THE INTERNATIONAL UNIVERSITY(IU) - VIETNAM NATIONAL UNIVERSITY - HCMC
MIDTERM EXAMINATION
Semester 2, 2014-15 • Date: March 31, 2015 • Duration: 90 minutes
Department of Mathematics
CALCULUS I
Lecturer
Chair:
NV Thu, N Dinh, PHA Ngoc, NM Quan
TT Duong, TV Linh
Assoc. Prof. Nguyen Dinh
INSTRUCTIONS: Each student is allowed a scientific calculator and a maximum of two double-sided sheets of reference material (size A4 or similar),
stapled together and marked with their name and ID. All other documents
and electronic devices are forbidden.
1. (i) (15 marks) Find the value of the constant k so that the function
(
kx2
if x ≤ 2
f (x) :=
x + k if x > 2
is continuous on (−∞, ∞). √
.
(ii) (10 marks) Find limx→1 x2x−1
−1
2. (15 marks) Show that the equation x3 − 2015x2 + 2x + 3 = 0 has three
distinct real roots.
3. The height (in feet) of an object projected up into the sky is
h(t) = −16t2 + at,
where t is the time after projection measured in seconds and a is a constant.
Suppose an object was projected at t = 0 and dropped down on the ground
at t = 3 (three seconds after projection).
a. (5 marks) Find the number a.
b. (10 marks) Find the greatest height that the object has achieved.
c. (5 marks) What is the falling speed of the object when it hit the ground?
4. (i) (10 marks) Show that the function f (x) = |x − 2| is continuous but
not differentiable at x = 2.
(ii) (15 marks) Find the derivative of y = (sin x)cos x , 0 < x < π.
5. (15 marks) Let g be a differentiable function on (−∞, ∞). Assume that
g(x) + x sin(g(x)) = x2 ,
∀x ∈ (−∞, ∞).
Find g 0 (0) and write the equation of the tangent line to the graph of g at the
point (0, 0).
End.
Solutions:
1. (i) Clearly, f is continuous on (−∞, 2] and (2, ∞). Thus, f is continuous
on (−∞, ∞) if and only if it is continuous from the right at 2. That is,
lim f (x) = f (2),
x→2+
or equivalently,
2 + k = 4k.
This gives k = 2/3.
(ii)
√
√
x−1
x−1
1
√
= lim √
= lim √
= 1/4.
lim 2
x→1 ( x − 1)( x + 1)(x + 1)
x→1 ( x + 1)(x + 1)
x→1 x − 1
2. Let f (x) = x3 − 2015x2 + 2x + 3. Note that f (x) is a polynomial, so f (x)
is continuous everywhere. On the other hand, we have f (−1) = −2015 < 0,
f (0) = 3 > 0, f (1) = −2009 < 0, and f (2015) = 2 × 2015 + 3 = 4033 > 0.
By the intermediate value theorem, there exists a real root in each following
interval (−1, 0), (0, 1), and (1, 2015). Thus, the given equation has three
distinct real roots.
3. a. Since h(3) = 0, it follows that −144 + 3a = 0. Thus, a = 48.
b. The function h is now given by h(t) = −16t2 + 48t = −4(2t − 3)2 + 36.
48
= 32 .
Hence, the greatest height is 36 feet achieved at t = − −32
c. |h0 (3)| = | − 32(3) + 48| = 48 feet per second.
4. (i) Clearly, limx→2 f (x) = 0 = f (2). Thus, f is continuous at x = 2. On
the other hand,
f (2 + h) − f (2)
h
lim+
= lim+ = 1
h→0
h→0 h
h
and
f (2 + h) − f (2)
−h
lim−
= lim−
= −1.
h→0
h→0
h
h
Therefore, f is not differentiable at 2.
(ii) Note that
ln y = cos x ln(sin(x)),
0 < x < π.
This implies
1 dy
cos x
= − sin x ln(sin x) + cos x
.
y dx
sin x
Therefore,
dy
cos x
= (sin x)cos x [− sin x ln(sin x) + cos x
].
dx
sin x
5. Note that g(0) = 0, and so (0, 0) is on the graph of g. Using the implicit
differentiation method, we get
g 0 (x) + xg 0 (x) cos(g(x)) + sin(g(x)) = 2x,
and hence (when taking x = 0)
g 0 (0) + sin(g(0)) = 0.
Since g(0) = 0, the last equation gives g 0 (0) = 0. As (0, 0) is on the graph of
g and g 0 (0) = 0, the tangent line to the graph of g at (0, 0) is y = 0, which is
the Ox axis.