P3 Pure Mathematics 3

Example:Simplify x³ − 2x² − 3x4x² − 36.

Solution:First factorise top and bottom fully –

and now cancel all common factors, in this case (x – 3) to give

Example:Simplify 9x² − 43x² − 2x ÷ 3x² − x − 2x³ + x².

Solution:First turn the second fraction upside down and multiply

Example:Simplify 3xx² − 7x + 12 − 5x² − 4x + 3.

Solution:First factorise the denominators

Example:Solve xx + 1 − x − 1x = 12.

Solution:First multiply both sides by the Lowest Common Denominator

Example:Find the range of the function

f : x → 2x – 3 with domain x ∈ ℜ: –2 < x ≤ 4.

Solution:

First sketch the graph for values of x between – 2

and 4 (the domain), and we can see that we are only using

the y–axis from –7 to 5,

not including y = –7 (since x ≠ –2),

but including y = 5 (since x can equal 4)

and so the range is

y ∈ ℜ: –7 < y ≤ 5.

Example:Find the largest possible domain and the range for the function

f : x → (x − 3) + 1

Solution:

First notice that we cannot have the square root of a negative number and so

x – 3 cannot be negative

⇒ x – 3 ≥ 0

⇒ largest possible domain is x ∈ ℜ: x ≥ 3.

To find the range we first sketch the graph

and we see that the graph will cover all of the

y-axis from 1 upwards

and so the range is y ∈ ℜ: y ≥ 1.

Example:Find the largest possible domain and the range for the function

f : x → 2xx + 1. Give the equations of its asymptotes.

Solution:

The only problem occurs when the denominator is 0, and so x cannot be –1.

Thus the largest domain is x ∈ ℜ: x ≠ –1.

To find the range we sketch the graph

and we see that y can take any value

except 2,

so the range is y ∈ ℜ: y ≠ 2.

Asymptotes are x = –1 and y = 2

Examples:

1) The mapping x → √x where x ∈ ℜ is not a function as √–9 is not defined, but if we

restrict the domain to positive or zero real numbers then f : x → √ x where x ∈ ℜ, x ≥ 0

is a function.

2) x → 1x − 3 where x ∈ ℜ is not a function as the image of x = 3 is not defined,

but f : x → 1x − 3 where x ∈ ℜ, x ≠ 3 is a function.

Example:f : x → 3x – 2 and g : x → x² + 1. Find fg and gf.

Solution:

Think of f and g as ‘rules’

f is

g is

⇒ fg is

giving (x² + 1) × 3 – 2 = 3x² + 1

⇒ fg : x → 3x² + 1 or fg(x) = 3x² + 1.

gf is

giving (3x – 2)² + 1 = 9x² – 12x + 5

⇒ gf : x → 9x² – 12x + 5 or gf (x) = 9x² – 12x + 5.

Example:Find the inverse of f : x → 3x – 2.

Solution:

We have x → y = 3x – 2

(i)

interchanging x and y ⇒ x = 3y – 2

(ii)

solving for y ⇒ y = x + 23

⇒

f ^–1 : x → x + 23

Example:Find the inverse of g : x → x + 32x − 5.

Solution:

We have x → y = x + 32x − 5

(i)

interchanging x and y

⇒ x = y + 32y − 5

(ii)

solving for y

⇒ x(2y – 5) = y + 3

⇒

2xy – 5x = y + 3

⇒

2xy – y = 5x + 3

⇒

y(2x – 1) = 5x + 3

⇒

y = 5x + 32x − 1

⇒

g ^–1 : x → 5x + 32x − 1

Example:f (x) = (x – 3)² + 4, x ∈ ℜ, x ≤ 3.

(a)

Sketch the graph of y = f (x), and state its range.

(b)

Find the inverse function, f ^–1(x) and sketch its graph on the same diagram.

Show the line y = x on your diagram.

(c)

State the domain and range of f ^–1(x).

Solution:

(a)

As the domain of f is x ≤ 3, we only

have the ‘left’ part of the parabola.

The range is f (x) ≥ 4, f (x) ∈ ℜ.

(b)

To find the inverse, swap x and y, then

find y.

x = (y – 3)² + 4

⇒

y – 3 = ±√x − 4

From the reflection of y = f (x) in y = x,

we can see that we want the negative

sign

⇒

y = 3 – √x − 4

(c)

The domain of f ^–1(x) is x ≥ 4 (the range of f (x)).

The range of f ^–1(x) is f ^–1(x) ≤ 3 (the domain of f (x)).

Example:Sketch the graph of y = |x² – 3x|.

Solution:

First sketch the graph of y = x² – 3x Then reflect the portion below

the x-axis in the x–axis to give

Example:Sketch the graph of y = |x|² – 3|x|.

Solution:

f (x) = x² – 3x

⇒

f (|x|) = |x|² – 3|x|

First sketch the graph of y = x² – 3x for positive

values of x only.

Then reflect your graph in the y–axis to

complete the sketch.

