Differentiation

5 methods · 6 questions

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When a question gives a polynomial through a property of its roots (a repeated root, all roots positive, roots in a given relation) and asks you to prove something about the coefficients (or vice versa), do not try to find the roots. Name them and write down the elementary symmetric relations $\sum root$ , $\sum root \cdot root$ , $\prod root$ equal to the (signed) coefficients, then push the given condition through those relations.

Trigger: a polynomial described by its roots, with the unknowns living in the coefficients.

Instances:
(i) For $x^{3} - p x^{2} + q x - r$ read off $p = a + b + c$ , $q = ab + b c + c a$ , $r = ab c$ , then impose the extra condition.
(ii) A repeated root: set two named roots equal and eliminate to get a single relation between coefficients.
(iii) Symmetric functions of the roots ( $u + v + \frac{1}{uv}$ , $\frac{1}{u} + \frac{1}{v} + uv$ ) rewritten purely in terms of the coefficients $α, β$ .
(iv) Auxiliary symmetric quantities (such as $pq r, q r s, \dots$ ) built from the roots to feed a later inequality.

Linked questions (2)

STEP 3 2018 · Q1

The function $f$ is given by $f (β) = β - \frac{1}{β} - \frac{1}{β ^{2}} (β \neq = 0) .$ Find the stationary point of the curve $y = f (β)$ and sketch the curve. Sketch also the curve $y = g (β)$ , where $g (β) = β + \frac{3}{β} - \frac{1}{β ^{2}} (β \neq = 0) .$
Let $u$ and $v$ be the roots of the equation $x^{2} + α x + β = 0,$ where $β \neq = 0$ . Obtain expressions in terms of $α$ and $β$ for $u + v + \frac{1}{uv}$ and $\frac{1}{u} + \frac{1}{v} + uv$ .
Given that $u + v + \frac{1}{uv} = - 1$ , and that $u$ and $v$ are real, show that $\frac{1}{u} + \frac{1}{v} + uv ⩽ - 1$ .
Given instead that $u + v + \frac{1}{uv} = 3$ , and that $u$ and $v$ are real, find the greatest value of $\frac{1}{u} + \frac{1}{v} + uv$ .

Worked solution

Part (i).

To locate the stationary point of

y = f (β) = β - \frac{1}{β} - \frac{1}{β ^{2}}

, differentiate and set equal to zero:

f^{'} (β) = 1 + \frac{1}{β ^{2}} + \frac{2}{β ^{3}} = \frac{β ^{3} + β + 2}{β ^{3}} .

The numerator factors as

(β + 1) (β^{2} - β + 2)

. The quadratic factor has discriminant

1 - 8 = - 7 < 0

, so it is always positive. The only real root is therefore

β = - 1

, giving the unique stationary point

(- 1, f (- 1)) = (- 1, - 1)

.

Since

f^{'} (β) > 0

for

β > 0

(all three terms are positive) and the single stationary value occurs at

β = - 1

with

f^{''} (- 1) = - 4 < 0

, the point

(- 1, - 1)

is a local maximum on the branch

β < 0

. The curve has a vertical asymptote at

β = 0

and behaves like

y \sim β

for large

∣ β ∣

. Both branches tend to

- \infty

β \to 0

.

For

y = g (β) = β + \frac{3}{β} - \frac{1}{β ^{2}}

, differentiate:

g^{'} (β) = 1 - \frac{3}{β ^{2}} + \frac{2}{β ^{3}} = \frac{β ^{3} - 3 β + 2}{β ^{3}} = \frac{( β - 1 ) ^{2} ( β + 2 )}{β ^{3}} .

The roots are

β = 1

(double) and

β = - 2

. Since

(β - 1)^{2} ⩾ 0

always, the sign of

g^{'}

β < 0

is determined by

(β + 2) / β^{3}

: this gives a local maximum at

β = - 2

with

g (- 2) = - 2 - \frac{3}{2} - \frac{1}{4} = - \frac{15}{4}

. At

β = 1

, the sign of

g^{'}

does not change (the double root), so

(1, 3)

is a horizontal point of inflection on the positive branch. The sketch shows the same vertical asymptote and the two additional features compared with

f

.

Part (ii).

By Vieta's formulae for

x^{2} + α x + β = 0

u + v = - α

and

uv = β

. Direct substitution gives:

u + v + \frac{1}{uv} = - α + \frac{1}{β},

\frac{1}{u} + \frac{1}{v} + uv = \frac{u + v}{uv} + uv = - \frac{α}{β} + β .

