D-ERDW, D-HEST, D-USYS
Dr. Ana Cannas
Mathematics I
Fall 2015
MC-Sheet 2
For each question exactly one answer is correct. You are allowed to use a
formulary while solving the test.
1. Which of the following formulas is not a valid calculation rule for all
x, y, z > 1?
√
z
z
(a)
(xy ) = x(y ) .
(b)
xy
= xz .
xy−z
(c)
logx xyz = yz.
(d)
logx y z = z logx y.
It is valid for all x, y, z > 1,
xy
1
xy
=
= −z = xz ,
y−z
y
−z
x
x ·x
x
logx xyz = yz logx x = yz,
logx y z = z logx y
but
(xy )z = xyz
and therefore the first statement is generally wrong.
√
2. The inverse function g(y) of f (x) =
(a) g(y) = ln 1 + 2e−y .
1−y
(b) g(y) = ln
.
2y
(c)
g(y) = 1 + 2ey .
(d)
1
g(y) = 1 − e−y .
2
1
1+2ex
on the interval (0, 1) is:
We rewrite the equation f (x) = y such that x can be expressed as a function of
y:
1
1
1−y
= y ⇒ 1 + 2ex = ⇒ ex =
.
1 + 2ex
y
2y
applying the logarithm on both sides yields
1−y
x = ln
2y
It has to be noted that the inverse function is only defined on the interval (0, 1)
(This interval it the range of f (x)).
3. In the following image the red line is tangential to the blue curve in point
P . The curve corresponds to a function f : R → R. What is the value of the
derivative f 0 at point 1?
f HxL
P = H1,2L
-3
√
(a)
2
(b)
1
2
(c)
− 23
(d)
−2
The value f 0 (1) is the slope m which defines the tangent of the curve f in point
(1, f (1)). Therefore the slope is:
m=
∆y
2−0
2
1
=
= = .
∆x
1 − (−3)
4
2
4. The following three pictures show the graphs of three real valued functions
with real valued variables f (x), g(x) and h(x) where one is the derivative of the
other.
Which statement is true?
gHxL
fHxL
-3
3
x
x
-
√
(a)
f 0 = g.
(b)
g0 = f .
(c)
f 0 = h.
(d)
h0 = g.
hHxL
3
-3
3
x
3
The functions f (x) and h(x) both have two turning points on the interval [−3, 3]
and therefore there at those points the derivative has to be equal to 0. The
function f (x) is monotonic decreasing on (−∞, −3] (and [3, +∞)) and therefore
the derivative has negative values on those intervals. h(x) on the other hand is
monotonic increasing on (−∞, −3] (and [3, +∞)) and therefore the derivative of
h(x) has positive values on those intervals. Therefore the graph of g represents
the derivative of h(x).
π2
of the graph of
4
√
f (x) = cos x?
5. What is the slope of the tangent at x =
√
(a)
−1.
(b)
−π.
(c)
1
− .
π
(d)
−
π2
.
2
π2
of the graph f (x) corresponds to the first
4
π2
. One can calculate
derivative f 0 (x) evaluated at the point x =
4
The slope of the tangent at x =
√ 1
f 0 (x) = − sin x · √ .
2 x
And therefore
f0
π2
4
= − sin
π
2
·
1
1
1
=− .
π = −1 ·
22
π
π
6. Which of the following functions is strictly monotonic increasing on the interval ] − 1, 1[?
√
(a)
x 7→ x2
(b)
x 7→ |x| + x
(c)
x 7→ −e−x
(d)
x 7→ arccos x
On ] − 1, 0] x2 is strictly monotonic decreasing and |x| + x = 0 is constant therefore those can me eliminated. The function x 7→ arccos x is strictly monotonic
decreasing on the whole space. This means that only the exponential function
x 7→ −e−x is strictly monotonic increasing on this interval.
7. What is the derivative of
f (x) = 2x
√
2
+1
?
2
(a)
2x · 2x .
(b)
(x2 + 1) · 2x .
