b7.q)q:µ. We are given 7.µ~Á7 )µ~á7 :µ~á 7.q)µ~Á7.q:µ~Á7 )q:µ~á7.q)q:µ~. Then 7.r)r:µ~and 7. q) q:µ~~. Answer: D 2. For eah of the five possible answers at least on of the urves is a straight line. If the straight line was ²³, then ²³ ~. This is true sine a straight line would be of the form ²³ ~ b, and ²³ ~ Á ²³ ~. Therefore, the straight line must be ²³. We an eliminate E sine both urves are straight lines. We an eliminate C and D, sine the seond derivative of the onvex urve is ²³, but the two straight lines are negative for some values of. We see that for the urve in A is onave down, so that ²³ for, and we see that for the urve in A is onvex up, so that ²³ for. The straight line in A is for and it is for. This mathes with the seond derivative of the urve. As an alternative approah, for eah answer, the straight line goes through the origin, and must be of the form ²³ ~ (slope ). Therefore, ²³ ~ b b, sine the seond derivative of b is 0. This rules of answers C, D and E, sine there are no ubi funtions (C and D have quadrati looking funtions and E has a linear funtion). Sine ²³ also goes through the origin for eah answer, it follows that ~, so that ²³ ~ b and ²³ ~. If (negative slope for the straight line ²³ ~ ) and then lim ²³ ~ b ~ B. This does not our in A or B. B Therefore it must be true that (positive slope). This ours in answer A. Answer: A

2 3. lim ~ ~ ~ b. Answer: E ²³²³ ²³²b³ ²³²³ 4. The exponential time until failure random variable ; has density funtion of the form ²!³~!, and had distribution funtion -²!³~7 ;!µ~! for!. The median of the distribution is the time point that satisfies the relationship -²³ ~ ; in other words, is the time point for whih there is a 50 probability of failure by time. We are given that ~, and therefore -²³~ ~, from whih it follows that ~. We are asked to find 7 ; µ~-² ³~. Using the relationship ~² ³, we get 7 ; µ~ ~² ³ ~. Notie that we ould solve for from the equation ~, but it is not neessary. Answer: D 5. We identify the following events: ( - the poliyholder insures exatly one ar (so that ( is the event that the poliyholder insures more than one ar), and : - the poliyholder insures a sports ar. We are given 7 (µ~ (from whih it follows that 7 (µ~), and 7 :µ~ (and 7 :µ~). We are also given the onditional probability 7 :O(µ~ ; "of those ustomers who insure more than one ar", means that we are looking at a onditional event given (. We are asked to find 7 (q: µ. We reate the following probability table, with the numerals in parentheses indiating the order in whih alulations are performed. ( Á ( Á : Á ²³ 7 : q (µ ²³ 7 : q ( µ ~ 7 :O( µ h 7 ( µ ~ 7 :µ 7 : q ( µ ~ ² ³²³ ~ ~ ~ :Á ²³ 7 (q:µ ~ 7 (µ 7 ( q :µ ~ ~ Answer: B

3 We,?µ~ h²á&³&. Sine the region of probability is defined with &, we apply double integration in the & order. It would be possible to reverse the order, but that would not make the solution any more µ and,?@ µ are found in a similar way.,?µ ~ h h & & ~ h & & &~ & e &~ &~ & e &~ &~ & e &~ ~ h µ ~ h µ ~ h h µ ~ & h h & & ~ h & & ~ h µ ~ h µ ~ h h ~,?@ µ ~ & h h & & ~ h & & ~ h µ ~ h µ ~ h h ~ Then ³ ~ ² ³² ³ ~. Answer: A 7. To find ²³ we apply the hange of variable " ~, so that " ~, or equivalently, ~ ". Then ²³ ~ ²"³ " ~ h ²"³ " (sine "~, we must adjust the limits of integration in the transformed integral). Variables of integration are "dummy variables", and therefore ²"³" ~ ²!³! ~ ²³ ~ ²³b ²³ ~ b (it doesn't matter what letter is used for the integration variable). Finally, ²³~ h ²"³"~ h²b ³~. Answer: C

