Course

Cambridge A Level Mathematics: Pure 3 (9709 Paper 3)

Name: Cambridge A Level Mathematics: Pure 3 (9709 Paper 3)
Price: 2499 BDT

Paper 3 of 9709, the pure paper the full A Level adds: algebra, logarithms, trigonometry, calculus with every function, numerical methods, vectors in three dimensions, differential equations and complex numbers.

100 lessons19h 5mEnglish

What you'll learn

Solve modulus equations and inequalities, divide polynomials, and use the factor and remainder theorems
Split fractions into partial fractions and expand (1 + x)ⁿ for any rational n, stating when it is valid
Use logarithms, e and ln x to solve equations and to find the constants in y = kxⁿ and y = k(aˣ)
Use all six trigonometric functions, the compound and double angle formulae, and the R form
Differentiate and integrate every Paper 3 function, choosing between substitution, partial fractions and parts
Find a root to a given accuracy with a sign change and an iteration
Work with lines in three dimensions: equations, angles, intersections and the foot of a perpendicular
Form and solve differential equations by separating the variables, and explain what the answer means
Calculate with complex numbers in Cartesian and polar form, and draw loci on an Argand diagram
Answer Paper 3 questions set out the way the mark scheme expects

Requirements

Pure Mathematics 1, from our course or from school
Confident with the chain rule, basic integration and solving trigonometric equations at Pure 1 level

Course content

100 lessons · 19h 5m

3.1 · Algebra

The modulus as a distance, the graph of y = |ax + b|, modulus equations and inequalities, dividing one polynomial by another, and the factor and remainder theorems.

0/12

The modulus function

|x| is the distance of x from 0, so it is never negative: |−4| = 4 and |4| = 4.

Preview12m 2.

Graphs of y = |ax + b|

Draw y = ax + b, then reflect the part below the x-axis upwards. The corner sits where ax + b = 0.

Preview12m

Solving modulus equations

Square both sides when |a| = |b|, or split into two cases, and check each answer in the original.

13m

Solving modulus inequalities

|x − a| < b means x is within b of a, so a − b < x < a + b. Graphs show which side is true.

13m

Division of polynomials

Long division by a linear polynomial, like dividing numbers: find the quotient and the remainder.

13m

Division of polynomials: dividing by a quadratic

The same steps with a quadratic divisor. The remainder can now be linear, like 3x − 2.

12m

The factor theorem

If p(a) = 0 then (x − a) is a factor, and if p(b/a) = 0 then (ax − b) is a factor.

13m

The remainder theorem

Dividing p(x) by (x − a) leaves remainder p(a), so you can find a remainder without dividing.

12m

Using the factor and remainder theorems

Find unknown coefficients from two facts, then factorise a cubic fully and solve p(x) = 0.

14m

10.

Algebra: practice

A timed drill of mixed questions: modulus graphs, equations, inequalities, division and both theorems.

10m

11.

Algebra: exam practice

Paper 3 style questions with full worked solutions and mark schemes.

15m

12.

Algebra: summary

The key rules, methods and common mistakes of this chapter on one screen.

3.2 · Logarithmic and exponential functions

Logarithms as indices, the laws of logarithms, equations and inequalities with the unknown in an index, e and ln x as inverse functions, and changing a curve into a straight line to find its constants.

0/14

Logarithms to base 10

lg 1000 = 3 because 10³ = 1000. A logarithm is the index you need.

Preview12m 2.

Logarithms to base a

y = aˣ and x = logₐ y say the same thing. Switch between the two forms both ways.

Preview12m

The laws of logarithms

Each law is an index law written again: multiply means add the logs, and a power becomes a multiplier.

13m

Solving logarithmic equations

Combine the logs into one, change to index form, then reject any answer that makes a log undefined.

12m

Solving exponential equations

Take logs of both sides so the power law brings the unknown down from the index.

13m

Solving exponential inequalities

Take logs as for an equation, and turn the sign round when you divide by a negative log.

12m

Natural logarithms: e and ln x

e is about 2.718, and ln x is the logarithm to base e. It follows every law you already know.

13m

The graphs of eˣ and ln x

ln x and eˣ are inverse functions, so each graph is the other reflected in y = x.

13m

Solving equations with e and ln

Use ln to undo e and e to undo ln, and spot a quadratic in eˣ.

12m

10.

Transforming a relationship to linear form: y = kxⁿ

Plot ln y against ln x to get a straight line. The gradient is n and the intercept is ln k.

14m

11.

Transforming a relationship to linear form: y = k(aˣ)

Plot ln y against x instead. The gradient is ln a and the intercept is ln k.

14m

12.

Logarithmic and exponential functions: practice

A timed drill of mixed questions: the laws, equations, inequalities, e and ln, and linear form.

10m

13.

Logarithmic and exponential functions: exam practice

Paper 3 style questions with full worked solutions and mark schemes.

15m

14.

Logarithmic and exponential functions: summary

The laws, the graphs and the common mistakes of this chapter on one screen.

