38 A Note 1 1 2 f Take the natural logarithm of each side of the equation and

38 a note 1 1 2 f take the natural logarithm of each

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38. A Note ( )11.2f=Take the natural logarithm of each side of the equation and then differentiate. ()()222( )2ln( )(23 )ln1 ;(23 )3ln1( )1fxxf xxxxxf xx=+=++()()32111(1)(1) ( 1)3ln(2)(1)13ln 2lnln 2ln82222fffee==− −= −+= −39. D ()( )( )()2121121,0,0;sec112dydydudvxevuyuedxdudvdxxev⎞⎛======+==40. E One solution technique is to evaluate each integral and note that the value is 11n+for each. Another technique is to use the substitution ()()1010101;1nnnuxxdxuduudu===. Integrals do not depend on the variable that is used and so 10nuduis the same as 10nxdx. 41. D ()()233232233112121118827333fx dxxdxx dxxxx=+=+=
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AP Calculus Multiple-Choice Question Collection171Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 1969 Calculus BC Solutions42. B Use the technique of antiderivatives by parts to evaluate 2cosxx dx2222cossin2( )2 sincossin2 sin( )sindvxdxuxvxdux dxf xxx dxxx dxxxxx dxCf xxxC======+=+43. E ()222411sec1secbbbaaadyLdxxdxx dxdx=+=+=+44. E 20,(0)2,(0)2yyyyy′′== −=; the characteristic equation is 220rr=. The solutions are 1,2rr= −=so the general solution to the differential equation is 221212with 2xxxxyc ec eyc ec e=+= −+. Using the initial conditions we have the system: 1212212and 220,2cccccc=+= −+==. The solution is 1( )2(1)2xf xefe==. 45. E The ratio test shows that the series is convergent for any value of xthat makes 11x+<. The solutions to 11x+=are the endpoints of the interval of convergence. Test 2x= −and 0x=in the series. The resulting series are ()211kkk=and 211kk=which are both convergent. The interval is 20x.
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AP Calculus Multiple-Choice Question Collection172Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 1973 Calculus AB Solutions1. E 34213(3 )42xx dxxxC=+2. E ()( )5( )5g xgf x==3. B 2222ln;xyxyxx===. At 2xe=, 22ye′ =. 4. A ( )sin; ( )1cosf xxxfxx=+=5. A lim00xxey→−∞==is a horizontal asymptote 6. D ()2(1)(1)(1)(1)21( ),(1)421xxfxfx+===+7. B Replace xwith ()xand see if the result is the opposite of the original. This is true for B. 555()3()3(3 )xxxxxx− −+== − −+. 8. B 22222333111117Distance(21 )333tdxtdtt=====9. A ()()()() ( )2cos3cos32cos3sin332cos3sin33ddyxxxxxxxdxdx′ ==⋅ −=⋅ −6sin3 cos3yxx′ = −10. C ()()()()45342324;;4441353xxxfxfxxfxxxxx′′====0 for 1 and 0 for 1fxfxf′′′′><<>has its maximum at 1x=.
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