4 8. We identify the following events: ( - the driver has an aident, ; (teen) - age of driver is (young) - age of driver is 21-30, 4 (middle age) - age of driver is 31-65, : (senior) - age of driver is The final olumn in the table lists the probabilities of 4 and :Á and the middle olumn gives the onditional probability of ( given driver age. The table an be interpreted as Age Probability of Aident Portion of Insured Drivers (O;µ~ 7 ;µ~ (O@µ~ (O4µ~ 7 4µ~ (O:µ~ 7 :µ~ We are asked to find 7 ;O(µ. We onstrut the following probability table, with numerals in parentheses indiating the order of the alulations. ; Á 4Á :Á ( ²³ 7 ( q ; µ ²³ 7 ( µ ²³ 7 ( q 4µ ²³ 7 ( q :µ ~7 (O;µh7 ;µ ~7 ~7 (O4µh7 4µ ~7 (O4µh7 4µ ~ ²³²³ ~ ²³² ³ ~ ²³²³ ~ ²³²³ ~ ~ ~ ~ ² ³ 7 (µ ~ 7 ( q ; µ b 7 ( µ b 7 ( q 4µ b 7 ( q :µ ~ 7 (q; µ 7 (µ ²³ 7 ; O(µ ~ ~ ~. Answer: B 9. To say that the payment is 20 less than the previous year's payment is the same as saying that the payment is 80 of the previous year's payments. The suessive yearly payments will be Á ²³ Á ²³ Á ²³ Á This forms a geometri series. The total laims paid in all years after the ompany stops selling malpratie insurane is b ²³ b ²³ b ²³ b Ä ~ b ²³ b ²³ b ²³ b ĵ ~ h ~. Answer: D

5 10. The range & is only valid for. This is true sine for and for. Therefore, the range for & is &, so that &. Also, the inequality & is equivalent to l &, so that & is equivalent to & l&. The marginal density funtion is found by integrating the joint density over the range for the other variable ; ²&³ ~ & ~ & e ~ &² l & &³ ~ ²& & ³ for &. & l & ~ & ~& l Note that it is true that for any partiular we have &. However, sine an be any number from 0 to 1, & an also be any number from 0 to 1. Answer: E 11. We apply the hain rule. In this ase, the derivative of an exponential funtion is equal to that exponential funtion multiplied by the derivative of the exponent.! ² ³! : ²!³ ~ h ²³² ³. ² ³ Then : ²³ ~ h ²³² ³ ~ Á. Answer: B OO 12.,?µ ~ h ²³ ~ h. For, OO~ and for, OO~. Then,,?µ~ h² ³ b h² ³~²³² b ³ ²³. Answer: D ~²³ < b = ~

6 13. The standard approximation to the sum (total) of a olletion of independent random variables is the normal approximation. The total ontribution is ; ~ * b* bäb*, the sum of the 2025 ontributions. * is the amount of the -th ontribution, the * 's are mutually independent, and eah has mean, * µ ~ and variane = * µ ~ ² ³. The mean and variane of ; are, ; µ ~, * µ ~ ² ³² ³ ~ Á Á and = ; µ ~ = * µ ~ ² ³² ³ ~ Á Á ~ ~. We will denote the 90th perentile of ; by. We find the approximate 90th perentile of ; by applying the normal approximation to ;. We wish to find so that 7 ; µ~. We standardize the probability: 7 ; µ~7 µ~. ; ÁÁ Á Á ; ÁÁ ÁÁ Á Á lá Á l is approximately standard normal (mean 0, variane 1), so that ÁÁ l is the 90-th perentile of the standard normal distribution. From the table for the Á Á standard normal distribution, we see that )( ³ ~. Therefore we have ÁÁ l ~, from whih we get ~ Á Á. Answer: C Á Á l 14. Using the produt rule for differentiation, we have ²³ ~ ²³µ ~ ²³ b ²³ ~ ²³ ²³µ, so that ²³ ~ ²³ ²³µ. We use the definition of derivative to find ²³~ lim ²b³ ²³ ~ lim b b b ~ lim bb bµ~ b. Then, ²³ ~ ² ³ b ~. Also, if we let ~ and ~, then ²b³²³ ~ ²b³²³ ~ ²³. However, ² b ³ ²³ ~ ² ³²³ b ²³²³ b b ²³ ~ Therefore, ²³ ~, so that ²³ ~. Finally, ²³ ~ µ ~. Answer: C

