01:640:151 Midterm: 01:640:151 Exam 1 2017 Fall SolutionExam
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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
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
cos(2x0) = cos2(x0)sin2(x0)=0:51 (0:7)2= 0:02:
sin(4x0) = 2 sin(2x0) cos(2x0) = 2((1:4)p0:51)(0:02) = 0:056p0:51
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)
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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
p1 + x2:
(4) Find all solutions of the equation
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
If xapproaches 5 from the right then x > 5 and x5>0, which implies jx5j=x5.
x!5+1 = 1:
If xapproaches 5 from the left then x < 5 and x5<0, which implies jx5j=(x5).
(x5) = lim
x!51 = 1:
(8) Find lim
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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