Examples:

1)

y = 2f (x – 3) is the image of y = f (x) under a stretch in the y-axis of factor 2

followed by a translation 03, or the translation followed by the stretch.

2)

y = 3x² + 6 is the image of y = x² under a stretch in the y-axis of factor 3

followed by a translation 60,

BUT these transformations cannot be done in the reverse order.

To do a translation before a stretch we have to notice that

3x² + 6 = 3(x² + 2) which is the image of y = x² under a translation of 20

followed by a stretch in the y-axis of factor 3.

3)

y = – sin(x + π) is the image of y = sin x under a reflection in the x-axis

followed by a translation of −π0−0π, or the translation followed by the reflection.

Example:Sketch the curve y= 4 − 2x, x > 0

Solution:

Note that the domain is x > 0, so no graph to the left of the y-axis.

x ≠ 0 ⇒ curve does not meet the x-axis

y = 0 ⇒ x = 1

Thinking of y = −2x translated up 4,

the horizontal asymptote is y = 4.

Do not forget that the y-axis, x = 0, is also an asymptote.

Example:Find cosec 35^o.

Solution:cosec 35^o = 1sin 35^o = 10⋅53576... = 1⋅743 to 4 S.F.

Example:Solve sec x = 3⋅2 for 0 ≤ x ≤ 2π

Solution:sec x = 3⋅2 ⇒ 1cos x = 3⋅2 ⇒ cos x = 13⋅2 = 0⋅3125

⇒ x = 1⋅25 or 2π – 1⋅25 = 5⋅03 radians to 3 S.F.

Example:Find the exact value of cos 15^o

Solution:

We know the exact values of sin 45^o, cos 45^o and sin 30^o, cos 30^o

so we consider cos 15 = cos (45 – 30) = cos 45 cos30 + sin 45 sin 30

= 12 × 32 + 12 × 12 = 3 + 122 = 6 + 24 .

Example:Given that A is obtuse and that B is acute, and sin A = 3/5 and cos B = 5/13 find

the exact value of sin (A + B).

Solution:

We know that sin (A + B) = sin A cos B + cos A sin B so we must first find

cos A and sin B.

Using sin² θ + cos² θ = 1

⇒ cos² A = 1 – 9/25 = 16/25

and sin² B = 1 – 25/169 = 144/169

⇒ cos A = ± 4/5 and sin B = ± 12/13

But A is obtuse so cos A is negative, and B is acute so sin B is positive

⇒ cos A = – 4/5 and sin B = 12/13

⇒ sin (A + B) = 3/5 × 5/13 + – 4/5 × 12/13 = –33/65

Example:Prove that

cos 2A + 11 - cos 2A = cot² A .

Solution:

L.H.S. =

cos 2A + 11 - cos 2A = 2 cos² A - 1 + 11 - (1 - 2 sin² A) = 2 cos² A2 sin² A = cot² A . Q.E.D.

Example:Eliminate θ from the parametric equations x = sec θ – 1, y = tan θ.

Solution:

We remember that tan²θ + 1 = sec²θ

⇒ sec²θ – tan²θ = 1.

secθ = x + 1 and tanθ = y

⇒ (x + 1)² – y² = 1

⇒ y² = (x + 1)² – 1 = x² + 2x.

Example:Solve the equation sec²A = 3 – tan A, for 0 ≤ A ≤ 360^o.

Solution:

We know that tan²A + 1 = sec²A

⇒ tan²A + 1 = 3 – tan A

⇒ tan²A + tan A – 2 = 0, factorising gives

⇒ (tan A – 1)(tan A + 2) = 0

⇒ tan A = 1 or tan A = –2

⇒ A = 45, 225, or 116⋅6, 296⋅6.

Example:Solve sin 3x − sin 5x = 0 for 0^o ≤ x ≤ 90^o.

Solution:

Using the formula sin P − sin Q = 2 cos

𝑃+𝑄2 sin 𝑃−𝑄2

⇒ 2 cos 4x sin(−x) = 0

⇒ cos 4x sin x = 0

⇒ cos 4x = 0,

or sin x = 0

⇒ 4x = 90, 270, (450), … or x = 0, (180), …

⇒ x = 0^o, 22⋅5^o or 67⋅5^o.

Example:Solve the equation 12 sin x – 5 cos x = 6 for 0^o ≤ x ≤ 360^o.

Solution:

First re–write in the above form:

notice that the sin x is positive and the cos x is negative so we need formula (4).

R sin (x – α) = R sin x cos α – R cos x sin α = 12 sin x – 5 cos x

Equating coefficients of sin x, ⇒ R cos α = 12 I

Equating coefficients of cos x, ⇒ R sin α = 5 II

Squaring and adding I and II ⇒ R² cos² α + R² sin² α = 12² + 5²

⇒ R² (cos² α + sin² α) = 144 + 25

but cos² α + sin² α = 1

⇒ R² = 169

⇒ R = ±13

But choosing the correct formula means that R is positive ⇒ R = +13

Substitute in I ⇒

cos α = 12/13

⇒ α = 22⋅620…^o or 337⋅379…^o or ……

and choosing the correct formula means that α is acute

⇒ α = 22⋅620…^o.

