Articles

1: Chapter 1 - Functions - Mathematics


  • 1.1: Relations and Functions
    • 1.1E: Exercises
  • 1.2: The Graph of a Function
    Descartes introduces his coordinate system, a method for representing points in the plane via pairs of real numbers. Indeed, the Cartesian plane of modern day is so named in honor of Rene Descartes, who some call the “Father of Modern Mathematics.” A Cartesian Coordinate System consists of a pair of axes, usually drawn at right angles to one another in the plane, one horizontal (labeled x) and one vertical (labeled y).
  • 1.3: Slope
    In the previous section on Linear Models, we saw that if the dependent variable was changing at a constant rate with respect to the independent variable, then the graph was a line. You may have also learned that higher rates led to steeper lines (lines that rose more quickly) and lower rates led to lines that were less steep. In this section, we will connect the intuitive concept of rate developed in the previous section with a formal definition of the slope of a line.
  • 1.4: Equations of Lines
    In this section we will develop the slope-intercept form of a line. When you have completed the work in this section, you should be able to look at the graph of a line and determine its equation in slope-intercept form.
  • 1.5: The Point-Slope Form of a Line
    In the last section, we developed the slope-intercept form of a line (y = mx + b). The slope-intercept form of a line is applicable when you’re given the slope and y-intercept of the line. However, there will be times when the y-intercept is unknown.
  • 1.6: Vertical Transformations
    In this section we study the art of transformations: scalings, reflections, and translations. We will restrict our attention to transformations in the vertical or y-direction. Our goal is to apply certain transformations to the equation of a function, then ask what effect it has on the graph of the function.
  • 1.7: Horizontal Transformations
    In the previous section, we introduced the concept of transformations. We made a change to the basic equation y = f(x), such as y = af(x), y = −f(x), y = f(x) − c, or y = f(x) + c, then studied how these changes affected the shape of the graph of y = f(x). In that section, we concentrated strictly on transformations that applied in th vertical direction. In this section, we will study transformations that will affect the shape of the graph in the horizontal direction.

Mathematics Part I Solutions for Class 10 Math Chapter 1 - Linear Equations In Two Variables

Mathematics Part I Solutions Solutions for Class 10 Math Chapter 1 Linear Equations In Two Variables are provided here with simple step-by-step explanations. These solutions for Linear Equations In Two Variables are extremely popular among Class 10 students for Math Linear Equations In Two Variables Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part I Solutions Book of Class 10 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part I Solutions Solutions. All Mathematics Part I Solutions Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

Page No 4:

Question 1:

Complete the following activity to solve the simultaneous equations.
5x + 3y = 9 -----(I)
2x + 3y = 12 ----- (II)

Answer:

Disclaimer: There is error in the Q. In (II) there should have been 2x - 3y = 12
5x + 3y = 9 -----(I)
2x - 3y = 12 ----- (II)
Add (I) and (II)
7x = 21
x = 3
Putting the value of x = 3 in (I) we get
5 3 + 3 y = 9 ⇒ 15 + 3 y = 9 ⇒ 3 y = 9 - 15 = - 6 ⇒ y = - 2
Thus, (x, y) = (3, - 6).

Page No 5:

Question 2:

Answer:

(1) 3a + 5b = 26 . (I)
a + 5b = 22 . (II)
Subtracting (II) from (I)
2a = 4
a = 2
Putting the value of a = 2 in (II)
5b = 22 - 2 = 20
b = 20 5 = 4
Thus, a = 2 and b = 4.

(3) 2x &ndash 3y = 9 . (I)
2x + y = 13 . (II)
Subtracting (II) from (I) we get
&ndash 3y &minus y = 9 &minus 13
⇒ - 4 y = - 4 ⇒ y = 1
Putting this value in (I) we get
2 x - 3 1 = 9 ⇒ 2 x = 9 + 3 = 12 ⇒ x = 12 2 = 6
Thus, (x, y) = (6, 1)

(4) 5m &ndash 3n = 19 . (I)
m &ndash 6n = &ndash7 . (II)
Multiplying (I) with 2 we get
10m &ndash 6n = 38 . (III)
m &ndash 6n = &ndash7 . (IV)
Subtracting (IV) from (III) we get
10 m - m - 6 n - - 6 n = 38 - - 7 ⇒ 9 m = 45 ⇒ m = 45 9 = 5
Putting the value of m = 5 in (II) we get
5 - 6 n = - 7 ⇒ - 6 n = - 7 - 5 ⇒ - 6 n = - 12 ⇒ n = - 12 - 6 = 2
Thus, (m, n) = (5, 2).

(5) 5x + 2y = &ndash3 . (I)
x + 5y = 4 . (II)
Multiply (II) with 5 we get
5x + 25y = 20 . (III)
Subtracting (III) from (I) we get
5 x - 5 x + 2 y - 25 y = - 3 - 20 ⇒ - 23 y = - 23 ⇒ y = - 23 - 23 = 1
Putting the value of y = 1 in (II) we get
x + 5 1 = 4 ⇒ x + 5 = 4 ⇒ x = 4 - 5 = - 1
Thus, (x, y) = (&minus1, 1)

x - y = 4 x + y = 10 ⇒ 2 x = 14 ⇒ x = 7
Putting the value of x = 7 in (IV) we get
7 + y = 10 ⇒ y = 10 - 7 ⇒ y = 3
Thus, (x, y) = (7, 3).

Page No 8:

Question 1:

Complete the following table to draw graph of the equations&ndash
(I) x + y = 3 (II) x &ndash y = 4

x + y = 3
x 3         0                 0        
y         0         5 3
(x, y) (3, 0)         0         (0, 3)
x &ndash y = 4
x         0         &ndash1 0
y 0         0         &ndash4
(x, y)         0                 0         (0, &ndash4)

Answer:

x + y = 3
x 3       - 2                 0        
y         0         5 3
(x, y) (3, 0)     - 2 ,   5     (0, 3)
x &ndash y = 4
x         4         &ndash1 0
y 0     - 5         &ndash4
(x, y)       4 , 0             - 1 , - 5     (0, &ndash4)

Page No 8:

Question 2:

Solve the following simultaneous equations graphically.
(1) x + y = 6 x &ndash y = 4
(2) x + y = 5 x &ndash y = 3
(3) x + y = 0 2x &ndash y = 9
(4) 3x &ndash y = 2 2x &ndash y = 3
(5) 3x &ndash 4y = &ndash7 5x &ndash 2y = 0
(6) 2x &ndash 3y = 4 3y &ndash x = 4

Answer:


Point of intersection of the two lines is (5, 1).

Point of intersection of the two lines is (4, 1)
(3) x + y = 0

​Point of intersection of the two lines is (3, &minus3).

​Point of intersection of the two lines is (&minus1, &minus5).

​Point of intersection of the two lines is (1, 2.5).

​Point of intersection of the two lines is (8, 4).

Page No 16:

Question 1:

Fill in the blanks with correct number

Answer:

Page No 16:

Question 2:

Find the values of following determinants.