7 15. The new amount paid to the surgeon is? ~? b, and the new amount of hospital harges We wish to find =? b@ µb*#²?á@ ³ =?µ~=?bµ~=?µ~ ÁÁ and µ ~ µ ~ ² µ ~ ²³²Á ³ ~ Á. *#²? ³ ~ *#²? b ³ ~ ³. We have used the ovariane rule *#²< b Á > b ³ ~ *#²< Á > ³. We still must know ³ to omplete the problem. We are given =? µ ~ Á, and we use the relationship ~ Á b Á b ³ S ³ ~ Á. Then *#²? ³ ~ ³ ~ Á. Finally, =? b@ µb*#²?á@ ³ ~ Á b Á b ²Á ³ ~ Á. Answer: C 16. The devie fails as soon as either omponent fails. The probability of failure within the first hour is 7 ²? ³ r ²@ ³µ. There are a ouple of ways in whih this an be found. We an use the probability rule 7 ²? ³r²@ ³µ~7? µ7 ²? ³q²@ ³µ, but this will require three separate double integrals (although the first two are equal beause of the symmetry of the distribution). Alternatively, we an use DeMorgan's rule, 7 (r)µ~7 ( q)µ, so that 7 ²? ³r²@ ³µ~7 ²? ³q²@ ³µ. Sine both? are between 0 and 3, we get b& ~ e ~ 7 ²? ³r²@ ³µ~ ~ ~ 7 ²? ³ q ²@ ³µ ~ ² ³ & ~ h ²& b & ³ e µ ~ h ²b³ ~ h² b³ ~ Then,. Answer: B &~ &~

8 ²³ 17. b bäb ~ h b b bäb µ ~ h ~ ~ We have used the formula for the sum of a geometri series: b bb bäb ~. In this ase, for the expression inside brakets, ~ and ~. We wish to find lim h b bäb µ ~ lim h µ. B B lim h² ³ B This will be equal to. To find lim h² ³, we let ~, so that the limit is equal to lim, B and using l'hospital's rule we get lim ~ lim ~. Therefore, lim h b bäb µ ~. Answer: C B 18. We define the following events 9 - renew at least one poliy next year ( - has an auto poliy, / - has a homeowner poliy A poliyholder with an auto poliy only an be desribed by the event (q/, and a poliyholder with a homeowner poliy only an be desribed by the event (q/. We are given 7 9O(q/µ~Á 7 9O( q/µ~ and 7 9O(q/µ~. We are also given 7 (µ~ Á7 /µ~ and 7 (q/µ~. We are asked to find 7 9µ. We use the rule 7 9µ~7 9q(q/µb7 9q( q/µb7 9q(q/µb7 9q( q/µ. Sine renewal an only our if there is at least one poliy, it follows that 7 9q( q/µ~; in other words, of there is no auto poliy (event ( ) and there is no homeowner poliy (event / ), then there an be no renewal. An alternative way of saying the same thing is that 9 is a subset (subevent) of (r/. (Note also that 7 (r/µ~7 (µb7 /µ7 (q/µ~ b ~, so this also show that 9 must be a subevent of (r/, and it also shows that 7 ( q/µ~7 (r/µ~~ so that ( q/ ~ ). This an be illustrated in the following diagram.

9 18. (ontinued) We find 7 9q(q/µ, 7 9q( q/µ and 7 9q(q/ µ by using the rule 7 * q +µ ~ 7 *O+µ h 7 +µ : 7 9 q ( q /µ ~ 7 9O( q /µ h 7 ( q /µ ~ ²³² ³ ~, 7 9 q ( q /µ ~ 7 9O( q /µ h 7 ( q /µ ~ ²³7 ( q /µ, 7 9 q ( q / µ ~ 7 9O( q / µ h 7 ( q / µ ~ ²³7 ( q / µ. In order to omplete the alulations we must find 7 ( q/µ and 7 (q/ µ. From the diagram above, or using the probability rule, we have 7 (µ ~ 7 ( q /µ b 7 ( q / µ S ~ b 7 ( q / µ S 7 ( q / µ ~, and 7 /µ ~ 7 ( q /µ b 7 ( q /µ S ~ b 7 ( q /µ S 7 ( q /µ ~. Then 7 9q( q/µ~²³² ³~and 7 9q(q/µ~²³² ³~. Finally, 7 9µ ~ b b ~. 53 of poliyholders will renew. Answer: D

