SMT Calculus Tes Soluions February 5,. Le f() = and le g() =. Compue f ()g (). Answer: 5 Soluion: We noe ha f () = and g () = 6. Then f ()g () =. Plugging in = we ge f ()g () = 6 = 3 5 = 5.. There is a unique posiive real number a such ha he angen line o y = + a = a goes hrough he origin. Compue a. Answer: Soluion: The slope of he angen line is a. The equaion for he angen line is (y(a +)) = a( a). Seing = y = gives us a = a, which has soluion a =. 3. Moor has $, and he is playing a gambling game. He ges o pick a number k beween and (inclusive). A fair coin is hen flipped. If he coin comes up heads, Moor is given 5k addiional dollars. Oherwise, Moor loses k dollars. Moor s happiness is equal o he log of he amoun of money ha he has afer his gambling game. Find he value of k ha Moor should selec o maimize his epeced happiness. Answer: 5 Soluion: Suppose ha Moor chooses a value of k. We wrie down he epeced value of Moor s happiness. If he coin comes up heads, Moor now has + (5k) = (5k + ) dollars. If he coin comes up ails, Moor now has k = ( k) dollars. Therefore, he epeced value of Moor s happiness is H(k) = log((5k + )) + log(( k)). We wan o maimize his. To do his, we differeniae, se he derivaive equal o zero, and look for criical values. Here, H (k) = ( ) 5 (5k + ) = ( 5 ( k) 5k + ) = k when 5k + = 5( k), so k =, and hence k = 5 is he only criical value. The maimal value of H(k) for k [, ] mus occur eiher a a criical value or an endpoin. Observe ha among he hree values H(), H(), and H( 5 ), he larges is H( 5 ). Therefore, Moor maimizes his happiness by selecing k = 5.. The se of poins (, y) in he plane saisfying /5 + y = form a curve enclosing a region. Compue he area of his region. Answer: 8 7 Soluion: The se of poins saisfying he equaion form a closed curve ha encloses a region. Observe ha his curve is preserved if we ransform or y y, so i is symmeric in all quadrans. In paricular, we can find he area in he firs quadran, where, y >. In he quadran, we can rewrie our equaion as y = /5. This curve inersecs he coordinae aes a (, ) and (, ), and i is coninuous, so he area is A = /5 d = 7.
SMT Calculus Tes Soluions February 5, The oal area is herefore A = 8/7. 5. Compue he improper inegral ( ) d. Answer: Soluion : Firs of all, we noe he many symmeries of he given epression. Specifically, we have and we subrac is reciprocal. We also recall ha square roos, when we ake heir derivaive, give us heir reciprocal. This inspires he guess ha he funcion f() = ( ) is somehow imporan o our inegral. Indeed, we find ha f () = so ha d = = ( ) =. Soluion : Alhough soluion is, perhaps, he preies way of solving his problem, i is no necessarily easy o noice. A more direc approach uses a rig subsiuion. Specifically, noing he imporance of and remembering he Pyhagorean ideniy sin + cos =, i makes sense o ry he subsiuion = sin θ. Then = cos sin = co. Also, d = 8 sin θ cos θ dθ, sin θ = when θ = and sin θ = when θ = π. The inegral becomes (co θ an θ) 8 sin θ cos θ dθ = 8 cos θ sin θ dθ = 8 This las inegral may be easily compued by he subsiuion θ θ: 8 cos θ dθ = cos θ dθ = (sin θ) π = ( ) =. cos θ dθ. Soluion 3: The simples way o solve his problem is perhaps o wrie he inegrand wih a common denominaor. This gives ( ) d = d = d. Subsiue u =, du = ( ) d. Then our inegral becomes d = du = u u =. 6. Compue [ lim ln ( + )]. Answer:
SMT Calculus Tes Soluions February 5, Soluion: We rewrie his limi in a form ha allows us o apply L Hôpial s Rule. Tha is, [ ( )] + lim ln ln ( ) + + ( ) 3 ( + ) by L Hôpial s Rule ( ) ( ) = + + ( ) =. 7. For a given >, le a n be he sequence defined by a = for n = and a n = a n for n. Find he larges for which he limi lim n a n converges. Answer: e /e Soluion: In order for lim a n o have a limi L, i mus be ha L = L, so ha = L /L. n Oherwise, we would be able o eend he recurrence and converge o a differen limiing value. Thus, we seek he maimum of he funcion f(l) = L /L. To do his, we solve df dl df dl = d ( dl e ln L L = L /L L ln L ) L, =. Since we see ha L = e. Thus, he maimum value for is f(e) = e /e. To be sure ha his is a maimum, we check as follows: d f dl = L L ( 3L + ln (L) + (L ) ln(l) + ) e = e e 3 <. e 8. Evaluae + + d. Answer: 3 Soluion: We subsiue he variable by and add he resuling inegral o he original inegral o ge I = = + + d + + d = + + + + ( + ) + d = + + + + + ( + ) ( + ) + d = d + d = + 6 3 = 8 3. So he given inegral is I = 3. Noe ha more generally, for even funcions f, a f() a +b d = a a f() d.