Answer:

(3) 7 3 5 3 3 2 1 2 = 7 3 × 1 2 - 5 3 × 3 2 = 7 6 - 5 2 = 7 - 15 6 = - 8 6 = - 4 3

Page No 16:

Question 3:

Solve the following simultaneous equations using Cramer&rsquos rule.
(1) 3x &ndash 4y = 10 4x + 3y = 5
(2) 4x + 3y &ndash 4 = 0 6x = 8 &ndash 5y
(3) x + 2y = &ndash1 2x &ndash 3y = 12
(4) 6x &ndash 4y = &ndash12 8x &ndash 3y = &ndash2
(5) 4m + 6n = 54 3m + 2n = 28
(6) 2 x + 3 y = 2     x - y 2 = 1 2

Answer:

(1) 3x &ndash 4y = 10
4x + 3y = 5
D = 3 - 4 4 3 = 3 × 3 - - 4 × 4 = 9 + 16 = 25 D x = 10 - 4 5 3 = 10 × 3 - - 4 × 5 = 30 + 20 = 50 D y = 3 10 4 5 = 3 × 5 - 10 × 4 = 15 - 40 = - 25
x = D x D = 50 25 = 2 y = D y D = - 25 25 = - 1 x , y = 2 , - 1

(2) 4x + 3y &ndash 4 = 0 6x = 8 &ndash 5y
D = 4 3 6 5 = 4 × 5 - 6 × 3 = 20 - 18 = 2 D x = 4 3 8 5 = 4 × 5 - 3 × 8 = 20 - 24 = - 4 D y = 4 4 6 8 = 4 × 8 - 6 × 4 = 32 - 24 = 8

x = D x D = - 4 2 = - 2 y = D y D = 8 2 = 4 x , y = - 2 , 4

(3) x + 2y = &ndash1 2x &ndash 3y = 12
D = 1 2 2 - 3 = 1 × - 3 - 2 × 2 = - 3 - 4 = - 7 D x = - 1 2 12 - 3 = - 1 × - 3 - 2 × 12 = 3 - 24 = - 21 D y = 1 - 1 2 12 = 1 × 12 - - 1 × 2 = 12 + 2 = 14
x = D x D = - 21 - 7 = 3 y = D y D = 14 - 7 = - 2 x , y = 3 , - 2

(4) 6x &ndash 4y = &ndash12 8x &ndash 3y = &ndash2

D = 6 - 4 8 - 3 = 6 × - 3 - - 4 × 8 = - 18 + 32 = 14 D x = - 12 - 4 - 2 - 3 = - 12 × - 3 - - 4 × - 2 = 36 - 8 = 28 D y = 6 - 12 8 - 2 = 6 × - 2 - - 12 × 8 = - 12 + 96 = 84
x = D x D = 28 14 = 2 y = D y D = 84 14 = 6 x , y = 2 , 6

(5) 4m + 6n = 54 3m + 2n = 28
D = 4 6 3 2 = 4 × 2 - 6 × 3 = 8 - 18 = - 10 D x = 54 6 28 2 = 54 × 2 - 6 × 28 = 108 - 168 = - 60 D y = 4 54 3 28 = 4 × 28 - 54 × 3 = 112 - 162 = - 50
x = D x D = - 60 - 10 = 6 y = D y D = - 50 - 10 = 5 x , y = 6 , 5

(6) 2 x + 3 y = 2     x - y 2 = 1 2
D = 2 3 1 - 1 2 = 2 × - 1 2 - 3 × 1 = - 1 - 3 = - 4 D x = 2 3 1 2 - 1 2 = 2 × - 1 2 - 3 × 1 2 = - 1 - 3 2 = - 5 2 D y = 2 2 1 1 2 = 2 × 1 2 - 2 × 1 = 1 - 2 = - 1
x = D x D = - 5 2 - 4 = 5 8 y = D y D = - 1 - 4 = 1 4 x , y = 5 8 , 1 4

Page No 19:

Question 1:

Solve the following simultaneous equations.

1   2 x - 3 y = 15   8 x + 5 y = 77 2   10 x + y + 2 x - y = 4   15 x + y - 5 x - y = - 2 3   27 x - 2 + 31 y + 3 = 85   31 x - 2 + 27 y + 3 = 89 4   1 3 x + y + 2 3 x - y = 3 4   1 2 3 x + y - 1 2 3 x - y = - 1 8

Answer:

1   2 x - 3 y = 15   8 x + 5 y = 77
Let   1 x = u   and   1 y = v
So, the equation becomes
2 u - 3 v = 15                                       . . . . . I 8 u + 5 v = 77                                       . . . . . II
Multiply (I) with 4 we get
8 u - 12 v = 60                               . . . . . III
(II) &minus (III)
8 u - 8 u + 5 v - - 12 v = 77 - 60 ⇒ 17 v = 17 ⇒ v = 1 Putting   the   value   of   v   in   I 2 u - 3 1 = 15 ⇒ 2 u = 15 + 3 = 18 ⇒ u = 9
Thus,
1 x = u = 9 ⇒ x = 1 9 1 y = v = 1 ⇒ y = 1 x , y = 1 9 , 1

2   10 x + y + 2 x - y = 4   15 x + y - 5 x - y = - 2
Let 1 x + y = u   and   1 x - y = v
So, the equation becomes
10 u + 2 v = 4                             . . . . . I 15 u - 5 v = - 2                     . . . . . II  
Multiplying (I) with 5 and (II) with 2 we get
50 u + 10 v = 20                                           . . . . . III 30 u - 10 v = - 4                                       . . . . . IV
Adding (III) and (IV) we get
u = 16 80 = 1 5
Putting this value in (I)
10 × 1 5 + 2 v = 4 ⇒ 2 + 2 v = 4 ⇒ v = 1

1 x + y = 1 5   and   1 x - y = 1 ⇒ x + y = 5   and   x - y = 1 Solving   these   equations   we   get x = 3   and   y = 2

3   27 x - 2 + 31 y + 3 = 85   31 x - 2 + 27 y + 3 = 89
Let   1 x - 2 = u   and   1 y + 3 = v
27 u + 31 v = 85                               . . . . . I 31 u + 27 v = 89                               . . . . . II Adding   I   and   II 58 u + 58 v = 174 u + v = 3                                                     . . . . . III Subtracting   II   from   I 4 u - 4 v = 4 ⇒ u - v = 1                                           . . . . . IV        
Adding (III) and (IV) we get
2 u = 4 ⇒ u = 2
Putting the value of u in III
2 + v = 3 ⇒ v = 1
1 x - 2 = u = 2 ⇒ x - 2 = 1 2 ⇒ x = 5 2
1 y + 3 = 1 ⇒ y + 3 = 1 ⇒ y = - 2
x , y = 5 2 , - 2

4   1 3 x + y + 2 3 x - y = 3 4   1 2 3 x + y - 1 2 3 x - y = - 1 8
Let 1 3 x + y = u   and   1 3 x - y = v
u + 2 v = 3 4   and   1 2 u - 1 2 v = - 1 8
So, the equations become
4 u + 4 v = 3                               . . . . . I 4 u - 4 v = 1                               . . . . . II
Adding (I) and (II)
8 u = 4 ⇒ u = 1 2
Putting the value of u in (I)
1 2 + 2 v = 3 4 ⇒ v = 1 4
1 3 x + y = u   and   1 3 x - y = v ⇒ 1 3 x + y = 1 2 3 x + y = 2                                   . . . . . III Also ,   1 3 x - y = 1 4 ⇒ 3 x - y = 4                         . . . . . IV                      
(III) + (IV) we get
6 x = 6 ⇒ x = 1 y = - 1

Page No 26:

Question 1:

Two numbers differ by 3. The sum of twice the smaller number and thrice the greater number is 19. Find the numbers.