10 19. The darkened border in the graph below is the perimeter of the retangle. The perimeter is 7²³ ~ ²b ³, and we are told that the interval for is. As S, the perimeter approahes ² b ³ ~, and as SBB, the perimeter approahes. The ritial point(s) for 7 ²³ are the solution(s) of 7 ²³ ~ ² ³ ~. The solution ours at ~ ² ³ ~. If ~ ² ³ ~ ² l ³, the perimeter is 2 l ² b ³ ~ ² b ³ ~. The funtion 7 ²³ has an absolute minimum at ~, but there is no absolute maximum sine lim 7 ²³ ~ B. Answer: A B 20. Let? denote the two loss amounts ( not payment amounts). We onsider the following ombinations of? that result in the total benefit payment not exeeding 5. Case 1:? (so loss? results in no payment) (so that results in a maximum payment of 5 after applying the dedutible of 2). Case 2:? (so loss? results in a maximum payment of 5 after the dedutible of 1 is applied) (so results in no payment). Case 3:? and ²?³b²@ ³ (? is paid for loss? is paid for The last ondition is equivalent to?b@. The probability that the total benefit paid does not exeed 5 is the sum of the probabilities for Cases 1, 2 and 3. 7 Case 1µ ~ 7 ²? ³ q ³µ ~7? µ~² ³² ³~ (we have used the independene of? to find the probability of the intersetion)

11 20. (ontinued) 7 Case 2µ ~ 7 ²? ³ q ³µ ~7? µ~² ³² ³~. 7 Case 3µ ~ ²Á &³ & ~? ²³ ²&³ & ~ ² ³² ³ & ~ ²³ & ~ ²³ µ ~ ²³² e ³ ~ (note that ²Á &³ ~? ²³ ²&³ beause? are independent). The total probability is b b ~. One we have identified Cases 1, 2 and 3, this problem ould be approahed from a graphial point of view. Sine? are independent and uniform, the joint distribution of? is uniform on the square, &, with joint density ²³²³ ~. Sine the joint distribution is uniform, the probability of any event involving? is equal to the onstant density (.01 in this ase) multiplied by the area of the region representing the event. The three regions for Cases 1, 2 and 3 are indiated in the graph below. The 10 d 10 square is the full region for the joint distribution. The retangular area for Case 1 is d ~ for a probability of d ~. The retangular area for Case 2 is d ~ for a probability of d ~. The triangular are for Case 3 is d d ~ for a probability of d ~. The total probability for Cases 1, 2 and 3 ombined is again. ~ ~ Answer: C

12 3 *!! 21. Labor and apital are both funtions of time, with ~ and ~. We apply the produt and hain rule to get 7 3 *!!!! ~ ² ³ h ²³ ² ³²³ b ²³ ² ³²³ ² ³µ ~ ~ ² 3 * ³ ~ ² ³ 3 h h * b 3 h h * h µ. Answer: D 22. The 30-th perentile of?, say, is the point for whih -²³~7? µ~. ²³ ²³ ~ ~ Therefore, ~ ~ e ~ ² ³ ~. Solving for results in ~. The 70-th perentile of?, say, is the point for whih 7? µ~. ²³ ²³ ~ ~ Therefore, ~ ~ e ~ ² ³ ~. Solving for results in ~. Then ~. Answer: B 23. The density funtion is ²&³. If we an find - ²&³, the umulative funtion then ²&³ ~ - ²&³. We an find - ²&³ from the relationship and ; and from - ²!³ (the df of ; ³ ; - &µ~7 ; &µ~7 ; l (the desription of - ²!³ indiates that ; is defined for only positive numbers). ; Therefore, - ²&³~- ² l &³~² ³ ; The density funtion is ²&³ ~ - ²&³ ~ Answer: 24.,?µ ~ h ² & ³ &. The "inside" integral is h ² & ³ & ~ h ² & ³ & ~ h µ. The omplete integral is ~ h µ ~ h ² ³ e µ ~. ~ Note that we ould have found ²³, the marginal density funtion of? first and the have found,?µ. This would be done as follows: ²³ ~ ² & ³ & ~ ² ³, and then??? ~ ~,?µ ~ h ²³ ~ h ² ³ ~ h ² ³ ~ h² ³ e µ~ (as in the first approah). Answer: C