3.3 · Trigonometry

Secant, cosecant and cotangent and their graphs, two new Pythagorean identities, the compound and double angle formulae, proving identities, and writing a sin θ + b cos θ as one wave R sin(θ ± α) or R cos(θ ± α).

0/13

The cosecant, secant and cotangent ratios

sec θ = 1/cos θ, cosec θ = 1/sin θ and cot θ = 1/tan θ, with exact values from the ones you know.

Preview12m 2.

Graphs of sec θ, cosec θ and cot θ

Where cos θ is 0, sec θ has an asymptote. Where cos θ is 1 or −1, the two graphs touch.

Preview13m

The identities sec²θ = 1 + tan²θ and cosec²θ = 1 + cot²θ

Divide sin²θ + cos²θ = 1 by cos²θ or by sin²θ, and two new identities appear.

12m

Solving equations with sec, cosec and cot

Use an identity to leave one function, solve the quadratic, then find every angle in the range.

13m

Compound angle formulae: sin(A ± B) and cos(A ± B)

sin(A + B) is not sin A + sin B. The formulae give exact values such as sin 75°.

14m

Compound angle formulae: tan(A ± B)

Divide the sin formula by the cos formula, then use it for exact values and equations.

12m

Double angle formulae

Put B = A in the compound angle formulae. cos 2A has three forms: choose the one that helps.

14m

Further trigonometric identities

Prove an identity by working on one side only, choosing the identity that fits.

14m

Expressing a sin θ + b cos θ in the form R sin(θ ± α) or R cos(θ ± α)

Two waves of the same period add to one wave. R = √(a² + b²) and tan α comes from a and b.

14m

10.

Using the R form: greatest values and equations

R sin(θ + α) lies between −R and R, and 3 cos θ + 2 sin θ = 1 becomes a one-function equation.

13m

11.

Trigonometry: practice

A timed drill of mixed questions: the six functions, the identities, exact values and the R form.

10m

12.

Trigonometry: exam practice

Paper 3 style questions with full worked solutions and mark schemes.

15m

13.

Trigonometry: summary

Every identity and formula of this chapter, with the common mistakes, on one screen.

3.4 · Differentiation

The product and quotient rules, the derivatives of eˣ, ln x, sin x, cos x and tan x, and the first derivative of a curve given implicitly or by parameters, used for tangents, normals and stationary points.

0/12

The product rule

For y = uv, dy/dx = u dv/dx + v du/dx. A growing rectangle gains a strip on each side.

Preview13m 2.

The quotient rule

For y = u/v, dy/dx = (v du/dx − u dv/dx)/v². The order on top matters.

Preview13m

Derivatives of exponential functions

The gradient of eˣ equals its height everywhere, and e to the power f(x) differentiates to f′(x) times itself.

13m

Derivatives of natural logarithmic functions

ln x differentiates to 1/x, and ln f(x) to f′(x)/f(x). Use the laws of logs first to make it easier.

13m

Derivatives of trigonometric functions: sin x and cos x

The gradient of sin x traces cos x, and the gradient of cos x traces −sin x, with x in radians.

13m

Derivatives of trigonometric functions: tan x

Write tan x as sin x / cos x and use the quotient rule: the answer is sec²x.

12m

Implicit differentiation

Differentiate every term with respect to x. Each y term gains a dy/dx, then collect and factorise.

14m

Parametric differentiation

When x and y both depend on t, dy/dx = (dy/dt) ÷ (dx/dt), and the answer is in terms of t.

13m

Tangents and normals to implicit and parametric curves

Find the gradient at a point, then the tangent and the normal. Set dy/dx = 0 for stationary points.

14m

10.

Differentiation: practice

A timed drill of mixed questions: which rule, which derivative, and gradients of implicit and parametric curves.

10m

11.

Differentiation: exam practice

Paper 3 style questions with full worked solutions and mark schemes.

15m

12.

Differentiation: summary

Every derivative and rule of this chapter, with the common mistakes, on one screen.

3.5 · Integration

Reversing the new derivatives: e to the power ax + b, 1/(ax + b), sin, cos and sec² of ax + b, then using the double angle formulae to integrate sin²x, cos²x and more.

0/8

Integration of exponential functions

eˣ integrates to itself, and e to the power ax + b integrates to that power divided by a.

Preview12m 2.

Integration of 1/(ax + b)

1/x integrates to ln|x|, which fills the gap in the power rule. 1/(ax + b) gives (1/a) ln|ax + b|.

Preview13m

Integration of sin(ax + b), cos(ax + b) and sec²(ax + b)

Read each trigonometric derivative backwards, watch the minus sign, and divide by a.

13m

Further integration of trigonometric functions: sin²x and cos²x

There is no rule for sin²x, so rewrite it with cos 2x first. The same works for cos²(2x).

13m

Further integration of trigonometric functions: tan²x and other identities

tan²x = sec²x − 1 and sin x cos x = ½ sin 2x turn hard integrals into standard ones.

13m

Integration: practice

A timed drill of mixed questions: the standard integrals, exact definite values and identities.

Integration: exam practice

Paper 3 style questions with full worked solutions and mark schemes.