Answer:

Let the smaller number be x and the larger number be y.
Given that the two numbers differ by 3 so,
y - x = 3 . (I)
Also, sum of twice the smaller number and thrice the greater number is 19
So, 2 x + 3 y = 19 . (II)
The two equations obtained are
y - x = 3
2 x + 3 y = 19
Multiplying (I) by 3 we get
3 y - 3 x = 9 . (III)
Adding (III) and (II) we have
4y = 28
⇒ y = 28 4 = 7
Putting the value of y = 7 in (I) we get
7 - x = 3 ⇒ - x = 3 - 7 ⇒ - x = - 4 ⇒ x = 4
Thus, the two numbers are 4 and 7.

Page No 26:

Question 2:

Answer:

The length of the given rectangle is 2 x + y + 8 and 4 x - y
2 x + y + 8 = 4 x - y ⇒ y + y + 8 = 4 x - 2 x ⇒ 8 + 2 y = 2 x ⇒ 2 x - 2 y = 8 Dividing   by   2 x - y = 4                                                                     . . . . . I
Breadth of the rectangle is 2y and x + 4.
2 y = x + 4 ⇒ x - 2 y = - 4                                                         . . . . . II  
Subtracting (II) from (I)
x - x - y - - 2 y = 4 - - 4 ⇒ - y + 2 y = 8 ⇒ y = 8 Putting   the   value   of   y = 8   in   ( I )   we   get x - 8 = 4 ⇒ x = 4 + 8 = 12
Length = 4 x - y = 4 12 - 8 = 40
Breadth = 2 × 8 = 16
Perimeter = 2 length + breadth = 2 40 + 16 = 112 units
Area = length × breadth = 40 × 16 = 640   unit 2

Page No 26:

Question 3:

The sum of father&rsquos age and twice the age of his son is 70. If we double the age of the father and add it to the age of his son the sum is 95. Find their present ages.

Answer:

Let the father's age be x years and son's age be y years.
Sum of father&rsquos age and twice the age of his son is 70 so,
x + 2 y = 70 . (I)
Double the age of the father added to the age of his son the sum is 95
2 x + y = 95 . (II)
Adding (I) and (II) we get
3 x + 3 y = 165 Dividing   by   3 x + y = 55                                                       . . . . . III
Subtracting (I) from (II)
2 x - x + y - 2 y = 95 - 70 ⇒ x - y = 25                                               . . . . . IV
Adding (III) and (IV) we get
2 x = 80 ⇒ x = 40 Putting   the   value   of   x = 40   in   III 40 + y = 55 ⇒ y = 55 - 40 ⇒ y = 15
Thus, the age of the father is 40 years and age of his son is 15 years.

Page No 26:

Question 4:

The denominator of a fraction is 4 more than twice its numerator. Denominator becomes 12 times the numerator, if both the numerator and the denominator are reduced by 6. Find the fraction.

Answer:

Let the fraction be x y .
Denominator of a fraction is 4 more than twice its numerator.
So,
y = 4 + 2 x ⇒ 2 x - y = - 4                                                     . . . . . I
Also, denominator becomes 12 times the numerator, if both the numerator and the denominator are reduced by 6.
So,
y - 6 = 12 x - 6 ⇒ y - 6 = 12 x - 72 ⇒ 12 x - y = 72 - 6 = 66 ⇒ 12 x - y = 66                                                           . . . . . II
Subtracting (I) from (II)
12 x - 2 x - y - - y = 66 - - 4 ⇒ 10 x = 70 ⇒ x = 70 10 = 7 ⇒ x = 7
Putting the value of x = 7 in (I)
2 7 - y = - 4 ⇒ 14 - y = - 4 ⇒ y = 14 + 4 = 18
Thus, the fraction obtained is 7 18 .

Page No 26:

Question 5:

Two types of boxes A, B are to be placed in a truck having capacity of 10 tons. When 150 boxes of type A and 100 boxes of type B are loaded in the truck, it weighes 10 tons. But when 260 boxes of type A are loaded in the truck, it can still accommodate 40 boxes of type B, so that it is fully loaded. Find the weight of each type of box.

Answer:

Let the weight of box A be x and that of box B be y.
When 150 boxes of type A and 100 boxes of type B are loaded in the truck, it weighes 10 tons i.e 10000 kg.
So,
150 x + 100 y = 10000 ⇒ 15 x + 10 y = 1000 ⇒ 3 x + 2 y = 200                                                                               . . . . . I
When 260 boxes of type A are loaded in the truck, it can still accommodate 40 boxes of type B, so that it is fully loaded.
260 x + 40 y = 10000 ⇒ 26 x + 4 y = 1000 ⇒ 13 x + 2 y = 500                                                                       . . . . . II
Subtracting (I) from (II) we get
13 x - 3 x + 2 y - 2 y = 500 - 200 ⇒ 10 x = 300 ⇒ x = 30 Putting   the   value   of   x = 30   in   I 3 30 + 2 y = 200 ⇒ 90 + 2 y = 200 ⇒ 2 y = 200 - 90 = 110 ⇒ y = 110 2 = 55
Thus, weight of box A = 30 kg and that of box B = 55 kg.

Page No 26:

Question 6:

Out of 1900 km, Vishal travelled some distance by bus and some by aeroplane. Bus travels with average speed 60 km/hr and the average speed of aeroplane is 700 km/hr. It takes 5 hours to complete the journey. Find the distance, Vishal travelled by bus.

Answer:

We know speed = distance time
Average speed of bus = 60km/h.
Let the time taken in bus be x hours.
Average speed of bus = 700km/h.
Let the time taken in bus be y hours.
Total distance covered = 1900 km
60 x + 700 y = 1900 ⇒ 6 x + 70 y = 190 ⇒ 3 x + 35 y = 95                                 . . . . . I
It takes 5 hours to complete the journey so,
x + y = 5                                     . . . . . II
Multiplying (II) with 3
3 x + 3 y = 15                                             . . . . . III
Subtracting (III) from (I) we get
3 x - 3 x + 35 y - 3 y = 95 - 15 ⇒ 32 y = 80 ⇒ y = 2 . 5
Putting the value of y = 2.5 in (II) we get
x + 2 . 5 = 5 ⇒ x = 2 . 5
Distance travelled by Vishal by bus = speed × time = 60 × 2 . 5 = 150   km .

Page No 27:

Question 1:

Answer:

(1) 4 x +5 y = 19
When x = 1, then y will be
4 1 + 5 y = 19 ⇒ 4 + 5 y = 19 ⇒ 5 y = 19 - 4 = 15 ⇒ 5 y = 15 ⇒ y = 15 5 = 3
Hence, the correct answer is option (B).

(2) x = D x D = 49 7 = 7
Hence, the correct answer is option (A).

