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KOC¸ UNIVERSITY
MATH 106
SECOND EXAM MAY 7, 2012
Duration of Exam: 105 minutes
INSTRUCTIONS: No calculators may be used on the test. No questions, and talking 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 10 2 15 3 15 4 15 5 15 6 15 7 15 TOTAL 100
Problem 1 (10 pts) Does there exist a differentiable function f : R → R such that f (0) = −1, f (2) = 4, and f ′(x) ≤ 2 for all x? Justify your answer. SOLUTION:
Claim. There is no differentiable function f : R → R such that f (0) = − 1 , f (2) = 4, and f ′(x) ≤ 2 for all x ∈ R.
Proof. Suppose that there exists a differentiable function f : R → R such that f (0) = − 1 , f (2) = 4, and f ′(x) ≤ 2, for all x ∈ R. Then f is differentiable on (0, 2) and continuous on [0, 2]. So, by the Mean Value Theorem there exists a number c ∈ (0, 2) such that
f ′(c) = f^ (2) 2 −−^ f 0 (0)=^52 > 2
This contradicts to the assumption that f ′(x) ≤ 2 for all x ∈ R which proves our claim.
Problem 2 (15 pts) Find lim x→∞(ex^ + x)^1 /x^ if it exists. SOLUTION: Let f (x) = (ex^ + x)^1 /x^ ⇒ ln(f (x)) = (^1) x · ln(ex^ + x) = ln(e
x (^) + x) x ⇒ (^) xlim→∞ ln(f (x)) = (^) xlim→∞^ ln(e
x (^) + x) x.^ This is the indeterminate form^
∞∞. We apply
L’Hospital’s Rule to calculate
xlim→∞^ ln(e
x (^) + x) x = lim^ x→∞
ex^ + 1 ex^ + x = lim^ x→∞
ex ex^ + 1 = 1. Then lim x→∞ f (x) = lim x→∞ eln(f^ (x))^ = e^1 = e.
Problem 4 (15 pts) Sketch the region enclosed by the curves y = cos(πx) and y = 4x^2 − 1, and find its area. SOLUTION:
In the figure above, the red curve is the graph of y = cos(πx), while the magenta curve is the graph of y = 4x^2 − 1. They intersect with each other at (− 1 / 2 , 0) and (1/ 2 , 0), hence the integral giving the area of the region is as follows:
A =
x=− 1 / 2
(cos(πx) − (4x^2 − 1))dx = 2
x=
(cos(πx) − (4x^2 − 1))dx
This is true because (cos(πx) − (4x^2 − 1)) is an even function. It follows that
A = 2
(sin(πx) π −^
4 x^3 3 +^ x
1 / 2 x=
=^23 + π^2
.
Problem 5 (15 pts) Find the volume of the solid obtained by rotating the region bounded by y = x^2 and x = y^2 about the line y = 1.
SOLUTION:
volume =
∫^1
0
π(1 − x^2 )^2 dx −
∫^1
0
π(1 − √x)^2 dx
∫^1
0
π(1 − 2 x^2 + x^4 )dx −
∫^1
0
π(1 − 2 √x + x)dx
= π
x − 2 x
3 3 +^
x^5 5
1 0
− π
x − 4 x
√x 3 +^
x^2 2
1 0
= π
1 − 23 +^15 − (0 − 0 + 0)
− π
1 − 43 +^12 − (0 − 0 + 0)
=^1130 π