Part (iii).

The condition

u + v + \frac{1}{uv} = - 1

gives

- α + \frac{1}{β} = - 1

, so

α = 1 + \frac{1}{β}

. Substituting into the expression from part (ii):

\frac{1}{u} + \frac{1}{v} + uv = - \frac{α}{β} + β = - \frac{1}{β} (1 + \frac{1}{β}) + β = β - \frac{1}{β} - \frac{1}{β ^{2}} = f (β) .

The task reduces to showing

f (β) ⩽ - 1

whenever

u, v

are real.

For

u, v

real the discriminant must be non-negative:

α^{2} - 4 β ⩾ 0

. Substituting

α = 1 + 1/ β

(1 + \frac{1}{β})^{2} ⩾ 4 β .

For

β < 0

the right-hand side is negative while the left is non-negative, so the inequality holds automatically for all

β < 0

. For

β > 0

, multiply through by

β^{2} > 0

β^{2} + 2 β + 1 ⩾ 4 β^{3} ⟹ 4 β^{3} - β^{2} - 2 β - 1 ⩽ 0.

Factoring:

4 β^{3} - β^{2} - 2 β - 1 = (β - 1) (4 β^{2} + 3 β + 1)

. The quadratic

4 β^{2} + 3 β + 1

has discriminant

9 - 16 < 0

and positive leading coefficient, so it is always positive. Therefore the reality condition for

β > 0

β ⩽ 1

.

The admissible values are thus

β < 0

0 < β ⩽ 1

. In each region, compare

f (β)

with

- 1

:

- For

β < 0

: the unique stationary point on this branch is the local maximum

(- 1, - 1)

from part (i), so

f (β) ⩽ - 1

throughout.
- For

0 < β ⩽ 1

f^{'} (β) = (β^{3} + β + 2) / β^{3} > 0

(all terms positive for

β > 0

), so

f

is strictly increasing on this interval with

f (1) = 1 - 1 - 1 = - 1

. Thus

f (β) ⩽ - 1

for

0 < β ⩽ 1

.

In both cases

\frac{1}{u} + \frac{1}{v} + uv = f (β) ⩽ - 1

, as required.

Part (iv).

Now the condition

u + v + \frac{1}{uv} = 3

gives

α = \frac{1}{β} - 3

, and the same substitution yields:

\frac{1}{u} + \frac{1}{v} + uv = β + \frac{3}{β} - \frac{1}{β ^{2}} = g (β) .

The problem is now to maximise

g (β)

subject to the reality condition

α^{2} ⩾ 4 β

, i.e.,

(1/ β - 3)^{2} ⩾ 4 β

.

For

β < 0

the right-hand side is again negative, so the condition holds for all

β < 0

. For

β > 0

, multiply by

β^{2}

1 - 6 β + 9 β^{2} ⩾ 4 β^{3} ⟹ 4 β^{3} - 9 β^{2} + 6 β - 1 ⩽ 0.

Factoring:

4 β^{3} - 9 β^{2} + 6 β - 1 = (β - 1)^{2} (4 β - 1)

. This is

⩽ 0

precisely when

4 β - 1 ⩽ 0

(using

(β - 1)^{2} ⩾ 0

) or

β = 1

. So the admissible positive values are

0 < β ⩽ \frac{1}{4}

together with

β = 1

.

Evaluate

g

on each admissible region:
-

β < 0

g

has a local maximum at

β = - 2

with

g (- 2) = - \frac{15}{4}

.
-

0 < β ⩽ \frac{1}{4}

g^{'} (β) = (β - 1)^{2} (β + 2) / β^{3} > 0

throughout, so

g

is increasing and attains its maximum at

β = \frac{1}{4}

g (\frac{1}{4}) = \frac{1}{4} + 12 - 16 = - \frac{15}{4}

.
-

β = 1

g (1) = 1 + 3 - 1 = 3

.

By reference to the sketch of

g

from part (i), the point

(1, 3)

is a horizontal inflection sitting above all other values on the admissible set. The greatest value of

\frac{1}{u} + \frac{1}{v} + uv

is therefore

3

STEP 2 2022 · Q8

Let

M = (a c b d)

be a real matrix with

a \neq = d

. The transformation represented by

M

has exactly two distinct invariant lines through the origin.

