Chapter 12 Sequences; Induction; the Binomial Theorem

Chapter 12: Sequences; Induction; the Binomial Theorem
Chapter 12 Cumulative Review
1.
2. a.
2
x = 9
2 2
x = 9 or x =−9
x =± 3 or x =± 3i
The solution set is { −3,3, − 3,3} i i .
b.
x
2 2
+ y = 100 and
2
y = 3x
.
2 2
⎪
⎧ x + y =
3
⎯⎯→
2
x +
2
y =
2 2
100 3 3 300
⎨
⎪⎩
y = 3 x ⎯⎯→− 3 x + y = 0
2
3y
+ y = 300
2
2
3y
+ y− 300=
0
( )( )
( )
− 1± 1 −4 3 −300 − 1±
3601
y = =
23 6
Substitute and solve for x:
− 1+ 3601 2 − 1+
3601
y = ⇒ 3x
=
6 6
2 − 1+ 3601 − 1+
3601
x = ⇒ x =±
18 18
or
−1−
3601 2 −1−
3601
y = ⇒ 3x
=
6 6
2 −1−
3601 −1−
3601
x = ⇒ x = ±
18 18
−1−
3601
which is not real since < 0
18
Therefore, the system has solutions:
⎧⎛
⎪ − 1+ 3601 − 1+
3601
⎞
⎨ ⎜
, ⎟ ,
⎪ ⎜ 18 6 ⎟
⎩ ⎝
⎠
⎛
⎜−
⎜
⎝
− 1+ 3601 − 1+
3601 ⎞⎫
, ⎟ ⎪ ⎬
18 6 ⎟
⎠⎭ ⎪
c. The graphs of the circle and parabola
intersect at the points
⎛ − 1+ 3601 − 1+
3601 ⎞
⎜
, ⎟≈ 1.81,9.84
⎜ 18 6 ⎟
⎝
⎠
⎛
−
⎜
⎝
x
3. 2e = 5
x 5
e =
2
x ⎛5
⎞
ln ( e ) = ln ⎜ ⎟
⎝ 2 ⎠
⎛5
⎞
x = ln ⎜ ⎟≈0.916
⎝2
⎠
− 1+ 3601 − 1+
3601 ⎞
, ⎟≈ −
18 6 ⎟
⎠
⎧ ⎛5
⎞⎫
The solution set is ⎨ln ⎜ ⎟⎬
⎩ ⎝ 2 ⎠⎭ .
( )
( 1.81,9.84)
4. slope = m = 5 ; Since the x-intercept is 2, we
know the point ( 2,0 ) is on the graph of the line
and is a solution to the equation y = 5x+ b.
y = 5x+
b
0= 5( 2)
+ b
0= 10+
b
− 10 = b
Therefore, the equation of the line with slope 5
and x-intercept 2 is y = 5x− 10.
5. Given a circle with center (–1, 2) and containing
the point (3, 5), we first use the distance formula
to determine the radius.
( ( )) ( )
2 2 2 2
r = 3− − 1 + 5− 2 = 4 + 3
= 16 + 9 = 25 = 5
Therefore, the equation of the circle is given by
2 2 2
x− − 1 + y− 2 = 5
( ( )) ( )
( x ) ( y )
2 2 2
+ 1 + − 2 = 5
2 2
x + 2x+ 1+ y − 4y+ 4=
25
2 2
x + y + 2x−4y− 20=
0
1284
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