(3) 5 3 - 7 - 4 = 5 × - 4 - 3 × - 7 = - 20 + 21 = 1
Hence, the correct answer is option (D).

(4) x + y = 3 3 x &ndash 2 y &ndash 4 = 0
D = 1 1 3 - 2 = 1 × - 2 - 1 × 3 = - 2 - 3 = - 5
Hence, the correct answer is option (C).

(5) ax + by = c and mx + ny = d
D = a b m n = a n - b m
an
&ne bm
So, D &ne 0.
So, the given equations have a unique solution or only one common solution.
Hence, the correct answer is option A.

Page No 27:

Question 2:

Complete the following table to draw the graph of 2x &ndash 6y = 3

Answer:

Page No 27:

Question 3:

Answer:


The solution of the given equations is the point of intersection of the two line i.e(3, 2).


The solution of the given equations is the point of intersection of the two line i.e ( - 2 , - 1 ).


The solution of the given equations is the point of intersection of the two line i.e (0, 5).

The solution of the given equations is the point of intersection of the two line i.e (2, 4).

The solution of the given equations is the point of intersection of the two line i.e (3, 1).

Page No 27:

Question 4:

Find the values of each of the following determinants.

Answer:

(1) 4 3 2 7 = 4 × 7 - 3 × 2 = 28 - 6 = 22

(2) 5 - 2 - 3 1 = 5 × 1 - - 2 × - 3 = 5 - 6 = - 1

(3) 3 - 1 1 4 = 3 × 4 - - 1 × 1 = 12 + 1 = 13

Page No 28:

Question 5:

Solve the following equations by Cramer&rsquos method.
(1) 6x &ndash 3y = &ndash10 3x + 5y &ndash 8 = 0
(2) 4m &ndash 2n = &ndash4 4m + 3n = 16
(3) 3x &ndash 2y = 5 2 1 3 x + 3 y = - 4 3
(4) 7x + 3y = 15 12y &ndash 5x = 39
(5) x + y - 8 2 = x + 2 y - 14 3 = 3 x - y 4

Answer:

(1) 6x &ndash 3y = &ndash10 3x + 5y &ndash 8 = 0
D = 6 - 3 3 5 = 6 × 5 - - 3 × 3 = 30 + 9 = 39 D x = - 10 - 3 8 5 = - 10 × 5 - - 3 × 8 = - 50 + 24 = - 26 D y = 6 - 10 3 8 = 6 × 8 - - 10 × 3 = 48 + 30 = 78 x = D x D = - 26 39 = - 2 3 y = D y D = 78 39 = 2 x , y = - 2 3 , 2

(2) 4m &ndash 2n = &ndash4 4m + 3n = 16
D = 4 - 2 4 3 = 4 × 3 - - 2 × 4 = 12 + 8 = 20 D x = - 4 - 2 16 3 = - 4 × 3 - - 2 × 16 = - 12 + 32 = 20 D y = 4 - 4 4 16 = 4 × 16 - - 4 × 4 = 64 + 16 = 80 x = D x D = 20 20 = 1 y = D y D = 80 20 = 4 x , y = 1 , 4
(3) 3x &ndash 2y = 5 2 1 3 x + 3 y = - 4 3
D = 3 - 2 1 3 3 = 9 + 2 3 = 29 3 D x = 5 2 - 2 - 4 3 3 = 15 2 - 8 3 = 29 6 D y = 3 5 2 1 3 - 4 3 = - 4 - 5 6 = - 29 6 x = D x D = 29 6 29 3 = 1 2 y = D y D = - 29 6 29 3 = - 1 2 x , y = 1 2 , - 1 2

(4) 7x + 3y = 15 12y &ndash 5x = 39
D = 7 3 - 5 12 = 7 × 12 - - 5 × 3 = 84 + 15 = 99 D x = 15 3 39 12 = 15 × 12 - 39 × 3 = 180 - 117 = 63 D y = 7 15 - 5 39 = 7 × 39 - - 5 × 15 = 273 + 75 = 348 x = D x D = 63 99 = 7 11 y = D y D = 348 99 = 116 33 x , y = 7 11 , 116 33

(5) x + y - 8 2 = x + 2 y - 14 3 = 3 x - y 4
x + y - 8 2 = x + 2 y - 14 3 ⇒ 3 x + 3 y - 24 = 2 x + 4 y - 28 ⇒ x - y = - 4                                                       . . . . . I and   x + 2 y - 14 3 = 3 x - y 4 ⇒ 4 x + 8 y - 56 = 9 x - 3 y ⇒ 5 x - 11 y = - 56                                   . . . . . II

From (I) and (II)
D = 1 - 1 5 - 11 = - 11 × 1 - - 1 × 5 = - 11 + 5 = - 6 D x = - 4 - 1 - 56 - 11 = - 11 × - 4 - - 1 × - 56 = 44 - 56 = - 12 D y = 1 - 4 5 - 56 = - 56 × 1 - - 4 × 5 = - 56 + 20 = - 36 x = D x D = - 12 - 6 = 2 y = D y D = - 36 - 6 = 6 x , y = 2 , 6

Page No 28:

Question 6:

Solve the following simultaneous equations.
(1) 2 x + 2 3 y = 1 6     3 x + 2 y = 0
(2) 7 2 x + 1 + 13 y + 2 = 27     13 2 x + 1 + 7 y + 2 = 33
(3) 148 x + 231 y = 527 x y     231 x + 148 y = 610 x y
(4) 7 x - 2 y x y = 5     8 x + 7 y x y = 15
(5) 1 2 3 x + 4 y + 1 5 2 x - 3 y = 1 4     5 3 x + 4 y - 2 2 x - 3 y = - 3 2

Answer:

(2) 7 2 x + 1 + 13 y + 2 = 27     13 2 x + 1 + 7 y + 2 = 33
Let 1 2 x + 1 = u   and   1 y + 2 = v
7 u + 13 v = 27                         . . . . . I 13 u + 7 v = 33                         . . . . . II
(I) + (II)
20 u + 20 v = 60 u + v = 3                                     . . . . . III
(II) &minus (I)
6 u - 6 v = 6                   u - v = 1                                   . . . . . IV
(III) + (IV)
2 u = 4 ⇒ u = 2 Putting   the   value   of   u   in   ( IV )   2 - v = 1 ⇒ v = 1
1 2 x + 1 = u = 2   ⇒ 2 x + 1 = 1 2 ⇒ x = - 1 4 and   1 y + 2 = v = 1 ⇒ y + 2 = 1 ⇒ y = - 1 x , y = - 1 4 , - 1

(3) 148 x + 231 y = 527 x y     231 x + 148 y = 610 x y
Multiply by xy
148 y + 231 x = 527                 . . . . . I       231 y + 148 x = 610                 . . . . . II Adding   I   and   II     379 y + 379 x = 1137 ⇒ x + y = 3                                             . . . . . III II - I 83 y - 83 x = 83 ⇒ y - x = 1                                           . . . . . IV III + IV 2 y = 4 ⇒ y = 2