(i) Show that, if neither invariant line is the

y

-axis, then the gradients of the invariant lines are the roots of the equation

b m^{2} + (a - d) m - c = 0.

If one invariant line is the

y

-axis, what is the gradient of the other?

(ii) Show that, if the angle between the two invariant lines is

4 5^{\circ}

, then

(a - d)^{2} = (b - c)^{2} - 4 b c .

(iii) Find a necessary and sufficient condition, on some or all of

a

b

c

and

d

, for the two invariant lines to make equal angles with the line

y = x

.

(iv) Give an example of a matrix which satisfies both the conditions in parts (ii) and (iii).

Worked solution

Part (i). Suppose neither invariant line is the

y

-axis. Any line through the origin (other than the

y

-axis) has the form

y = m x

, so we look for points

(x, m x)

that map to other points on the same line, say

(x^{'}, m x^{'})

. The condition is:

(a c b d) (x m x) = (x^{'} m x^{'})

This gives

x^{'} = a x + bm x

and

m x^{'} = c x + d m x

. Substituting the first into the second:

c x + d m x = m (a x + bm x) = ma x + b m^{2} x

Since the line is not the

y

-axis,

x \neq = 0

, so we may divide by

x

and rearrange:

b m^{2} + (a - d) m - c = 0

as required. If instead the

y

-axis is invariant, then any point

(0, y)

must map to

(0, y^{'})

, forcing

b = 0

. For the second line

y = m x

(with

x \neq = 0

), the same elimination gives

c + d m = ma

, so

m = \frac{a - d}{c}

Part (ii). The directions of the two lines can be encoded as vectors

(1 m_{1})

and

(1 m_{2})

. Since the angle between the lines is

4 5^{\circ}

, using the dot product formula and squaring both sides:

(1 + m_{1} m_{2})^{2} = \frac{1}{2} (1 + m_{1}^{2}) (1 + m_{2}^{2})

2 (1 + m_{1} m_{2})^{2} = 1 + m_{1}^{2} + m_{2}^{2} + m_{1}^{2} m_{2}^{2}

Writing

m_{1}^{2} + m_{2}^{2} = (m_{1} + m_{2})^{2} - 2 m_{1} m_{2}

2 (1 + m_{1} m_{2})^{2} = (m_{1} + m_{2})^{2} - 2 m_{1} m_{2} + 1 + (m_{1} m_{2})^{2} (*)

From part (i),

m_{1}, m_{2}

are roots of

b m^{2} + (a - d) m - c = 0

, so by Vieta's formulae:

m_{1} + m_{2} = \frac{d - a}{b}, m_{1} m_{2} = - \frac{c}{b}

Substituting into

(*)

and multiplying through by

b^{2}

2 (b - c)^{2} = (d - a)^{2} + 2 c b + b^{2} + c^{2}

2 (b - c)^{2} = (a - d)^{2} + 2 b c + (b - c)^{2} + 2 b c

∴ (a - d)^{2} = (b - c)^{2} - 4 b c

The special case where one line is the

y

-axis (so

b = 0

) gives the other line gradient

\pm 1

from

4 5^{\circ}

, hence

(a - d)^{2} = c^{2}

, which matches the formula with

b = 0

.

Part (iii). The key geometric insight: two lines through the origin make equal angles with

y = x

precisely when they are reflections of each other in

y = x

. If a point

(p, q)

lies on one line, its reflection

(q, p)

lies on the other, so the gradients satisfy

m_{1} m_{2} = 1

.

Using Vieta's formulae,

m_{1} m_{2} = - c / b

, so the condition becomes

b + c = 0

.

Special case: if one line is the

y

-axis and the other makes equal angles with

y = x

, the other must be the

x

-axis (

m = 0

). This forces

b = 0

and

c = 0

, which is consistent with

b + c = 0

.

The necessary and sufficient condition is therefore

b + c = 0

.

Part (iv). We need both

b + c = 0

and

(a - d)^{2} = (b - c)^{2} - 4 b c

. Setting

c = - b

(a - d)^{2} = (2 b)^{2} + 4 b^{2} = 8 b^{2}

Choosing

b = 2

gives

c = - 2

and

(a - d)^{2} = 16

, e.g.

a = 5

d = 1

. One valid matrix is:

M = (5 - 2 21)