MCR3U FINAL EXAM

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FUNCTIONS MCR3U
FINAL EXAM
[DATE]
[Company name]
[Company address]

Question 1 – D
Question 2 - C
Question 3 - C
Question 4 - B
Question 5 - A
Question 6 – A
Question 7 – B
Question 8 – C
Question 9 – D
Question 10 – C
Question 11 – D
Question 12 – B
Question 13 – B
Question 14 – D
Question 15 – C
Question 16 – C
Question 17 – C
Question 18 – B
Question 19 – C
Working
Question 20 - C
Question 21- B
Question 22 - C
Question 23 - B
1

Question 24
Length of rectangle = ( 6 x−29 )
Width of rectangle ¿ ( 5 x−17 )
Perimeter of rectangle plot =?
Perimeter of rectangle plot ¿ 2(Length+Width)
¿ 2 { ( 6 x−29 ) + ( 5 x−17 ) }
¿ 2 ( 11 x−46 )
¿ 22 x−92
Question 25
2 x2 −15 x +7
¿ 2 x2 − ( 14 x+ x ) +7
¿ 2 x2 −14 x−x+ 7
¿ 2 x ( x−7 )−1 ( x −7 )
¿ ( x−7 ) ( 2 x−1 )
Question 26
f ( n )=n3 −2 n2+4 n−8
Factor n2 from the first two terms and factor 4 from the last 2 terms.
f ( n )=n2 ( n−2 )+4 ( n−2 )
You now have two terms both containing a factor of (n−2). Factor(n−2) from both of these
terms.
f ( n )= ( n−2 ) ( n2 +4 )
Question 27
5 √192
¿ 5 √26∗3
¿ 5∗23 √ 3
¿ 40∗√3
¿ 69.28
2

Question 28
−7 x2−23 x=−10
Hence, x=−3.67449 , 0.38878
Question 29
43
160 ( 52∗5−3 ) −2
¿ 43
1 ( 52
53 )−2
¿ 43
( 1
5 )
−2
Question 30
g ( x ) =−2 ( 32 x−6 ) +9
f ( x )=3x
Here,
Sequence of transformation is described below.
3

g ( x ) =−2 ( 32 x−6 ) +9
f ( x )=3x
Square of f(x)
f 1 ( x )=32 x
Multiply by 3−6
4

f 2 ( x )=32 x−6
Multiply by -2
f 3 ( x )=−2 (32 x−6 )
Add -9
f 4 ( x )=−2 ( 32 x−6 ) +9
Hence,
g ( x ) =−2 ( 32 x−6 ) +9
Question 31
Sequence
349,321,293,265….. ?
It is an arithmetic progression
Where,
a 1=349
d=321−349=−28
Now,
an=a 1+ ( n−1 ) d
¿ 349+ ( n−1 ) (−28 )
¿ 349−28 n+ 28
¿ 377−28 n
Hence,
an=377−28 n ( n≥ 1 )
Question 32
5 th term=64
14 th term=−1
Now,
an=a+ ( n−1 ) d
a5=a+ ( 5−1 ) d
64=a+4 d
5

Similarly,
−1=a+ ( 14−1 ) d
−1=a+13 d
Hence,
a= 836
9 , d =−65
9
Sum of 20 terms
Sn= n
2 [ 2 a+ ( n−1 ) d ]
¿ 20
2 [ 2∗836
9 + (20−1 ) (−65
9 ) ]
¿−75240
577
Question 33
(a) Graph
f ( x ) =−4 ( x−2 )2 +11
6

(b) f (4 )
From The graph, it is apparent that when the value of x is 4 then the graph is turned to
negative y direction and crossed at -5 and thus, the value of function at x = 4 would be -5.
(c) f ( 4 )+ f ( 1 )
f ( 4 )=−4 ( 4−2 )2 +11=−4 ( 4 ) +11=−16 +11=−5
f ( 1 ) =−4 (1−2 )2 +11=−4 (−1 )2 +11=−4+11=7
f ( 4 )+ f ( 1 )=−5+7=2
Question 34
4 x2−2 x
x2 +5 x +4 ÷ 2 x2 +2 x
x2 +2 x +1
7

Apply divide fraction rule here
a
b
c
d
= ad
bc
Hence,
¿
( 4 x2−2 x )
2 x2 +2 x ∗x2 +2 x+ 1
( x2+5 x + 4 )
¿
2 x ( 2 x−1 )
2 x (x +1) ∗x2+ 2 x +1
( x2 +5 x +4 )
¿
( 2 x−1 )
( x +1) ∗x2+ 2 x +1
( x2 +5 x +4 )
¿
( 2 x−1 )
( x +1) ∗( x+1)( x+ 1)
( x2 +5 x+ 4 )
¿
( 2 x−1 )
1 ∗( x+1)
( x2+ 5 x + 4 )
¿
( 2 x−1 )
1 ∗( x+1)
( x+1 ) ( x+ 4 )
¿ 2 x−1
x +4 Correct answer
Here, Julian has taken wrong common factor which makes her answer wrong.
Question 35
Sides and angles of tringle =?
8

Sin Law
a
sinA = b
sin B = c
sinC
20.1
sin A = 28.6
sin101 = c
sin C
sin A=20.1
28.6 ∗sin 101= 20.1
28.6 ∗0.98162=0.68988
A=sin−1 (¿ 0.68988)=43.6 20 ¿
Angle C = 180 – (A+B) = 180 – (43.62 + 101) = 35.3 70
Now,
c
sin C = 20.1
0.68988 =29.135
c
sin 35.37 =29.135
c=0.57885∗29.135=16.8649
a
sinA = c
sin C
a=0.68988∗16.8649=11.6346
Hence,
a=11.63 , b=28.6 ,c =16.86
A=43.6 20 , B=10 10 , C=35.3 70
Question 36
9

 Let the base function is cox ( x )
Amplitude of the function ¿ 5− ( −3 )
2 =4 vertical
Period ¿ 15
Regular period ¿ 2 π
Now,
B= Regular period
Period =2 π 1
15
Middle axis=5+ −3
2 =1
Reflection about x axis = -f(x)
Phase shift = 0
Hence, the equation can be drawn based on the graph as
y= A cos ( Bx ± C )+ D
y=−4 cos ( 2 π
15 x )+1
Now,
 Let the base function is sin ( x )
Amplitude of the function ¿ 5− ( −3 )
2 =4 vertical
Period ¿ 15
10

Regular period ¿ 2 π
Now,
B= Regular period
Period =2 π 1
15
Middle axis=5+ −3
2 =1
Reflection about x axis = -f(x)
Phase shift = 2 units
Hence, the equation can be drawn based on the graph as
y= A sin ( Bx ± C ) + D
y=−4 sin ( 2 π
15 x ± π )+1
11