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National MAΘ Convention 2012 Limits and Derivatives Solutions 1. B definition of e . 2. D or E The series looks like the expansion of e x except the powers of x are decreased by 1 and there is no 1 term. dr 12 cos 3 sin 4sin 3 cos 3. C . Evaluating at gives the correct answer. 4 d 12 cos 3 sin 4sin 3 cos x 1 1 1 4. C lim x 0 3sin x cos x 3 3 d dy dx dx 5. B The second derivative of parametric is . Therefore the 2nd derivative is dx dt 2 2t 2 6t 6t , which simplifies to . 6 9t 5 27t 1 6. B Using chain rule, our derivative is . Evaluating at x 2 yields the desired x x2 1 result. 7. C L’Hopital once we get lim x x 2 5 x x . Multiply by the conjugate and the limit is approaching 5 . 2 8. D Using product rule (don’t forget to apply the chain rule for arcsin3x ), yields the desired result. 9. A By definition of an inverse f g x x . Therefore the numerator is 0 and so is the limit. a sin x b cos x 2 10. E L’Hopital once gives lim . Plugging in 0 gives a b . x 0 sec 2 x 11. A 3ln 1 4 x 6 x 2 4 12 x ln y lim ln y lim 3 ln y 12 y e12 2 x 0 x 0 x 1 4x 6x 12. Changed to C from B Taking the derivative of both, the only solution that satisfies the given conditions is B. sin 2 x tan 2 x . 13. C Using chain rule for the natural log derivative we get cos 2 x 14. Changed to D or E B y 1 2 . The graph crosses the x axis at x 1 . The slope 2 x is 1 , which is also the tangent of the angle. 4 Limits and Derivatives Solutions National MAΘ Convention 2012 3x 2 y y 3 dy and when evaluated at 1,1 is – 1. 3 2 dx x 3xy x3 3xy 2 6 xy 3x 2 y 3 y 2 y 3x 2 y y 3 3x 2 3 y 2 6 xyy d2y and when 3 2 2 dx 2 x 3xy 15. A evaluated at the point 1,1 and y 1 gives 0. 16. C Basic limit 17. A y 12 x 2 6 x 6 2 x 2 x 1 0 x 1 , 1 . Substituting these values into 2 the derivative yield the desired answers. dx dz dx dx 7.5 . 18. D x y 40 300 dt dt dt dt dy x 19. B . Take the reciprocal to get the desired result. dx 25 x 2 20. B We need the 2nd derivative here so y 6 x 6 x 1 . The slope is increasing over the interval where x 1 . 1 cos x cos x 1 sin x sin x . Plug in to get – 2. 21. C f x 2 2 1 cos x 22. E The function can only have a derivative if g c 0 . 23. B The sine function oscillates while the denominator grows without bound thus the limit approaches 0. 24. A Use the general form of Taylor Polynomials. 25. D 5 n n 1 2n 1 1 n n 1 3 2n2 3n 1 3 n 1 7 lim 2 lim 5 n n 6 n 2 n n 2n2 2 2 3 6 26. A 2 2x x x 27. A Based off of the given criteria, the derivative must be a linear function and the second derivative is the constant function thus either always concave up or concave down. The only choice that fits this description is a parabola. 28. E If continuous then b 3 2a b a 3 . If differentiable then 2 2 b 4a 2 b 6 b 4 . Their sum is 11 2 2 x 2 29. C Multiplying by the conjugate yields lim . Direct substitute to get the x 2 8 x2 x correct answer. Limits and Derivatives Solutions National MAΘ Convention 2012 x3 3x 2 x 4 5 3x 3 6 x 2 x x 3 3x 2 x 1 2 x 3 3x 2 1 x 2 which factors to 2 x 1 x 1 , so the x values will be 1 and 1 . Plugging these 2 values into the derivative and summing those values gives you the desired answer. 30. C 3x 2 6 x 1