CIE Pure Maths P1 June 2013

(1)

Video worked solution to 1

(2)

Video worked solution to 2

(3)

Video worked solution to 3

(4)

Video worked solution to 4

(5)

Video worked solution to 5(i) | 5(ii)

(6)

Video worked solution to 6

(7)

Video worked solution to 7(i) | 7(ii)

(8)

Video worked solution to 8(i)(ii)(iii) | 8(iv)

(9)

Video worked solution to 9(i) | 9(ii) | 9(iii)

(10)

Video worked solution to 10(i) | 10(ii)(iii)

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Rate of Change

Problem 1: A rectangular water tank (see figure below) is being filled at the constant rate of 20 liters / second. The base of the tank has dimensions w = 1 meter and L = 2 meters. What is the rate of change of the height of water in the tank?(express the answer in cm / sec). 

 

Solution to Problem 1:

• The volume V of water in the tank is given by.
  
  V = w*L*H
  
• We know the rate of change of the volume dV/dt = 20 liter /sec. We need to find the rate of change of the height H of water dH/dt. V and H are functions of time. We can differentiate both side of the above formula to obtain
  
  dV/dt = W*L*dH/dt
  
• note W and L do not change with time and are therefore considered as constants in the above operation of differentiation.
  
• We now find a formula for dH/dt as follows.
  
  dH/dt = dV/dt / W*L
  
• We need to convert liters into cubic cm and meters into cm as follows
  
  1 liter = 1 cubic decimeter
  = 1000 cubic centimeters
  = 1000 cm ³
  
  and 1 meter = 100 centimeter.
  
• We now evaluate the rate of change of the height H of water.
  
  dH/dt = dV/dt / W*L
  
  = ( 20*1000 cm ³ / sec ) / (100 cm * 200 cm)
  
  = 1 cm / sec.

Problem 2: An airplane is flying in a straight direction and at a constant height of 5000 meters (see figure below). The angle of elevation of the airplane from a fixed point of observation is a. The speed of the airplane is 500 km / hr. What is the rate of change of angle a when it is 25 degrees? (Express the answer in degrees / second and round to one decimal place). 

Solution to Problem 2: 

• The airplane is flying horizontally at the rate of dx/dt = 500 km/hr. We need a relationship between angle a and distance x. From trigonometry, we can write
  
  tan a = h/x
  
• angle a and distance x are both functions of time t. Differentiate both sides of the above formula with respect to t.
  
  d(tan a)/dt = d(h/x)/dt
  
• We now use the chain rule to further expand the terms in the above formula
  
  d(tan a)/dt = (sec ² a) da/dt
  
  d(h/x)/dt = h*(-1 / x ²) dx/dt.
  
  (note: height h is constant)
  
• Substitute the above into the original formula to obtain
  
  (sec ² a) da/dt = h*(-1 / x ²) dx/dt
  
• The above can be written as
  
  da/dt = [ h*(-1 / x ²) dx/dt ] / (sec ² a)
  
• We now use the first formula to find x in terms of a and h follows
  
  x = h / tan a
  
• Substitute the above into the formula for da/dt and simplify
  
  da/dt = [ h*(- tan ²a / h ²) dx/dt ] / (sec ² a)
  
  = [ (- tan ²a / h) dx/dt ] / (sec ² a)
  
  = (- sin ²a / h) dx/dt
• Use the values for a, h and dx/dt to approximate da/dt with the right conversion of units: 1km = 1000 m and 1 hr = 3600 sec.
  
  da/dt = [- sin ²(25 deg)/5000 m]*[500 000 m/3600 sec]
  
  = -0.005 radians/sec
  
  = -0.005 * [ 180 degrees / Pi radians] /sec
  
  = -0.3 degrees/sec

Problem 3: If two resistors with resistances R1 and R2 are connected in parallel as shown in the figure below, their electrical behavior is equivalent to a resistor of resistance R such that 

1 / R = 1 / R1 + 1 / R2

If R1 changes with time at a rate r = dR1/dt and R2 is constant, express the rate of change dR / dt of the resistance of R in terms of dR1/dt, R1 and R2. 

