MCR3U MATHS Chapter 4 Test Solution - Complete Answers

This document presents a complete solution to a MATHS homework assignment, specifically a Chapter 4 test from the MCR3U course. The solution addresses a variety of problems, including simplifying expressions with exponents, evaluating expressions, solving exponential equations, describing transformations of functions, modeling population growth, calculating half-life, and determining the domain, range, and asymptotes of a function. Each question is answered with detailed steps, explanations, and relevant mathematical principles. The document utilizes power rules, Euler's formula, and laws of indices to provide comprehensive answers. The assignment covers topics such as exponents, functions, and their transformations, exponential growth and decay, and the analysis of functions including their domains, ranges, and asymptotes. Reference to a textbook is also provided.

Mathematics 1
MATHS
by [NAME]
Course
Professor’s Name
Institution
Location of Institution
Date

Mathematics 2
Question 1
a) 4
√ √ y5
Written in power form;
[ [ y5 ] 1
2 ] 1
4
Applying power rules;
y5 × 1
2 × 1
4
¿ y
5
8
b) (−4 x−5 y2 ) ( 18 x10 y−3 )
−3 x−3 y
¿ (−4 x−2 y2 ) ( 6 x10 y−4 )
−1
¿ ( 4 x−2 y2 ) ( 6 x10 y−4 )
¿ ( 2 4 x8 y−2 )
c) ( 36 m−6 n−1
16 m2 n−5 ) 3
2
¿ ( 9 m−8 n4
4 )3
2
¿ ( 27 m−12 n6
8 )
d) ( 54 x−5 y6
6 x2 y−3 ) 1−2
¿ ( 9 x−7 y9
1 )−2

Mathematics 3
¿ 9 x14 y−18
Question 2
a) (−81 )
1
4
Assume that −81=a+bi where i= √−1
Plotting polar coordinates we find that;
a=r cos θ
b=r sin θ
Therefore, a+ bi=r (cos θ+isin θ)
Using Euler’s formula;
cos θ+ isin θ=eiθ
Therefore, a+ bi=r eiθ ⟹−81=r eiθ
So if r =81then eiθ=−1
For eiθ=−1 thenθ= ( 2 k−1 ) π wherek =1,2,3,4 … .
Assume that ¿
Therefore, α4 =81 ei ( 2 k−1 ) π
α2=± 9 e
i ( 2 k−1 ) π
2
¿
¿
α =± 3 e
i ( 2 k−1 ) π
4
¿
∨α=± i 3 e
i ( 2k−1 ) π
4
¿
¿
¿
Going back to Euler’s method
α =± 3 ¿
Starting with the positive value first, α = (3 (± √2
2 )+ ± √ 2
2 )=1.5 √2+1.5i √2

Mathematics 4
For negative value, α =− ( 1.5 √2+1.5 i √2 )
For α =± i3 e
i (2 k−1 ) π
4
¿
¿
Starting with the positive value first, α =−1.5 √ 2+1.5 i √2
For negative value, α = ( 1.5 √ 2−1.5 i √ 2 )
Therefore (−81 )
1
4 has four possible values .
1.5 √2+1.5 i √2
− ( 1.5 √ 2+1.5i √ 2 )
−1.5 √ 2+ 1.5i √2
( 1.5 √ 2−1.5 i √ 2 )
b) ( 16
49 )
−1.5
¿ 1
( 16
49 ) 3
2
= ( 49
16 ) 3
2
This is like finding the square root then cubing the result.
¿ ( 7
4 )
3
¿ ( 343
64 )
c) 5−7 × ( 52 ) 4
5−3
¿ 5−7 ×58
5−3 According to the law of indices,
¿ 5−7+ 8
5−3 = 51
5−3 =54

Mathematics 5
¿ 625
Question 3
a) 3x= 1
243
Factorizing the denominator,
3x= 1
35
3x=3−5
Because both are in base 3
x=−5
b) 322+2 x=4−2 x
Factorization to base 2
25 (2+ 2 x)=22 (−2 x)
Equating the indices
10+10 x=−4 x
10=−14 x
x=−10
14
x=−5
7
Question 4
g ( x )=0.4 ( 6−0.5 x−7 )−1 is the transformation for f ( x )=6x

Mathematics 6
The stretch factor 0.4 leads to a compression of the curve and a vertical reflection. The curve
also shifts down vertically by 1 step thus moving the horizontal asymptote to y=−1.
Additionally, another compression factor of -0.5 leads to a horizontal reflection while the
horizontal shift of -7 leads to a shift to the right by 7 units.
Question 5
a) The population doubles in 2 hours. This implies that the population increases by 50%
after every hour. Taking the initial population as a and time as t, we can find the equation
at t hours as;
P ( t ) =120 (2
t
2 )
b) After 210 minutes
210 minutes=3.5 hours
Therefore;
P ( t ) =120 ( 2
3.5
2 )
¿ 403.63
But since cells can only be in whole numbers, total number after 210 minutes
¿ 404
Question 6
Sodium-24 has a half-life of 15 hours.
f ( x )=a ( 1
2 ) x
H
a) a=4 g , x=30 hours∧H =15 hours
f ( x ) =4 ( 1
2 ) 30
15

Mathematics 7
¿ 4 ( 1
2 )
2
=4 × 0.25
¿ 1 g
b) After 7.5 hours
f ( x ) =4 ( 1
2 ) 7.5
15
¿ 4 ( 1
2 )
0.5
=4 × 0.70711
¿ 2.828 g
To the nearest 10th of a gram
¿ 2.8 g
c) Decay to 0.5g
0.5=4 ( 1
2 ) x
15
0.125=( 1
2 ) x
15
log 0.125=log ( 1
2 ) x
15
log 0.125= x
15 log 0.5
−0.903089987= x
15 (−0.301029995)
3= x
15
x=45 hours
Question 7

Mathematics 8
f ( x ) =abx
Where b=1+r
Therefore,
f ( x )=a(1+r )x
Where;
f ( x)=final population
a=initial population
r =percentage growthrate
Therefore,
19000=12600 ( 1+r )12
1.507936508= ( 1+r ) 12
12
√1.507936508=1+ r
1.034821=1+r
r =0.034821
Question 8
f ( x ) =2 ( 4 ) x+3−5
a) Domain and range
Domain is all real numbers.
Range: f ( x ) ∈ R where f ( x ) ≥−5
b) y-intercept
Substitute x with 0.
2 ( 4 )3−5=123

Mathematics 9
c) asymptotes
Vertical asymptote: none
Horizontal asymptote: none
This is because the degree of the numerator is not equal to the degree of the denominator.
d) Sketch

Mathematics 10
Reference

Mathematics 11
Bird J. Engineering mathematics. 7th ed. Oxford: Routledge; 2010.