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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  e12
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 