⇒ 12 sin x – 5 cos x = 13 sin(x – 22⋅620…)

To solve

12 sin x – 5 cos x = 6

⇒ 12 sin x – 5 cos x = 13 sin(x – 22⋅620…) = 6

⇒ sin(x – 22⋅620…) = 6/13

⇒ x – 22⋅620… = 27⋅486… or 180 – 27⋅486… = 152⋅514…

⇒ x = 50⋅1^o or 175⋅1^o .

Example:Find the maximum value of 12 sin x – 5 cos x and the smallest positive value

of x for which it occurs.

Solution:

From the above example 12 sin x – 5 cos x = 13 sin(x – 22.6).

The maximum value of sin(anything) is 1 and occurs when the angle is 90, 450, 810

etc. i.e. 90 + 360n

⇒ the max value of 13 sin(x – 22⋅6) is 13

when x – 22⋅6 = 90 + 360n ⇒ x = 112⋅6 + 360n^o,

⇒ smallest positive value of x is 112⋅6^o.

Example:Solve e^{2x + 3} = 5.

Solution:

Take the natural logarithm of each side,

remember that ln x is the inverse of e^x

⇒ ln (e^{2x + 3}) = ln 5 ⇒ 2x + 3 = ln 5

⇒ x = ln 5 − 32 = –0⋅695 to 3 S.F.

Example:Solve ln (3x – 5) = 4.

Solution:

(i) From the definition of logs log_a x = y ⇔ x = a^y

ln (3x – 5) = 4

⇒ 3x − 5 = e⁴

⇒ x = e⁴ + 53 = 19⋅9 to 3 S.F.

OR

(ii) Raise both sides to the power of e, .

⇒ e^{ln (3x – 5)} = e⁴ ⇒ (3x – 5) = e⁴

remember that e^x is the inverse of ln x

⇒ x = e⁴ + 53 = 19⋅9 to 3 S.F.

Example:y = √(x³ – 2x). Find dydx .

Solution:y = √(x³ – 2x) = (x³ – 2x)¹₂. Put u = x³ – 2x

Example:Differentiate y = x² × √(x – 5).

Solution:y = x² × √(x – 5) = x² × (x – 5)¹₂

so put u = x² and v = (x – 5)¹₂

Example:Differentiate y = 2x − 3x² + 5x

Solution:y = 2x − 3x² + 5x, so put u = 2x – 3 and v = x² + 5x

Example:If y = 3x − 2√x − 1 , find dydx, expressing your answer as a single algebraic fraction in

its simplest form.

Solution:y = 3x − 2(x − 1)¹²

Example:Find the derivative of f (x) = x³ – 5e^x at the point where x = 2.

Solution:f (x) = x³ – 5e^x

Example:Differentiate the function f (x) = ln 3x – ln x⁵

Solution:f (x) = ln 3x – ln x⁵ = ln 3 + ln x – 5 ln x = ln 3 – 4 ln x

Example:Find the derivative of f (x) = log₁₀ 3x.

Solution:f (x) = log₁₀ 3x = ln 3xln 10using change of base formula

Example:y = e^x2. Find dydx .

Solution:y = e^u, where u = x²

Example:y = ln 7x³. Find dydx .

Solution:y = ln u, where u = 7x³

Example:x = sin² 3y. Find dydx .

Solution:First find dxdy as this is easier.

Example:If y = 5^{x + 2}, find dydx .

Solution 1:y = 5^x × 5² = 25 × 5^x

Solution 2: y = 5^{x + 2} = 5^u

Example:y= 7x², find dydx

Solution:y = 7^u ⇒ dydx = 7^u ln 7 × dudx = 7x²ln 7 × 2xusing the chain rule

Example:y = sin⁴ x. Find dydx .

Solution:y = sin⁴ x. Put u = sin x

Example:y = e^{sin x}. Find dydx .

Solution:y = e^u, where u = sin x

Example:y = ln (sec x). Find dydx .

Solution:y = ln u, where u = sec x

Example:Differentiate y = x² × cosec 3x.

Solution:dydx = u dvdx + v dudx

y = x² × cosec 3x

Put u = x² and v = cosec 3x

Example:Differentiate y = tan 2x7x³

Solution:y = uv ⇒ dydx = v dudx − u dvdxv²

y = tan 2x7x³

Put u = tan 2x and v = 7x³

Example:

(a) Show that a root, α, of the equation f (x) = x³ – 8x – 7 = 0 lies between 3 and 4.

(b) Show that the equation x³ – 8x – 7 = 0 can be re–arranged as x = ³8x + 7 .

(c) Starting with x₁ = 3, use the iteration x_{n + 1} = ³8x_n + 7 to find the first four iterations for x.

(d) Show that your value of x₄ is correct to 3 S.F.

Solution:

(a) f (3) = 27 – 24 – 7 = –4, and f (4) = 64 – 32 – 7 = +25

Thus f (x) changes sign and f (x) is continuous ⇒ there is a root between 3 and 4.