Solution to Problem 3: 

• We start by differentiating, with respect to time, both sides of the given formula for resistance R, noting that R2 is constant and d(1/R2)/dt = 0
  
  (-1/R ²)dR/dt = (-1/R1 ²)dR1/dt
  
• Arrange the above to obtain
  
  dR/dt = (R/R1) ²dR1/dt
  
• From the formula 1 / R = 1 / R1 + 1 / R2, we can write
  
  R = R1*R2 / (R1 + R2)
  
• Substitute R in the formula for dR/dt and simplify
  
  dR/dt = (R1*R2 / R1*(R1 + R2)) ²dR1/dt
  
  = (R2 / (R1 + R2)) ²dR1/dt

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Paper 1 & Paper 2 Checkpoint

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Algebra

Algebra IGCSE

On this tutorial you will learn

Here another link for IGCSE course here

Worksheet Further algebra download here

Expanding brackets

• Expanding a single bracket
• A quick way to square a bracket

Solving Equations (Linear type)

• Before you start - What is a term?
• Linear equations with a positive x-term
• Linear equations with a negative x-term
• Equations with two x terms and 2 constants
• Equations with brackets
• Fractional equations (1)
• Fractional equations (2)
• Fractional equations (3)
• Fractional equations (4)
• Fractional equations (5)

Indices

• Introduction
• Negative indices
• Fractions raised to a negative index
• Rational (fractional) indices

Factorising

• Introduction
• HCF types
• Grouping types
  • Quadratic expressions
    • Introduction
    • HCF (Highest Common Factor) types
    • Difference of two squares type
    • Trinomials : (grouping method)
    • Trinomials : (inspection method) 1 | 2 | 3 | 4 | 5 | 6 | 7
• Miscellaneous Exercise
  • Exercise

Quadratic Equations

• Equations
  • Solve by factorising
  • Solve by completing the square
  • Solve by the quadratic formula

Quadratic Graphs

• Sketching quadratic graphs

Simultaneous Equations

• Elimination method for linear types
  • Elimination method for linear equations
• Substitution Method for linear and non linear types
  • Substitution method

Inequalities

• Linear inequalities
  • introduction
  • rules for reversing the inequality sign
  • solving a linear type
  • solving a double inequality

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Quadratic Equations

On this session we will learn how to solve quadratic equation by 

1. Factorising

2. Completing the square

3. Formula

1. Quadratic Equations - Solution by factorising

All quadratic equations have the form

If a quadratic equation factorises then we can solve it by using a factorising method as shown in this video.

Summary Exercise

2. Quadratic Equations - Completing the square

In this tutorial you are shown how to solve a quadratic equation by using the method of completing the square. This is often used when a quadratic equation does not factorise.

3. Quadratic Equations - ABC Formula 

Some quadratic equations cannot be solved by factorising and so an alternative method is the quadratic formula. It is based on a method called completing the square

The first tutorial shows you when to use it, clues that are given in a question and common mistakes that are often made.

In the second video its show you how to derive it.

and Here the video how to derive the formula

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Transformation

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Become second is always forgotten

Alhamdulillah, Alhamdulillah, tsumma alhamdulillah, washsholatu wassalamu 'alaa rosulillah wa 'alaa aalihi washohbihi wa mawwaalah

My dear students

When we talk about the history about energy and electricity they will only mention the man who come up with, Not the man who developed

When we talking about phones, they only mention the man who came up with, Not the man who developed,

when we talking about fuqoha we always start with Imam Syafi'i because he is the first place

There is always no place for second person!

So as a muslim, if we say "i am happy, i am satisfied with the second place"
then i will tell YOU!

THAT IS MENTALITY OF A LOSER!

Because our Messenger SAW said only one person and he aim that

He SAW said " i want to be the one who has the most followers" so He want to be first!

He SAW want US as His followers to be the first and the best nations!

Do you know why today we muslims all over the world became followers or looser Not became a leader?

Its Because we NO longer AIM HIGH!
Remember Big Person always start from a Big Dream

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