# math-124-2000-final - U UNIVERSITY OF SASKATCHEWAN...

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U n i v e r s i t y o f S a s k a t c h e w a n DEO ET PAT- RIE UNIVERSITY OF SASKATCHEWAN MATHEMATICS 124.3 — Final Examination(April 15, 2000) Solutions NO BOOKS, NOTES OR CALCULATORS ALLOWED. Show all of your work. No credit will be given for unsubstantiated correct answers. (1)(4 × 2%) Evaluate (a) lim x →+∞ e 1 x Solution: lim x →+∞ e 1 x = e lim x →+∞ 1 x = e 0 = 1 (b) lim x 0 xe 1 x Solution: lim x 0 xe 1 x = lim x 0 e 1 x 1 x = (by L’Hopital’s Rule) lim x 0 e 1 x 1 x 2 1 x 2 = lim x 0 e 1 x = −∞ (c) lim x 0 + x x Solution: lim x 0 + x x = lim x 0 + e ln x x = lim x 0 + e (x ln x) = e lim x 0 + x ln x = e lim x 0 + ln x 1 x = ( by L’Hopital’s Rule) e lim x 0 + 1 x 1 x 2 = e lim x 0 + x = e 0 = 1 1
U n i v e r s i t y o f S a s k a t c h e w a n DEO ET PAT- RIE (d) lim x 0 tan x sin x x 3 Solution: Repeatedly using L’Hopital’s Rule, we get: lim x 0 tan x sin x x 3 = lim x 0 sec 2 x cos x 3 x 2 = lim x 0 2 sec 2 x tan x + sin x 6 x = lim x 0 ( 2 sec 3 x + 1 ) sin x 6 x = lim x 0 2 sec 3 x + 1 6 sin x x = lim x 0 2 sec 3 x + 1 6 lim x 0 sin x x = 2 + 1 6 ( 1 ) = 1 2 (2)(3 × 3%) Find the derivative f (x) if f (x) = (a) cos x sin x ln ( 2 + t)dt Solution: Recalling the formula d dx h(x) g(x) k(t)dt = k(h(x))h (x) k(g(x))g (x) , we let k(t) = ln ( 2 + t) , g(x) = sin x and h(x) = cos x and have f (x) = ln ( 2 + cos x)( cos x) ln ( 2 + x)( sin x) = ln ( 2 + cos x)( sin x) ln ( 2 + x)( cos x) = sin x ln ( 2 + cos x) cos x ln ( 2 + x) 2
U n i v e r s i t y o f S a s k a t c h e w a n DEO ET PAT- RIE (b) tanh ln 1 x + coth ln 1 x . Solution: Simplify first: Remembering that tanh x = e x e x e x + e x , and that coth x = e x + e x e x e x we have tanh ln 1 x + coth ln 1 x = 1 x x 1 x + x + 1 x + x 1 x x = 1 x 2 1 + x 2 + 1 + x 2 1 x 2 = ( 1 x 2 ) 2 + ( 1 + x 2 ) 2 1 x 4 = 1 2 x 2 + x 4 + 1 + 2 x 2 + x 4 1 x 4 = 2 1 + x 4 1 x 4 , so f (x) = 2 ( 1 x 4 )( 4 x 3 ) ( 1 + x 4 )( 4 x 3 ) ( 1 x 4 ) 2 = 8 x 3 1 x 4 + 1 + x 4 ( 1 x 4 ) 2 = 16 x 3 ( 1 x 4 ) 2 (c) sin 1 e x 2 . Solution: Using d dx sin 1 x = 1 1 x 2 and the Chain Rule, we get: d dx sin 1 e x 2 = 1 1 ( e x 2 ) 2 e x 2 = 1 1 e 2 x 2 e x 2 x 2 = 2 xe x 2 1 e 2 x 2 (3)(4%) Use differentials to estimate the value of sin 31 . Solution: If y = sin x , dy = cos xdx , so sin 31 sin 30 + cos 30 π 180 = 1 2 + 3 2 π 180 = 1 2 + 3 π 360 3
U n i v e r s i t y o f S a s k a t c h e w a n DEO ET PAT- RIE (4)(2 × 4%) Evaluate (a) 5 i = 1 csc π 6 i Solution: 5 i = 1 csc π 6 i = csc π 6 ( 1 ) + csc π 6 ( 2 ) + csc π 6 ( 3 ) + csc π 6 ( 4 ) + csc π 6 ( 5 ) = 1 sin π 6 + 1 sin π 3 + 1 sin π 2 + 1 sin 2 π 3 + 1 sin 5 π 6 = 1 1 2 + 1 3 2 + 1 1 + 1 3 2 + 1 1 2 = 2 + 2 3 + 1 + 2 3 + 2 = 5 + 2 2 3 = 5 + 4 3 3 (b) 55 j = 1 4 j + 1 + sin π 2 j Solution: 55 j = 1 4 j + 1 + sin π 2 j = 4 55 j = 1 j + 55 j = 1 1 + ( 1 + 0 + ( 1 ) + 0 ) + ( 1 + 0 + ( 1 ) + 0 ) + · · ·+ ( 1 + 0 + ( 1 )) = 4 55 ( 55 + 1 ) 2 + 55 + 0 = 2 ( 55 )( 56 ) + 55 = 55 [ 112 + 1 ] = 55 ( 113 ) = 6215 4
U n i v e r s i t y o f S a s k a t c h e w a n DEO ET PAT- RIE (5)(5%) The interval [ 0 , π] is partitioned into the intervals 0 , π 3 , π 3 , 2 π 3 and 2 π 3 , π .