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Midterm

01:640:151 Midterm: 01:640:151 Exam 1 2017 Fall SolutionExam


Department
Mathematics
Course Code
01:640:151
Professor
All
Study Guide
Midterm

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Solutions to Review Sheet for Exam 1, Math 151, Fall 2017
These problems are presented in order to help you understand the material that is listed
prior to the rst exam in the syllabus. DO NOT assume that your rst midterm exam will
resemble this set of problems. The following 20 problems are not meant to be a sample
exam. These problems are just a study aid. Since we have not covered L'H^opital's Rule
yet, this rule should not be used to answer any of these questions.
(1) Simplify sin1(sin(7=4)) and sin1(sin(5=6)).
Recall that the formula sin1(sin(x)) = xis valid only when =2x=2.
sin(7=4) = sin(7=42) = sin(=4) and =2<=4< =2. Therefore,
sin1(sin(7=4)) = sin1(sin(=4)) = =4:
sin(5=6) = sin(5=6) = sin(=6) = sin(=6) and =2< =6< =2. Therefore,
sin1(sin(5=6)) = sin1(sin(=6)) = =6:
(2) Assume x0has the properties sin(x0) = 0:7 and cos(x0)<0. Evaluate sin(4x0) and
cos(4x0).
We know cos(x0) = q1sin2(x0) = p1(sin(x0))2=p1(0:7)2=p0:51 and
cos(x0) is negative. Therefore, the correct choice of is and cos(x0) = p0:51. Now
sin(2x0) = 2 sin(x0) cos(x0) = 2(0:7)(p0:51) = (1:4)p0:51
and
cos(2x0) = cos2(x0)sin2(x0)=0:51 (0:7)2= 0:02:
Then
sin(4x0) = 2 sin(2x0) cos(2x0) = 2((1:4)p0:51)(0:02) = 0:056p0:51
and
cos(4x0) = cos2(2x0)sin2(2x0) = (0:02)2((1:4)p0:51)2
= 0:0004 (1:96)(0:51) = 0:9992:
(3) Simplify cot(sin1x) and cos(tan1x).
We know (cos(sin1x))2= cos2(sin1x)=1sin2(sin1x)=1(sin(sin1x))2= 1 x2,
hence cos(sin1x) = p1x2. Since sin1xis in the interval [=2; =2], we must have
cos(sin1x)0, and this implies that the must be +. This says cos(sin1x) = p1x2.
We also know sin(sin1x) = x. This implies
cot(sin1x) = cos(sin1x)
sin(sin1x)=
p1x2
x:
1

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We know (sec(tan1x))2= sec2(tan1x) = 1 + tan2(tan1x) = 1 + (tan(tan1x))2= 1 +
x2, hence sec(tan1x) = p1 + x2. Since tan1xis in the interval (=2; =2), we must
have sec(tan1x)>0, and this implies that the must be +. This says sec(tan1x) =
p1 + x2. This implies
cos(tan1x) = 1
sec(tan1x)=1
p1 + x2:
(4) Find all solutions of the equation
e6+x2=22
e7x= 1.
Since (et)2=e2t, this can rewritten as e2(6+x2=2)
e7x= 1. This equation is equivalent to
e2(6+x2=2) =e7x. Since eu=evimplies u=v, we now know 2(6 + x2=2) = 7x, which is
x27x+ 12 = 0. The quadratic equation gives us the solutions x= 3 and x= 4.
(5) Solve for xin the equation log27 x= 1=3.
From log27 x= 1=3 we get 27log27 x= 271=3= 3. Since 27log27 x=x, we get x= 3.
(6) The position of a particle at time tseconds is s=t2+ 1=t meters. Find the average
velocity over the time interval [1;5].
The average velocity over the time interval [1;5] is
(52+ 1=5) (12+ 1=1)
51= 5:8 meters=second:
(7) Find lim
x!5+
x5
jx5jand lim
x!5
x5
jx5j.
If xapproaches 5 from the right then x > 5 and x5>0, which implies jx5j=x5.
This gives
lim
x!5+
x5
jx5j= lim
x!5+
x5
x5= lim
x!5+1 = 1:
If xapproaches 5 from the left then x < 5 and x5<0, which implies jx5j=(x5).
This gives
lim
x!5
x5
jx5j= lim
x!5
x5
(x5) = lim
x!51 = 1:
(8) Find lim
x!4+
10 x2
(x4)5and lim
x!4
10 x2
(x4)5.
As xapproaches 4 from the right, x4 approaches 0 and remains positive. This implies
that (x4)5approaches 0 and remains positive as xapproaches 4 from the right. This
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