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KOC¸ UNIVERSITY
MATH 106
FIRST EXAM MARCH 26, 2012
Duration of Exam: 90 minutes
INSTRUCTIONS: No calculators may be used on the test. No questions, and talk- ing allowed. You must always explain your answers and show your work to receive full credit. Use the back of these pages if necessary. Print (use CAPITAL LETTERS) and sign your name. GOOD LUCK!
Solutions by Ali Alp Uzman and Candan G¨ud¨uc¨u
(Check One): (Emre Alkan) :(Burak Ozbaˇ¨ gcı):^ —–—–
PROBLEM POINTS SCORE 1 20 2 10 3 10 4 15 5 15 6 15 7 15 TOTAL 100
Problem 1a (5 pts) Find the limit
xlim→ 2 +
x^2 − 2 x − 8 x^2 − 5 x + 6
ANSWER:
Notice that
xlim→ 2 +
x^2 − 2 x − 8 x^2 − 5 x + 6 = lim x→ 2 + (x + 2)(x − 4) (x − 2)(x − 3)
since the numerator approaches to −8 and the denominator approaches to 0 with
negative values. Hence this limit does not exist.
Problem 1b (5 pts) Find the limit
lim t→ 0
t
1 + t
t
ANSWER:
lim t→ 0
t
1 + t
t
= lim t→ 0
t
1 + t
1 + t t
1 + t
= lim t→ 0
1 + t t
1 + t
= lim t→ 0
1 + t t
1 + t
1 + t 1 +
1 + t
= lim t→ 0
−t t
1 + t(1 +
1 + t)
Problem 1c (5 pts) Find the limit, if it exists. If the limit does not exist, explain
why.
xlim→− 6
2 x + 12 |x + 6| ANSWER:
Since lim x→− 6 +
2 x+ |x+6| = 2^6 =^ −2 =^ x→−lim 6 − 2 x+ |x+6| , we conclude that^ xlim→− 6 2 x+ |x+6| does not exist.
Problem 1d (5 pts) Using the Squeeze theorem, find the limit
xlim→ 0
x^3 + x^2 sin π x
Problem 3 (10 pts) Is there a number that is exactly 1 more than its cube? Justify
your answer.
ANSWER:
We need to find a real number x such that x = x^3 + 1 ⇒ x^3 − x + 1 = 0.
Let f (x) := x^3 − x + 1. Observe that f (−2) = −5, while f (−1) = 1. So
f (−2) = − 5 < 0 < 1 = f (−1). Then by Intermediate Value Theorem, there is
c ∈ (− 2 , −1) such that f (c) = c^3 − c + 1 = 0, and hence c = c^3 + 1.
Problem 4 (15 pts) Find the horizontal and vertical asymptotes of the curve
y = x^2 − x x^2 − 6 x + 5.
ANSWER:
x^ lim→∞
x^2 − x x^2 − 6 x + 5 = lim x→∞
x^2 (1 − 1 x
x^2 (1 −
x
x^2
x→−∞lim^ x
(^2) − x x^2 − 6 x + 5 = (^) x→−∞lim
x^2 (1 −
x
x^2 (1 −
x
x^2
Thus y = 1 is the only horizontal asymptote of the curve.
lim x→ 5 −
x^2 − x x^2 − 6 x + 5 = lim x→ 5 −
x(x − 1) (x − 1)(x − 5) = lim x→ 5 −
x (x − 5)
lim x→ 5 +
x^2 − x x^2 − 6 x + 5 = lim x→ 5 +
x(x − 1) (x − 1)(x − 5) = lim x→ 5 +
x (x − 5)
Thus x = 5 is the only vertical asymptote of the curve.
xlim→ 1 −
x^2 − x x^2 − 6 x + 5 = lim x→ 1 − x(x − 1) (x − 1)(x − 5) = lim x→ 1 − x (x − 5)
lim x→ 1 +
x^2 − x x^2 − 6 x + 5 = lim x→ 1 +
x(x − 1) (x − 1)(x − 5) = lim x→ 1 +
x (x − 5)
= −^1
x=1 isn’t a vertical asymptote of the curve.
Problem 5 (15 pts) Find an equation of the tangent line to the curve y = x
x
that is parallel to the line y = 3x + 1.
ANSWER:
Let (a, a
a) be a point on the graph of y = x
x where the slope of the tangent line
is 3. The only solution to the equation
a 2 = 3 is^ a^ = 4, and thus^ y^ =^ a
a = 8.
Therefore the tangent line at (4, 8) is given by y − 8 = 3(x − 4) or equivalently
y = 3x − 4.
Problem 6 (15 pts) Find dydx by implicit differentiation from the equation
ey^ cos x = 1 + sin(xy).
ANSWER:
d(ey^ cos x) dx
d(1 + sin(xy)) dx
ey^ cos x dy dx +^ e
y(− sin x) = y cos(xy) + x cos(xy) dy dx
(ey^ cos x − x cos(xy)) dy dx = y cos(xy) + ey^ sin x
dy dx = y^ cos(xy) +^ e
y (^) sin x ey^ cos x − x cos(xy)
Problem 7 (15 pts) Two sides of a triangle have lengths 12m. and 15m. The
angle between them is increasing at a rate 2 degrees per minute. How fast is the
length of the third side increasing when the angle between the sides of fixed length
is 60 degrees.
ANSWER:
Let θ(t) denote the angle depending on time t, between two sides of fixed length.
Also we denote the length of the third side with c(t) which is time dependent. We
are given the rate of change of θ(t) with respect to time, that is,
d dt θ(t) = 2 degree/minute = π 90 radian/minute