Putting the value of y in (IV)
2 - x = 1 ⇒ x = 1 x , y = 1 , 2

(4) 7 x - 2 y x y = 5     8 x + 7 y x y = 15
⇒ 7 y - 2 x = 5   and   8 y + 7 x = 15
Let 1 x = u , 1 y = v
7 v - 2 u = 5                         . . . . . I 8 v + 7 u = 15                     . . . . . II        
Multiply (I) with 7 and (II) with 2
49 v - 14 u = 35                         . . . . . III 16 v + 14 u = 30                         . . . . . IV
Adding (III) and (IV)
65 v = 65 ⇒ v = 1 And   1 y = v = 1 ⇒ y = 1
Putting the value of v in (I)
7 1 - 2 u = 5 ⇒ u = 1 1 x = u = 1 ⇒ x = 1 x , y = 1 , 1

(5) 1 2 3 x + 4 y + 1 5 2 x - 3 y = 1 4     5 3 x + 4 y - 2 2 x - 3 y = - 3 2
1 3 x + 4 y = u , 1 2 x - 3 y = v 1 2 u + 1 5 v = 1 4                 ⇒ 10 u + 4 v = 5                                     . . . . . I 5 u - 2 v = - 3 2 ⇒ 10 u - 4 v = - 3                             . . . . . II
(I) + (II)
20 u = 2 ⇒ u = 1 10
Putting the value of u in (II)
10 × 1 10 - 4 v = - 3 ⇒ 1 + 3 = 4 v ⇒ v = 1
1 3 x + 4 y = u = 1 10 ⇒ 3 x + 4 y = 10                             . . . . . III 1 2 x - 3 y = v = 1 ⇒ 2 x - 3 y = 1                                 . . . . . IV
Multiply (III) with 2 and (IV) with 3
6 x + 8 y = 20                       . . . . . V 6 x - 9 y = 3                           . . . . . VI
(V) &minus (VI)
17 y = 17 ⇒ y = 1
Putting the value of y in (VI)
6 x - 9 = 3 ⇒ 6 x = 12 ⇒ x = 2 x , y = 2 , 1

Page No 28:

Question 7:

Solve the following word problems.
(1) A two digit number and the number with digits interchanged add up to 143. In the given number the digit in unit&rsquos place is 3 more than the digit in the ten&rsquos place. Find the original number.
(2) Kantabai bought 1 1 2 kg tea and 5 kg sugar from a shop. She paid Rs 50 as return fare for rickshaw. Total expense was Rs 700. Then she realised that by ordering online the goods can be bought with free home delivery at the same price. So next month she placed the order online for 2 kg tea and 7 kg sugar. She paid Rs 880 for that. Find the rate of sugar and tea per kg.
(3) To find number of notes that Anushka had, complete the following activity.

(4) Sum of the present ages of Manish and Savita is 31. Manish&rsquos age 3 years ago was 4 times the age of Savita. Find their present ages.
(5) In a factory the ratio of salary of skilled and unskilled workers is 5 : 3. Total salary of one day of both of them is Rs 720. Find daily wages of skilled and unskilled workers.
(6) Places A and B are 30 km apart and they are on a st raight road. Hamid travels from A to B on bike. At the same time Joseph starts from B on bike, travels towards A. They meet each other after 20 minutes. If Joseph would have started from B at the same time but in the opposite direction (instead of towards A) Hamid would have caught him after 3 hours. Find the speed of Hamid and Joseph.

Answer:

(1) Let the number at the unit's place be x and the digit at the ten's place be y.
The number is thus 10y + x
After interchanging the digits the number becomes 10x + y.
Given that two digit number and the number with digits interchanged add up to 143.
So, 10y + x + 10x + y = 143
⇒ 11 x + 11 y = 143 ⇒ x + y = 13                                           . . . . . I
Also, in the given number the digit in unit&rsquos place is 3 more than the digit in the ten&rsquos place.
So, x - y = 3                                   . . . . . II
Adding (I) and (II) we get
2 x = 16 ⇒ x = 8
Putting the value of x in (I) we get
8 + y = 13 ⇒ y = 13 - 8 = 5
Thus, the number is 58.

(2) Let the rate of tea be x Rs per kg and that of sugar be y Rs per kg.
When Kantabai bought the items by going to the shop,
3 2 x + 5 y + 50 = 700 ⇒ 3 x + 10 y = 1300                     . . . . . I
When Kantabai bought the items online then
2 x + 7 y = 880                         . . . . . II
Multiplying (I) with 2 and (II) with 3 we get
6 x + 20 y = 2600                           . . . . . III 6 x + 21 y = 2640                           . . . . . IV
(IV) - (III)
y = 40
Putting the value of y = 40 in (II)
2 x + 7 40 = 880 ⇒ 2 x = 880 - 280 = 600 ⇒ x = 300
Thus, tea is at 300 Rs per kg and sugar is 40 Rs per kg.

(3) Disclaimer: There is error in the question given. Instead of Rs 10 notes there should be Rs 100 notes.
Let the number of notes of Rs 100 be x and that of Rs 50 be y.
100 x + 50 y = 2500 ⇒ 2 x + y = 50                                 . . . . . I
When number of notes is interchanged so,
50 x + 100 y = 2000 ⇒ x + 2 y = 40                                 . . . . . II
Multiply (I) with 2
4 x + 2 y = 100                         . . . . . III
Subtracting (III) from (II) we get
3 x = 60 ⇒ x = 20 3 x = 60 ⇒ x = 20
Putting the value of x in (I) we get
y = 10
Thus, there are 20 Rs 100 notes and 10 Rs 50 notes.

(4) Let the present age of Manish be x years and that of Savita be y years.
Sum of their present ages = 31
x + y = 31                               . . . . . I
Their age 3 years ago was
Manish's age = x - 3
Savita's age = y - 3
Manish&rsquos age 3 years ago was 4 times the age of Savita.
x - 3 = 4 y - 3 ⇒ x - 3 = 4 y - 12 ⇒ x - 4 y = - 9                           . . . . . II
(I) - (II) we get
5 y = 40 ⇒ y = 8
Putting the value of y in (I) we get
x + 8 = 31 ⇒ x = 23
Thus, age of Manish is 23 years and age of Savita is 8 years.

(5) Ratio of salary of skilled to unskilled workers = 5 : 3
Let one day salary of skilled person be x and that of unskilled person be y.
Their total one day salary Rs 720
x + y = 720                         . . . . . I
Also,
x y = 5 3 ⇒ 3 x = 5 y ⇒ 3 x - 5 y = 0                         . . . . . II
Multiplying (I) with 3 we get
3 x + 3 y = 2160                             . . . . . III
(III) - (II)
8 y = 2160 ⇒ y = 270
Putting value of y in (I) we get
x = 450
One day salary of skilled person Rs 450 and that of unskilled person Rs 270.

(6) Let the speed of Hamid be x km/h and that of Joseph be y km/h.
When both travel in same direction so, the distance covered by them together will be 30 km.
We know Speed = Distance Time
They meet each other after 20 min = 20 60 = 1 3 hours
x 3 + y 3 = 30 ⇒ x + y = 90                         . . . . . I
When Joseph started from point B but moved in opposite direction.
Distance travelled by Hamid - Distance travelled by Joseph = 30
⇒ 3 x - 3 y = 30 ⇒ x - y = 10                               . . . . . II
Adding (I) and (II) we get
2 x = 100 ⇒ x = 50
Putting the value of x in (II) we get
50 - y = 10 ⇒ y = 40
Thus, speed of Hamid is 50 km/h and that of Joseph is 40 km/h.