15m

Integration: summary

Every standard integral of this chapter, with the common mistakes, on one screen.

3.6 · Numerical solutions of equations

Find roughly where a root is with a sketch or a change of sign, then improve it with an iteration xₙ₊₁ = F(xₙ) to a set accuracy, and see why some iterations fail.

0/10

Finding a starting point: using graphs

Some equations have no exact method. Sketch two graphs and count where they cross.

Preview12m 2.

Finding a starting point: a change of sign

If f(a) and f(b) have opposite signs and f is continuous, a root lies between a and b.

Preview13m

Improving your solution: iterative formulae

Put each answer back into xₙ₊₁ = F(xₙ). The values x₁, x₂, x₃ get closer and closer to the root.

13m

How an iterative formula relates to the equation

If the values settle on α, then α = F(α). Rearrange that to find the equation being solved.

13m

Using an iteration to a given accuracy

Keep going until two values agree to the accuracy asked, then show it with a change of sign.

13m

When an iteration fails to converge

A staircase or cobweb diagram shows the values moving away. Another rearrangement may work.

13m

Using iteration in problems from other topics

A sector area or a stationary point can lead to an equation with no exact answer. Iterate it.

14m

Numerical solutions of equations: practice

A timed drill of mixed questions: sign changes, iterations, accuracy and the equation being solved.

10m

Numerical solutions of equations: exam practice

Paper 3 style questions with full worked solutions and mark schemes.

15m

10.

Numerical solutions of equations: summary

The methods, the wording examiners want and the common mistakes of this chapter on one screen.

3.1 · Further algebra

Split a fraction into partial fractions in the three syllabus forms, expand (1 + x)ⁿ and (a + x)ⁿ when n is negative or a fraction, say when the expansion is valid, and combine the two.

0/11

Improper algebraic fractions

When the top has the same degree as the bottom, divide first to get a number plus a proper fraction.

Preview12m 2.

Partial fractions: distinct linear factors

One constant over each linear factor. Substitute the value that makes a factor zero to find each constant.

Preview13m

Partial fractions: a repeated linear factor

(cx + d)² needs a term over (cx + d) and a term over (cx + d)². Three unknowns, three equations.

13m

Partial fractions: a quadratic factor

Over cx² + d the top is Bx + C, not a single constant. Compare coefficients to finish.

13m

Binomial expansion of (1 + x)ⁿ for values of n that are not positive integers

The same formula as before, but the series never stops, so it is only true when |x| < 1.

14m

Expanding (1 + bx)ⁿ and the values of x for which it is valid

Replace x by bx everywhere, brackets included. The expansion is valid when |bx| < 1.

13m

Binomial expansion of (a + x)ⁿ for values of n that are not positive integers

Take out aⁿ first to get aⁿ(1 + x/a)ⁿ. Now it is valid when |x| < |a|.

13m

Partial fractions and binomial expansions

Split into partial fractions, expand each piece, add, and give the range where both are valid.

14m

Further algebra: practice

A timed drill of mixed questions: choosing the form, finding constants, expansions and validity.

10m

10.

Further algebra: exam practice

Paper 3 style questions with full worked solutions and mark schemes.

15m

11.

Further algebra: summary

The three partial fraction forms, the binomial series and the common mistakes on one screen.

3.4-3.5 · Further calculus

The derivative of tan⁻¹ x and the integral of 1/(x² + a²), then the integration methods: k f′(x)/f(x), a given substitution, partial fractions and integration by parts, and how to choose between them.

0/12

Derivative of tan⁻¹ x

Write x = tan y, differentiate, and use sec²y = 1 + x². The answer is 1/(1 + x²).

Preview12m 2.

Integration of 1/(x² + a²)

Read the tan⁻¹ derivative backwards: the integral is (1/a) tan⁻¹(x/a) + c.

Preview12m

Integration of k f′(x)/f(x)

When the top is a multiple of the derivative of the bottom, the answer is a multiple of ln|f(x)|.

13m

Integration by substitution

Use the given u. Replace x and dx together, integrate in u, then write the answer in x again.

14m

Integration by substitution: definite integrals

Change the limits to values of u as well, and you never need to go back to x.

13m

The use of partial fractions in integration

Split first. Each linear piece gives a ln, a squared piece gives a power, and a quadratic piece may give tan⁻¹.

14m

Integration by parts

The product rule backwards. Choose u so that du/dx is simpler, such as x in x sin 2x.

14m

Integration by parts: ln x, tan⁻¹ x and using it twice

Write ln x as 1 × ln x and take u = ln x. For x²e⁻ˣ, use the method two times.

14m

Further integration: choosing a method

Look at the shape of the integrand first: standard form, f′/f, substitution, partial fractions or parts.

14m

10.

Further calculus: practice

A timed drill of mixed integrals: name the method, then carry it out.

10m

11.

Further calculus: exam practice

Paper 3 style questions with full worked solutions and mark schemes.

15m

12.

Further calculus: summary

Every method of this chapter, when to use it and the common mistakes, on one screen.

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