(c)
(ln 2)2x
(d)
4x(ln 2)2x .
2
2
+1
.
2
Using the chain rule one calculates f (x) = 2h(x) with h(x) = x2 + 1
f 0 (x) = h0 (x)(ln(2))2h(x)
this results in
f 0 (x) = 2x(ln 2)2x
2
+1
.
8. Which of the following statements is wrong?
The derivative of the function
(a) x(t) = sin e2t is ẋ(t) = 2e2t cos e2t .
Correct. This follows form the application of the chain rule
0
ẋ(t) = e2t cos e2t = 2e2t cos e2t .
√
(b)
x(t) =
2
1
+ t ln t, t > 0, is ẋ(t) = − 2 + ln t + t.
2
t
t
Wrong. Using the product rule the result is ẋ(t) = −
(c)
2
+ ln t + 1.
t3
2
2
x(t) = eln t+t , t > 0, is ẋ(t) = 1 + 2t2 et .
2
2
2
2
Correct. x(t) can be rewritten as x(t) = eln t et = tet , and therefore ẋ(t) = et + 2t2 et .
(d)
1
sin2 (t2 )
2
x(t) =
is ẋ(t) = 2t sin(t ) 1 +
.
cos(t2 )
cos2 (t2 )
1 − cos2 (t2 )
1
=
− cos(t2 ), and therecos(t2 )
cos(t2 )
−2t sin(t2 )
1
2
2
ẋ(t) = −
+
2t
sin(t
)
=
2t
sin(t
)
1
+
.
cos2 (t2 )
cos2 (t2 )
Correct. x(t) can be rewritten as x(t) =
fore
9. The equation
y 2 = x2 − sin(xy) + 1
defines y as differentiable function of x in the surroundings of (x, y) = (0, 1).
dy
What is the derivative
of this function?
dx
√
(a)
2y + x cos(xy)
.
2x − y cos(xy)
(b)
2y + x cos(xy)
.
−2x + y cos(xy)
(c)
2x − y cos(xy)
.
2y + x cos(xy)
(d)
−2x + y cos(xy)
.
2y + x cos(xy)
y(x) satisfies
(
y(x)2 − x2 + sin(xy(x)) − 1 = 0
y(0) = 1.
Differentiating the equation with respect to y(x) and using the chain rule yields
the result
⇐⇒
⇐⇒
d
y 2 − x2 + sin(xy) − 1 = 0
dx
dy
dy
2y
− 2x + cos(xy) x
+y =0
dx
dx
dy
(2y + x cos(xy))
= 2x − y cos(xy)
dx
and because 2y + x cos(xy) 6= 0 in the surroundings of (1, 0) it follows that
2x − y cos(xy)
dy
=
.
dx
2y + x cos(xy)
10. Let f (x) = π + 2 arctan(x) and g(x) the inverse function of f (x).
Which of the following statements is wrong?
(a)
f (x) is defined and differentiable for all real values x.
The domain of arctan is R, this does not change after scaling and displacement. Because
arctan is differentiable everywhere this is also true for f (x) and
f 0 (x) = 2 ·
(b)
1
.
1 + x2
The range of f (x) is (0, 2π).
The range of arctan is − π2 ,
π
2
. Therefore the range of 2 arctan is the interval (−π, π)
and the range of f is the interval (0, 2π).
√
(c)
g(x) is defined and differentiable for all real values x.
Because the domain of f is the interval (0, 2π) the inverse function g(x) is only defined
on this interval.
(d)
x
g(x) = cot π −
2
We rewrite the equation f (x) = y:
π + 2 arctan(x) = y
⇔
arctan(x) =
y−π
2
π
2
⇔
x = tan
− x we find that
π y
π y
tan(x) = cot
−
−
= cot(π − ),
2
2
2
2
Using the relation tan(x) = cot
this means that
respectively g(x) = cot π −
y
g(y) = cot π −
2
x
2
.
y
2
−
π
.
2