1: Chapter 1 - Functions - Mathematics

For problems 1 – 4 the given functions perform the indicated function evaluations.

  1. (fleft( x ight) = 3 - 5x - 2 ) Solution
    1. (fleft( 4 ight) )
    2. (fleft( 0 ight))
    3. (fleft( < - 3> ight) )
    1. (fleft( <6 - t> ight) )
    2. (fleft( <7 - 4x> ight))
    3. (fleft( ight) )
    1. (gleft( 0 ight) )
    2. (gleft( < - 3> ight))
    3. (gleft( <10> ight) )
    1. (gleft( <> ight) )
    2. (gleft( ight))
    3. (gleft( <- 3t + 1> ight) )
    1. (hleft( 0 ight) )
    2. (hleft( < - frac<1><2>> ight))
    3. (hleft( <2>> ight) )
    1. (hleft( <9z> ight) )
    2. (hleft( <- 2z> ight) )
    3. (hleft( ight) )
    1. (Rleft( 0 ight) )
    2. (Rleft( 6 ight))
    3. (Rleft( < - 9> ight) )
    1. (Rleft( ight))
    2. (Rleft( <- 3> ight))
    3. (Rleft( - 1> ight) )

    The difference quotient of a function (fleft( x ight) ) is defined to be,

    For problems 5 – 9 compute the difference quotient of the given function.

    1. (fleft( x ight) = 4x - 9 ) Solution
    2. (gleft( x ight) = 6 - ) Solution
    3. (fleft( t ight) = 2 - 3t + 9 ) Solution
    4. (displaystyle yleft( z ight) = frac<1><> ) Solution
    5. (displaystyle Aleft( t ight) = frac<<2t>><<3 - t>> ) Solution

    For problems 10 – 17 determine all the roots of the given function.

    1. (fleft( x ight) = - 4 - 32 ) Solution
    2. (Rleft( y ight) = 12 + 11y - 5 ) Solution
    3. (hleft( t ight) = 18 - 3t - 2 ) Solution
    4. (gleft( x ight) = + 7 - x ) Solution
    5. (Wleft( x ight) = + 6 - 27 ) Solution
    6. (fleft( t ight) = <3>>> - 7<3>>> - 8t ) Solution
    7. (displaystyle hleft( z ight) = frac<> - frac<4><> ) Solution
    8. (displaystyle gleft( w ight) = frac<<2w>><> + frac<><<2w - 3>> ) Solution

    For problems 18 – 22 find the domain and range of the given function.

    1. (Yleft( t ight) = 3 - 2t + 1 ) Solution
    2. (gleft( z ight) = - - 4z + 7 ) Solution
    3. (fleft( z ight) = 2 + sqrt <+ 1> ) Solution
    4. (hleft( y ight) = - 3sqrt <14 + 3y>) Solution
    5. (Mleft( x ight) = 5 - left| ight| ) Solution

    For problems 23 – 32 find the domain of the given function.

    1. (displaystyle fleft( w ight) = frac <<- 3w + 1>><<12w - 7>> ) Solution
    2. (displaystyle Rleft( z ight) = frac<5><<+ 10 + 9z>> ) Solution
    3. (displaystyle gleft( t ight) = frac<<6t - >><<7 - t - 4>> ) Solution
    4. (gleft( x ight) = sqrt <25 - > ) Solution
    5. (hleft( x ight) = sqrt <- - 20> ) Solution
    6. (displaystyle Pleft( t ight) = frac<<5t + 1>><- - 8t> >> ) Solution
    7. (fleft( z ight) = sqrt + sqrt ) Solution
    8. (displaystyle hleft( y ight) = sqrt <2y + 9>- frac<1><>> ) Solution
    9. (displaystyle Aleft( x ight) = frac<4><> - sqrt <- 36> ) Solution
    10. (Qleft( y ight) = sqrt <+ 1> - sqrt[3]<<1 - y>> ) Solution

    For problems 33 – 36 compute (left( ight)left( x ight) ) and (left( ight)left( x ight) ) for each of the given pair of functions.


    2nd Year Maths Chapter 1 Functions and Limits MCQs with Answers

    Question# 1 : limx →0 e -1/x4 =?

    Answer

    Answer

    Question# 2 : limx→3 4x-8/x+2=?

    Answer

    Answer

    Question# 3 : limz→0 sinpz/mz=?

    Answer

    Answer

    Question# 4 : Inverse function of y= x/x+5 is:

    Answer

    Answer

    Question# 5 : The graph of the function y=logx has an asymptote at______.

    Answer

    Answer

    Question# 6 : limx→∞ (x/1+x) x =?

    Answer

    Answer

    Question# 7 : If f(x) = x 2 -1/x 2 then f(x) =?

    Answer

    Answer

    Question# 8 : Which of the following is an implicit function?

    Answer

    Answer

    Question# 9 : For a function if right hand limit = left hand limit then which of the following must be true?

    • Function is continuous
    • Function has the same domain and range
    • Function is defined
    • Its limit exists

    Answer

    Answer

    Question# 10 : The domain of y=√-x is:

    Answer

    Answer

    Question# 11 : f(x)= acosx where g(x)= x 3 , then f’og(x)=?

    Answer

    Answer

    Question# 12 : When x approaches to positive infinity then limit lnx/x is equal to:

    Answer

    Answer

    Question# 13 : The horizontal asymptote for f(x)= 3x+1/2x-1

    Answer

    Answer

    Question# 14 : f(x)= secx tanx is a:

    Answer

    Answer

    Question# 15 : If f(π)= π 2 -2 π-1, then f(a-1)=?


    1.1 Review of Functions

    In this section, we provide a formal definition of a function and examine several ways in which functions are represented—namely, through tables, formulas, and graphs. We study formal notation and terms related to functions. We also define composition of functions and symmetry properties. Most of this material will be a review for you, but it serves as a handy reference to remind you of some of the algebraic techniques useful for working with functions.

    Functions

    Definition

    The concept of a function can be visualized using Figure 1.2, Figure 1.3, and Figure 1.4.

    Media

    Visit this applet link to see more about graphs of functions.

    We can use similar notation if we want to include one of the endpoints, but not the other. To denote the set of nonnegative real numbers, we would use the set-builder notation

    The smallest number in this set is zero, but this set does not have a largest number. Using interval notation, we would use the symbol ∞ , ∞ , which refers to positive infinity, and we would write the set as

    refers to the set of all real numbers.

    Some functions are defined using different equations for different parts of their domain. These types of functions are known as piecewise-defined functions . For example, suppose we want to define a function f f with a domain that is the set of all real numbers such that f ( x ) = 3 x + 1 f ( x ) = 3 x + 1 for x ≥ 2 x ≥ 2 and f ( x ) = x 2 f ( x ) = x 2 for x < 2 . x < 2 . We denote this function by writing

    Example 1.1

    Evaluating Functions

    For the function f ( x ) = 3 x 2 + 2 x − 1 , f ( x ) = 3 x 2 + 2 x − 1 , evaluate

    Solution

    Substitute the given value for x in the formula for f ( x ) . f ( x ) .

