AP Calculus Multiple Choice: BC Edition Solutions


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1 AP Clculus Multiple Choice: BC Edition Solutions J. Slon Mrch 8, 04 ) 0 dx ( x) is A) B) C) D) E) Divergent This function inside the integrl hs verticl symptotes t x =, nd the integrl bounds contin this vlue. We cnnot integrte through symptotes like this nd the the integrl is divergent, so the nswer is E. ) Which of the following integrls gives the length of the grph y = sin( x) between x = nd x = b where 0 < < b? A) D) x + cos ( x) dx B) + 4x cos ( x) dx E) + cos ( x) dx C) + cos ( x) dx 4x sin ( x) + 4x cos ( x) dx The rc length of differentible function f(x) over the intervl (, b) is defined s s = + [f (x)] dx. We need f (x), which is given s f (x) = cos( x) x for our function f(x). Then piecing this together gives x = ( ) cos( x) + dx = x + 4x cos ( x) dx
2 nd thus the nswer is D. ) Which of the following integrls represents the re enclosed by the smller loop of the grph of r = + sin(θ)? A) ( + sin(θ)) dθ B) ( + sin(θ)) dθ C) ( + sin(θ)) dθ D) ( + sin(θ)) dθ E) π/6 ( + sin(θ)) dθ π/6 π/6 Use your grphing clcultor to grph this function in polr mode. You will see this. y r = + sin(θ) x Wht we wnt is the re of the tiny loop stemming off of the origin. We first need to find the ngles which bound tht loop. These will be ngles for which the rius r is 0, since this occurs t the origin. Thus we solve 0 = + sin(θ) = sin(θ) sin(θ) = θ = 7π 6, π 6. So these re our bounds. The re within generl polr curve r(θ) between two ngle bounds α nd β is given s A = β α r(θ) dθ. If we plug everything in to the bove, we get tht the re enclosed is ( + sin(θ)) dθ,
3 nd thus the nswer is A. 4) A curve is described by the prmetric equtions x = t + t nd y = t + t. An eqution of the line tngent to the curve t the point determined by t = is A) x y = 0 B) 4x 5y = C) 4x y = 0 D) 5x 4y = 7 E) 5x y = In order to write the eqution of line tngent to the prmetric curve, we need to find the slope, in other words the derivtive dy/dx t t =. For prmetric curves, dy dx = dy / dx dt dt. Esily enough, nd dx dt = t +, dy dt = t + t, dy dx = t + t t +, dy dx = () + () = 5 t= () + 4. In order to use point slope, we need n ctul point! Plugging t = into x(t) nd y(t) respectively gives (x, y) = (, ). Putting this ll together gives nd thus the nswer is D. y = 5 (x ) 4 4y 8 = 5x 5 5x 4y = 7, 5) The tble gives selected vlues for the derivtive of function g on the intervle x. If g( ) = nd Euler s method with step size of.5 is used to pproximte g(), wht is the resulting pproximtion? A) 6.5 B).5 C).5 D).5 E) x g (x) We re sked to use Euler s method on the intervl [, ] with step size.5. This mens tht we will need the derivtive vlues t x = nd x = 0.5, which re respectively nd s per the tble. Using Euler s method gives Thus the nswer is D. +.5() +.5() =.5.
4 6) Wht re ll vlues of x for which the series n n x n A) All x except x = 0 B) x = C) x converges? D) x > E) The series diverges for ll x Whenever we get power series like this nd re sked to find out where it converges, the thing to do is pply rtio test. Tking the limit of the rtio of ech term to the previous gives (n + ) n+ x n (n + ) lim n x n+ n = lim = n n xn x. The condition for convergence is tht the bsolute vlue of the bove limit is less thn. So x < = x >. This mkes sense, becuse plugging in prticulr vlue for x mkes the series semi geometric (if you ignore the fctor of n). As long s the bse of the fctor in the denomintor is greter thn, the rtio of terms decreses lmost geometriclly, nd is thus convergent. In light of this, the nswer is D. 7) Which of the following series converge to? I. II. III. n=0 n n + 8 ( ) n n A) I only B) II only C) II only D) I nd III only E) II nd III only n I. This series doesn t even converge, becuse lim n n + t ll, it certin doesn t converge to. = 0. If it doesn t converge II. This is geometric series with r = /. Using the stndrd sum formul, nd remembering to ccount for the strting index of, [ ] 8 ( ) = 8 n ( /) [ = ] = =. 4
5 III. This is geometric series tht is much simpler thn the lst: n=0 n = / =. Thus both series II nd III converge to, nd the nswer is E. 8) The third degree Tylor polynomil bout x = 0 of ln( x) is A) x x x B) x+ x C) x x + x D) +x x E) x+ x x It s useful to know the Mclurin expnsion for ln( x) off the top of your hed, but in cse you don t, it s not difficult to construct, so we ll do tht here. For nottionl purposes I will define f(x) = ln( x). Finding derivtive vlues t x = 0 is s follows: f(x) = ln( x) f(0) = ln() = 0 f (x) = x f (0) = f (x) = ( x) f (0) = f (x) = ( x) f (0) = Thus putting this together gives tht P (x) = 0! x +! x! x nd thus the nswer is E. = x + x x 5
6 9) For series S, let S = n +, if n is odd where n = (n )/ Which of the following sttements re true? (n + ) if n is even I. S converges becuse the terms of S lternte nd lim n n =. II. S diverges becuse it is not true tht n+ < n for ll n. III. S converges lthough it is not true tht n+ < n for ll n. A) None B) I only C) II only D) III only E) I nd III only Your first rection to this problem is probbly to wnt to choose nswer F) I hve no ide, but nonetheless this is doble if you pick it prt. This series cn relly be thought of s two series combined into one. One of the series is nd the other is The first series is geometric series with common rtio /, nd thus this converges. The second series cn be compred to pseries with p =, nd thus it too converges. The sum of two convergent series is lso convergent. This leves choices I nd III still vlid, but we need to exmine them individully to double check their explntions. It turns out tht I mkes no sense becuse in order for series to converge s stted, it bsolutely must be true tht lim n = 0, n wheres the nswer sys tht the limit is. Throw option I out. III is indeed true becuse we hve shown tht the series converges even though the bsolute vlue of ech term is not necessrily smller thn tht of the previous. III is the only correct sttement so the nswer is D. 6
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