myers (nmm698) – HW01 – Hamrick – (54868)
2
003
10.0 points
Which, if any, of the following statements are
true when
a, b
are real numbers?
A.
For all positive
a
and
b
,
√
a
+
√
b
=
r
a
+ 2
√
ab
+
b .
B.
For all
a
and
b
,
r
(
a
+
b
)
2
=
a
+
b .
C.
For all positive
a
and
b
,
a
−
b
√
a
+
√
b
=
√
a
−
√
b .
1.
A and C only
correct
2.
A and B only
3.
B and C only
4.
A only
5.
C only
6.
B only
7.
all of them
8.
none of them
Explanation:
A. TRUE: by the known product,
(
x
+
y
)
2
=
x
2
+ 2
xy
+
y
2
.
On the other hand,
r
(
x
+
y
)
2
=

x
+
y

,
so if
x
+
y >
0,
x
+
y
=
R
x
2
+ 2
xy
+
y
2
.
But if
a, b
are positive we can set
x
=
√
a
and
y
=
√
b
. The result follows since
x
and
y
are
then positive.
B. FALSE:
r
(
x
+
y
)
2
=

x
+
y

,
and since
R
(
·
) is always nonnegative, the
right hand side has to be nonnegative. But
if
a, b
can be positive or negative, an absolute
value sign is then needed on the right.
C.
TRUE: by the known di±erence of
squares factorization,
x
2
−
y
2
= (
x
−
y
)(
x
+
y
)
.
But if
a, b
are positive we can set
x
=
√
a
and
y
=
√
b
. The result follows after division.
keywords: square root, properties of square
root, PlaceUT, TrueFalse, T/F,
004
10.0 points
By removing absolute values, express
f
(
x
) =
4
−
x
2

x
−
2

as a piecewisede²ned function.
1.
f
(
x
) =
b
2
−
x ,
x >
−
2
,
x
−
2
,
x <
−
2
.
2.
f
(
x
) =
b
x
+ 2
,
x >
−
2
,
−
(
x
+ 2)
,
x <
−
2
.
3.
f
(
x
) =
b
x
+ 2
,
x >
2
,
−
(
x
+ 2)
,
x <
2
.
4.
f
(
x
) =
b
−
(
x
+ 2)
,
x >
2
,
x
+ 2
,
x <
2
.
cor
rect
5.
f
(
x
) =
b
x
−
2
,
x >
2
,
2
−
x ,
x <
2
.
6.
f
(
x
) =
b
x
−
2
,
x >
−
2
,
2
−
x ,
x <
−
2
.