MCR3U Functions Final Exam Solutions - [Company Name] - [Date]

This document contains the solutions to the MCR3U Functions Final Exam. The exam covers a wide range of topics including evaluating functions, transformations of functions, factoring polynomials, quadratic equations, inverse functions, simplifying expressions, the discriminant, trigonometric applications, Ferris wheel problems, sequences and series, and graphical analysis. The solutions provide answers to multiple-choice questions, short answer questions and working steps for the complete exam. The document includes detailed explanations, and calculations to guide students through the problem-solving process. Specific questions involve finding the perimeter of a rectangle, factoring polynomials by grouping, using the quadratic formula, and determining the recursive formula for a sequence, graphing functions, and applying the sine law to solve for unknown sides and angles of a triangle, and determining the equation of sinusoidal functions. The document offers a comprehensive review of the key concepts and problem-solving techniques required for success in a functions course.

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