13 25. The insurane payment H?.?* *? The insurane payment is less than.5 if the? *, or equivalently, if? * b. It must be true that *, beause if * then * b and then 7? * b µ ~ sine 7?µ~. b 7?*b µ~ ~²b ³. In order for this to be equal to.64 we must have ² b ³ ~ S b ~ (we ignore the negative square root sine? ) S ~. Answer: B b b b ²b³²³ 26. ²³ ~ ~. It is tempting to anel the b fator and write ²³~. This is not orret. ²³ is equal to, exept at the point ~. At that point ² ³ ~, whih is not defined. The graph of ²³ would be idential to the graph of, exept that there would be an empty spae at ~ ²³ is disontinuous at ~ sine ²³ is not defined, but it is possible to define ² ³ ~ ~, and then ²³ would be ontinuous at ~. Sine lim ²³ is infinite (atually, lim ²³ ~ b B²³ and lim ~ B), there is no way to define ²³ that would make ²³ ontinuous at ~. Answer: A b ( 27. The differential equation an be written in the form (! ~. This an be written as! (²!³ ~. For this differential equation, (²!³ ~ 2!. The initial investment of 5,000 is made at time 0 and we want the investment to grow to 20,000 in 24 years. Therefore (²³ ~ ~ 2 and (²³ ~ 2 ~ Á. Á 2 Á 2 Then ~ ~ S~ h²³~ h² ³~ h²³~ h²³. Answer: D

14 28. If we find the onditional density ²&O? ~ ³, then O?~ µ~ ²&O?~ ² ²&O? ~ ³ ~? ² ³ ² Á &³ ~ ² ³& ~ & Á &? & ²&O? ~ ³ ~ ~ ~ µ ~ & & ~ The onditional density is. The joint density is, and the marginal density of? at? ~ is ² ³ ~ ² ³& & ~ The onditional density is. The onditional probability is. Answer: C 9 # Á ²#³ ~ 7 = #µ ~ 7 Á #µ ~ 7 9 ² ³µ. If? has a uniform distribution on the interval ²Á³ then if, 7? µ ~. Sine 9 is uniform on ²Á ³ it follows that # ² # Á³ # # Á Á Á 7 9 ² ³µ ~ ~ ² ³ µ if ² ³. Answer: E ²b³ 30. The sequene for ~ Á Á Á has the following pattern Ä ²b³ Ä even odd odd even even odd odd even even odd Ä ²b³ ²³ Ä B The summation ²b³ ²³ an be broken into 4 summations ~ The first summation is for ²b³ ~ Á Á Á Á, so that ² ³ ~ ²b³ The seond summation is for ~ Á Á Á Á, ² ³ ~ ²b³ The third summation is for ~ Á Á Á Á, ² ³ ~ The fourth summation is for ~ Á Á Á Á, ² ³ ~ The first summation is b b b bä ~. ²b³ b b b bä ~ b b b bä ~ The seond summation is. The third summation is b b b bä ~. The fourth summation is. The total summation is the ombination of all four summations. This is B ²b³ ²³ ~ b ~ b ²³² ³ ²b ³² ³ b ~ ~ ~. Answer: E

15 31. We identify the following events 5 - non-smoker, 3 - light smoker, / - heavy smoker, + - dies during the 5-year study. We are given 7 5µ~ Á7 3µ~Á7 /µ~ We are also told that 7 +O3µ ~ 7 +O5µ ~ 7 +O/µ (the probability that a light smoker dies during the 5-year study period is 7 +O3µ ; it is the onditional probability of dying during the period given that the individual is a light smoker). We wish to find the onditional probability 7 /O+µ. We will find this probability from the basi definition of onditional probability, 7 /O+µ ~ 7 /q+µ 7 +µ. These probabilities an be found from the following probability table. The numerals indiate the order in whih the alulations are made. We are not given speifi values for 7 +O3µ, 7 +O5µ, or 7 +O/µ, so will let 7 +O5µ ~, and then 7 +O3µ~and 7 +O/µ~. 5 Á 3 Á / Á + ²³ 7 + q 5µ ²³ 7 + q 3µ ²³ 7 + q /µ ~ 7 +O5µ h 7 5µ ~ 7 +O3µ h 7 3µ ~ 7 +O/µ h 7 /µ ~ ²³² ³ ~ ~ ²³²³ ~ ~ ²³²³ ~ ²³ 7 +µ~7 +q5µb7 +q3µb7 +q/µ~ bb~. 7 /q+µ 7 +µ ² ³ 7 /O+µ ~ ~ ~. Answer: D 32. From the definition of -²³ we see that -²³ ~. This indiates that? has a point of probability at?~ with 7?~µ~. For, the density funtion for? is ²³~-²³~ We formulate the variane of? as =?µ~,? µ²,?µ³.,?µ~²³h7?~µb h²³~²³² ³b ²³~ b ~. Answer: C,? µ~²³h7?~µb h²³~²³² ³b ²³~ b ~ =?µ~ ² ³ ~