    Example 1.2

    Finding Domain and Range

    For each of the following functions, determine the i. domain and ii. range.

    Solution

    Find the domain and range for f ( x ) = 4 − 2 x + 5 . f ( x ) = 4 − 2 x + 5 .

    Representing Functions

    Typically, a function is represented using one or more of the following tools:

    We can identify a function in each form, but we can also use them together. For instance, we can plot on a graph the values from a table or create a table from a formula.

    Tables

    Functions described using a table of values arise frequently in real-world applications. Consider the following simple example. We can describe temperature on a given day as a function of time of day. Suppose we record the temperature every hour for a 24-hour period starting at midnight. We let our input variable x x be the time after midnight, measured in hours, and the output variable y y be the temperature x x hours after midnight, measured in degrees Fahrenheit. We record our data in Table 1.1.

    We can see from the table that temperature is a function of time, and the temperature decreases, then increases, and then decreases again. However, we cannot get a clear picture of the behavior of the function without graphing it.

    Graphs

    Given a function f f described by a table, we can provide a visual picture of the function in the form of a graph. Graphing the temperatures listed in Table 1.1 can give us a better idea of their fluctuation throughout the day. Figure 1.6 shows the plot of the temperature function.

    From the points plotted on the graph in Figure 1.6, we can visualize the general shape of the graph. It is often useful to connect the dots in the graph, which represent the data from the table. In this example, although we cannot make any definitive conclusion regarding what the temperature was at any time for which the temperature was not recorded, given the number of data points collected and the pattern in these points, it is reasonable to suspect that the temperatures at other times followed a similar pattern, as we can see in Figure 1.7.


    Chapter 1

    f ( 4 ) = 900 f ( 10 ) = 24 , 300 . f ( 4 ) = 900 f ( 10 ) = 24 , 300 .

    Section 1.1 Exercises

    x ≥ −2 y ≥ −1 x = −1 y = −1 + 2 x ≥ −2 y ≥ −1 x = −1 y = −1 + 2

    a. Yes, because there is only one winner for each year. b. No, because there are three teams that won more than once during the years 2001 to 2012.

    Section 1.2 Exercises

    Section 1.3 Exercises

    a. 0.550 rad/sec b. 0.236 rad/sec c. 0.698 rad/min d. 1.697 rad/min

    a. π/184 the voltage repeats every π/184 sec b. Approximately 59 periods

    Section 1.4 Exercises

    a. $31,250, $66,667, $107,143 b. ( p = 85 C C + 75 ) ( p = 85 C C + 75 ) c. 34 ppb

    Section 1.5 Exercises

    Domain: all real numbers, range: ( 2 , ∞ ) , y = 2 ( 2 , ∞ ) , y = 2

    Domain: all real numbers, range: ( 0 , ∞ ) , y = 0 ( 0 , ∞ ) , y = 0

    Domain: all real numbers, range: ( − ∞ , 1 ) , y = 1 ( − ∞ , 1 ) , y = 1

    Domain: all real numbers, range: ( −1 , ∞ ) , y = −1 ( −1 , ∞ ) , y = −1

    3 2 + 1 2 log 5 x + 3 2 log 5 y 3 2 + 1 2 log 5 x + 3 2 log 5 y

    Approximately $131,653 is accumulated in 5 years.

    i. a. pH = 8 b. Base ii. a. pH = 3 b. Acid iii. a. pH = 4 b. Acid

    333 million b. 94 years from 2013, or in 2107

    The San Francisco earthquake had 10 3.4 or

    2512 times more energy than the Japan earthquake.

    Review Exercises

    One-to-one yes, the function has an inverse inverse: f −1 ( x ) = 1 y f −1 ( x ) = 1 y

    x ≥ − 3 2 , f −1 ( x ) = − 3 2 + 1 2 4 y − 7 x ≥ − 3 2 , f −1 ( x ) = − 3 2 + 1 2 4 y − 7

    The population is less than 20,000 from December 8 through January 23 and more than 140,000 from May 29 through August 2

    As an Amazon Associate we earn from qualifying purchases.

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      Concept Chapter 1 Class 12 Maths RELATIONS AND FUNCTIONS Ncert Solutions

      Relation Types of Relation Reflexive Relation

      Symmetric Relation

      Transitive Relation Equivalence Relation

      Functions Types of Function One to One Function Many to one Function

      Many one onto Function One One onto Function(Injective)


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      2nd year honours
      Calculus2 … Ещё
      Lecture no.-01
      Mst.Lailatul kadri
      Assistant Professor,Mathematics
      Rajshahi College

      Department of Mathematics, Rajshahi College.

      Honours 4th Year
      Partial Differential Equation-243709 … Ещё
      Separation of Variables
      Professor Md. Shahidul Alam
      Head of the department
      Department of Mathematics
      Rajshahi College, Rajshahi

      Department of Mathematics, Rajshahi College.


      1: Chapter 1 - Functions - Mathematics


      Cambridge International
      Examinations

      Hialeah Gardens High School

      Textbooks at Cambridge.org webpage.

      Advanced Level Mathematics: Pure Mathematics 1
      ISBN: 9780521530118 at amazon.com

      Advanced Level Mathematics: Pure Mathematics 2 & 3
      ISBN: 978-0521530125 at amazon.com



      Selected Questions .
      Suggested solutions by C. Sotuyo:

      Pure Mathematics 1 by Hugh Neill and Douglas Quadling, Cambridge University Press, 2002.

      Miscellaneous exercises 1, page 15:
      1. Show that the triangle formed by the points (-2,5), (1, 3) and (5, 9) is right-angled.

      Miscellaneous exercises 4, page 62:
      4. For what values of k does the equation have a repeated root?

      Exercises 6D, page 87:
      12.
      A curve has equation , where a is a constant. Find the equations of the tangents to the graph at the points where it crosses the x-axis.

      Exercise 6E, page 93.

      2. Find the derivative of the function at x=p. (Use the difference of two squares formula as often as you can.)

      11. Show that the curves
      have exactly one point in common, and use differentiation to find the gradient of each curve at this point.

      Excercises 7C, page 110

      16. A circular cylinder is to fit inside a sphere of radius 10 cm. Calculate the maximum possible volume of the cylinder. (It is probably best to take as your independent variable the height, or half the height, of the cylinder.)

      Miscellaneous exercises 8, page 126 No. 8 & 9

      8. An arithmetic progression has first term a and common difference -1. The sum of the first n terms is equal to the sum of the first 3n terms. Express a in terms of n.

      9. Find the sum of the arithmetic progression 1, 4, 7, 10, 13, 16. 1000.
      Every third term of the above progression is removed, i.e. 7, 16, etc. Find the sum of the remaining terms.

      28. Prove that

      Miscellaneous exercises 11, page 173.
      Given that the function where x is a real number different from zero, find:


      Miscellaneous exercises 9, page 137
      Find an expression, in terms of n, for the coefficient of x in the expansion

      Miscellaneous exercises 10, No 5 page 153:
      Solve the equation 3cos2x=2, giving all the solutions in the interval 0<x<180.