SMT Calculus Tes Soluions February 5, 9. Le f saisfy = f()e f(). Calculae e f() d. Answer: e Soluion : Firs, we compue he aniderivaive. Make he subsiuion u = f(), so hence du = f () d. Noe ha f () = d d ef = e f f ()e f, so f () = e f + = f (+f). Thus, ( ) f()d = f() d = u du u (+u) = ( + u) du = ue u ( + u) du ue u ( + u)du = e u ( u u + ) = e f() ( f() f() + ) To conclude, when =, f() = and when = e, f() =. Thus, e f() d = e ( + ) e ( + ) = e. Soluion : Noe ha f is monoonically increasing and is he inverse of he funcion g(y) = ye y. Since f(e) =, he area under f() from o e is he area of he recangle wih verices (, ), (e, ), (, ), (e, ) minus he area o he lef of f() from o, and he laer is jus he inegral of g(y) from o. So we have e f()d = e = e [ye y ] + e y dy = g(y)dy = e ye y dy e y dy = [e y ] = e.. Given ha n= = π n 6, compue he sum n n. n= Answer: π ln Soluion: Firs of all, for he sake of clariy, I omi deails abou cerain calculaions which are jusifiable so here is a more clear focus on he acual compuaion. Specifically, I ake derivaives and inegrals of series wihou eplaining, and I inegrae funcions wih removable singulariies, bu he ordinary suden would no pay aenion o hese echnical issues anyway. I proceed now: The firs sep o obaining any insigh on his problem is o replace wih n. This allows us o n ake derivaives, geing rid of powers of n in he denominaor. Thus, we wrie f() = n= n n and wha we wan o find is f( π ) given ha f() = and f() = 6. As menioned before, we firs ake f () = and hen we ake (f ()) =. By reinegraing, n= n n n= n = f () = ln ( ) + C bu by plugging in = i is easy o check ha C =. Then f ln () () =. Because f() =, f() = ln () d. Thus, he answer we are looking for is equal o ln () d. This complees he firs par of he soluion. The second par consiss of compuing his inegral. We denoe I = ln d. There are wo hings we know abou his inegral: ha finding he aniderivaive, if i even eiss, would be eremely challenging, and also a relaed formula
SMT Calculus Tes Soluions February 5, ln () d = π 6 which is given. Noing ha is he midpoin of he inerval [, ] in which he inegral formula is relevan, we noe ha here are several ransformaions which give inegrals on he inerval [, ]. Specifically, he subsiuion of yields I = ln d. In addiion, I = ln () d ln () d = π 6 ln () d. Thus, we may wrie I = π 6 + ln ln d. The apparen symmery of he inegrand immediaely brings o mind he funcion g() = ln ln ( ) as a poenial aniderivaive: indeed, when we apply he produc rule, we easily ge g () = ln ln π. Thus, I = 6 + ln ln ( ). Because plugging in = is undefined, we resor o using limis and easily obain. Thus, I = π 6 ln = π 6 ln and I = π ln = π ln are boh correc and equally valid answers. Noe, wih only a lile more work (and some formalizing), we can obain he more general resul ha n= n + () n n n= = π n 6 ln ln ( ) when (, ).