16 . lim ²³ ~ lim ~ 3 B B lim lim B B 3 3b b b 33. We denote ²³ ~ ²²³³ ~ Á ²³ ~ ² ²³³ ~ Á Á ²³ ~ ² ²³³ ~ ² ³ ~ We wish to find. If the limit 3 exists, then 3 ~ ²³ ~ ² ³ ~ ²3³ ~. 3 3b Therefore, 3~, and the solutions to this equation are 3~and 3~. If then ²³~ ~; therefore if then ~² ³. This indiates that the limit annot be 0, sine anytime is. Therefore, the limit must be 1. Answer: B, the next term in the series 34. ²³ ~ ² b ³,. The total probability must be 1, so that ² b ³ ~ µ ~. Therefore, ~ and ²³ ~ ² b ³. ~ ~ Then, 7? µ ~ ²³ ~ ² b ³ ~ ² b ³ e ~. Answer: C 35. We an use a version of the seond derivative test for a funtion of two variables. Given the funtion ²Á&³, we define " : C C C C C& CC& " A C C C C& f If = ~, then ² Á& ³ C C C C 1. if " and, then has a relative minimum at ²Á&³ 2. if " and, then has a relative maximum at ²Á&³ 3. if ", then has neither a relative minimum or maximum at ²Á&³ C C C For the given funtion, C ~ & b Á C& ~ Á CC& ~. Then, " ~ ² & b ³²³ ² ³ ~ b & b. C C At the indiated ritial points, the values of and are " " C ²Á &³ C ² Á ³ ²Á³ ² Á ³ Answer: E Loal Max. or Min. No loal max. or min. Loal minimum No loal max. or min.

17 36. The probability funtion is 7? ~ µ Amount paid Expeted amount paid ~ ²³² ³ b ²³² ³ b ²³² ³ b ² ³² ³ b ² ³² ³ ~ Answer: D. 37. We define the following events., - the laim inludes emergeny room harges, 6 - the laim inludes operating room harges. We are given 7,r6µ~, 7,µ~ and, and 6 are independent. We are asked to find 7 6µ. We use the probability rule 7,r6µ~7,µb7 6µ7,q6µ. Sine, and 6are independent, we have 7,q6µ~7,µh7 6µ~² ³7 6µ (sine 7,µ~7,µ~ ~ ). Therefore, ~ 7, r 6µ ~ b 7 6µ 7 6µ. Solving for 7 6µ results in 7 6µ ~. Answer: D 38. Sine the inventory will be replenished when it drops to 1 and sine inventory is 19 at time!~, inventory will be replenished when the umulative number of units sold is 18. This ours " at time ", where :²"³ ~ ~ S " ~.!! The inventory being arried at time! for! is :²!³ ~ b ~.! The ost of arrying inventory at time! is ² ³. The total ost of arrying inventory from time! ~ to time! ~ is!!!~ ² ³! ~ ²! ³ e ~. Answer: C!~! > b! A! ²?b@ ³b! ²@?³ ²!! ³?b²! b! ³@ 39. 4²!Á!³~, µ~, µ~, µ ²!! ³? ²! b! ³@ ²!! ³? ²! b! ³@ ~, h µ ~, µ h, µ (this equality follows from the independene of? ²!!³ ²!b!³!b! ~4? ²! b!³~ h ~ Answer: E

18 &!!!!!! b!!~s!~ 40. The speed is m² ³ b² ³ ~ m²!³ b² ³ ~ m!b. The speed will be minimized where is minimized for!. The ritial point(s) our where. The speed at the ritial point!~ is l. The speed at the interval endpoint!~ is l. As!SBB the speed approahes. The minimum speed ours at! ~.! Also, sine the seond derivative of! b is, it follows from the seond derivative test that!~ is a loal minimum. Answer: B Answer: B