Calculus and Vectors Unit 1 Assessment Questions
2
x +3 x +10
(
)
1. f x = x2
( x+5 )( x2 )
(x 2)
f ( x )=
f ( x )=(x+5)
f ( x )=( 2 ) +5
f ( x )=7
The limit of the function is 7.
4 x27
lim
2. x 4 x3
4 ( 4 )27
lim
43
x4
lim
x4
lim
x4
4 ( 16 )7
43
647
43
l
Unit 2: Rates of change and derivatives (Part B)
Knowledge and Understanding questions
1. Evaluate the following limits algebraically:
lim 4 x 2 +1 (2 marks)
a)
x 3
x = 3 so we need to sub x = 3 into the equation.
3 2+1
f (3)=4
f (3)=4 (9)+1
f (3)=36+1
f
Unit 1: Rates of Change and derivatives (Part A) Assessment
Knowledge and Understanding
2
x +3 x10
1. Determine the limit of the function
as x approaches 2.
x2
x = 2 so we need to sub x = 2 into the equation but before we sub x = 2 into the
f (x) =
equati
Assignment 5
1. We will start by drawing a diagram of the two vectors, tail to tail, with an angle of 55
Using the parallelogram law of vector addition, completing the parallelogram and
drawing the diagonal that represents the sum of u and v
Finding Size
Assignment 6
a =(7,1,2)
1.
( 5,4,4 )
b=
cos=
We know that,
In 3-space, the dot product of
cos=
Hence,
u . v
|u |v|
u=( x1 , y 1 , z 1 )v =( x 2 , y 2 , z 2 ) is u v =x 1 x 2 + y 1 y 2 + z 1 z 2
( 7,1,2 ) . ( 5,4,4 )
7 + 1 +(2 ) 5 + 4 +4
2
1
( 7 )( 5 ) +
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MCV4U-B Lesson 2
Using Equation Editor in Microsoft Word 2007, 2010, and 2013
Using Equation Editor in Microsoft Word 2007, 2010, and 2013
To use Equatio
Assessment 3
4
x
x
1. a) f ( x )=x 3 +e 7
Using sum and differences rule,
x
Also for exponential functions, that is if g ( x ) =a where a 1 ,
'
x
g ( x )=k a where k=lna
Hence,
'
3
x
x
f ( x )=4 x ln 3(3 )+e
4 x 31.1 ( 3 x ) +e x
2
b) g ( x ) =5 ( sinx )
Task 1: Knowledge and Understanding questions
2
1.Determine the limit of the function
f(x )=
x +3 x10
x2
as x approaches 2
Solution: As x approaches 2, the value of the function,
22 +3(2)10 4 +610 0
f ( 2 )=
=
=
22
0
0
is indeterminate. Hence we will have
Assignment 2
Task 1
2
4 x +1
1) a. lim )
x 3
3
=4( 2 +1
36+1=37
2
x +2 x8
b . lim 2
x 2 x 7 x+10
When we substitute x=2 the denominator becomes 0, resulting in an indeterminate form.
Hence we will try to find common factors
lim
x 2
( x2 ) ( x + 4)
x 2 +4
Linear functions are polynomials with a degree of 1.
When the slope of a line is positive, we say the line is
increasing, which means it goes up from left to right.
When the slope of a line is negative, we say the line is
decreasing, which means it goe
Komal Patel
ILC no: -16295012
MCV4UC
Unit 2
Task 1: Knowledge and understanding functions
1. Find the first derivative of each of the following functions. (10 marks: 2
marks each)
a) f ( x )=x 4 3 x +e x 7
f ' ( x )=4 x 3 3x (ln 3)+e x
2
b)
g ( x ) =5 ( s
Unit Two Mark: 89 /100 = 89%
Komal Patel
ILC no: -16295012
MCV4UC
Unit 1
Task 1: Knowledge and Understanding questions
1. 11/11 Evaluate the following limits algebraically
2
4 x + 1 (2 marks)
a. lim
x 3
2
4 ( 3 ) +1
4 ( 9 ) +1
37
lim 4 x 2+ 1=37
x 3
x
Komal Patel
ILC no: -16295012
MCV4UC
Unit 1
Task 1: Knowledge and Understanding questions
2
x +3 x10
1. Determine the limit of the function f ( x )=
as x approaches 2.
x2
Solution:
x 2+3 x10
f ( x )=
x2
(x +5)( x2)
( x2)
f ( x )=x +5
f ( 2 )=2+5
f ( 2 )=7