      Miscellaneous exercises 10, No 19 page 155:
      The road to an island close to the shore is sometimes covered by the tide. When the water rises to the level of the road, the road is closed. (. ) See pdf doc for a solution to this problem.

      Excercise 13D, # 16 page 207:
      The roof of a house has a rectangular base of side 4 metres by 8 metres. The ridge line of the roof is 6 metres long, and centred 1 metre above the base of the roof. Calculate the acute angle between two opposite slanting edges of the roof.

      Answers to selected questions pdf doc here .

      Cambridge Math AS level

      Pure Mathematics 1

      Notice: Oct 20, 2014. I taught Pure 1 & 2 two years ago, 2012-2013. I kept the webpage, however. Reviewing the webpage stats I've learned that tens of people visit this page every day looking for resources . I am using La Tex (textmaker editor) to create the pdf file. It takes time the typesetting of math. Solutions of some problems are downloable for free.
      I'm working on the rest of the practices. Practices with answers appear in red, see below.

      The content of the practices --exercises, page number etc of Pure Mathematics 1, by Hugh Neill and Douglas Quadling, Cambridge University Press, 2002, are indicated below. The answers consists of step by step solutions. In some instances, like practice 4, the answers take 6 pages in practices 5 & 6, five pages each.

      Selected Questions : Suggested solutions by C. Sotuyo, here.

      Typesetting math with LaTex: in order to get this, type left(frac<1><2x>+x^3 ight)^8.
      Install MiKTeXfirst, and then, install the LaTex Editor, Texmaker. See here LaTex tutorial by Michelle Krummel.
      A quick guide to LaTex here, by Dickinson College. LaTex setup and tutorial, by Prof Elizabeth Arnold at James Madison Univ. And, finally, learn LaTeX in 30 minutes by ShareLaTeX.


      Notes, Practices & Assignments:

      1. Introduction to the course. Pretest. This diagnosis test is based on the Algebra II text book in use in Florida, USA.
      2. Geometry and coordinate geometry. Students should check online resources: Properties of quadrilaterals, pdf, 2 pages also Proofs in Coordinate Geometry.
      3. Chapter 1: Coordinates, points and lines. Practice 01page 15 exercises 1 to 12.Answers topractice 01.
      4. Aug 30th: review chapter 01. Next week, Tues 9/4, quiz chapter 1.
      5. Chapter 2. Surds and indices., Practice 02 here.
      6. Practice 03, ex 10 to 20 page 30 of the textbook.
      7. Quiz 02, surds and indices, based on practice 2 week 9/10-9/14.
      8. Week 09/17-9/21: Chapter 3. Functions and graphs & ex 3 to 19 page 49: Practice 04.
      9. Assignment chapter 01 & Assignment chapter 02 due 09/28.
        Assignment 01 is page 15 --textbook, except question 4.
        Assignment 02 is page 29 excercises 4 to 8, and miscellaneous from 1 tru 5, same page.
      10. Assignment 03, questions. Functions, domain, graphs. Assignment 03, answers.
      11. Online review: graphing polynomials, Paul's online math notes.
      12. Week 09/24-9/28: Chapter 4: QuadraticsPractice 05, Miscellaneous ex 4, page 62.
      13. Week 10/08-10/12. Quiz 3. Quadratics.
      14. Chapter 5, Inequalities. Practice 06, page 72 (ex 1 tru 11)
      15. Assignment 04. Question 4 page 72, from a) to l). Due 10/12.
      16. Chapter 6, Differentiation . Practice 07, 6D page 86-87, #1 tru 13. Solutionshere.
      17. Rules for differentiation, proofs. Practice 08 page 93, exercises 6E and miscellaneous ex 6, page 94, 1 tru 6.
      18. Assignment 05, page 86, chapter 6, exercise 6D # 5, 6, 7, 8, 9 and 14.
      19. Practice 09, differentiation, miscellaneous exercise 6 page 94, questions 7 tru 15. Questions 1 tru 6 included as part of Pratice08.
      20. Applications of differentiation. Practice 10a, Exercise 7C, from #6 to 16 page 110. Practice 10a, answers.
      21. Assignment 06, due Fri 11/2nd. Item c) for each question on exercise 7B.
      22. Derivatives as rates of change. Practice 10b, miscellaneous exercise 7, from question 1 to 17, pages 110 to 113.
        Practical Tips for Modeling Optimization Problems.
      23. Sequences, chapter 8 . Practice 11, section 8C (from 8 to 11, p125) and miscellaneous exc 8, p125-126, form 1 to 15. Sequences answers here.
      24. Assignment 07, Sequences, problems 10 & 11 page 125.
      25. Chapter 9, The binomial Theorem. Practice 12 Ex 9B and Miscellaneous 9 from number 1 to 33, pages 135 to 137. Solutions here. Exercises 9B #6 to 13 (page 135) & Miscellaneous excercise 9, #1 to 11.
      26. Chapter 10, Trigonometrypractice 13: excercises 10D, page 152 & miscellaneous 10, page 153 from 1 to 4.
      27. Trigonometry Practice 14, miscellaneous 10, page 153 to 155, # 5 tru 20. Solution here .
      28. Assignment 10, transformation of graphs of trig functions.
      29. Chapter 11, combining and inverting functions, Practice 15, miscellaneous exercise 11 page 172-173 #1 thru #20. Assignment 11, p172: 1 - 4. Solutionshere.
      30. Chapter 12, extending differentiation. Related rates problems ! Practice 16: p182-183. All of them, 1 tru 9. Practice 17, differentiation, related rates: miscellaneous exercise 12, pages 184-186.
      31. Vectors,Practice 18Miscellaneous exercise 13 p207-209. see also Michael Corral, vector Calculus, first 18 pages. Basic theory on vectors, recommended ! click here.
      32. Assignment 13, Vectors: exercise 13D: 12-16 page 20 7 questions & answers here.
      33. Geometric Sequences: Jan 7 & 9. Practice 19, exercises 14A p213-214 Practice 20, 14B p.217 & Practice 21 exercises 14C p 221. Quiz 7 Fri 1/11
        Assignment 14, Geometric seq: page 213, 14A. # 3, b) c) #4 c) #5 b) #6 c) and from 14B page 217 #2 a). Practice 22, geometric sequences miscellaneous exercise 14 pages 222-224.
      34. Second derivatives: Practice 23, 15A p228 & 229, also 15B p232.
      35. Second derivatives: Practice 24, miscellaneous exercise 15 pages 234-235.
      36. Integration: Practice 25, 16B p245 & p251, 16C. Assignment 15: p240 exer 10e, 14, 15e, & p244, 4e and #9. Integration, Practice 26, miscellaneous exercise 16 pages 255-257.
      37. Volumen of revolutions (Integration). Practice 27, ex 17 from 4 to 9 and miscellaneous exercise 17 pages 262-263. Assignment 16 p262 #2 d), #3 f), #6 & 7.
      38. Radians. Practice 28, miscellaneous exercise 18 p275-276. Assignment 17: Page 275 exercise 18D, question 2 a) b) c)

        Pure Mathematics 2

      Numberphile, Cambridge University: amazing youtube channel:


      Watch the video: PLUS TWO MATHEMATICSCHAPTER 1RELATIONS AND FUNCTIONScomposition of functions (December 2021).