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Functions & Graphs | The Maths Tailor — UK Methods

Functions & Graphs

7 methods · 26 questions

To find how many real roots $f (x) = 0$ has (or to fix a parameter so the count is a target value), differentiate, locate the stationary points, and evaluate $f$ there. The number of times the graph crosses the axis is dictated by how that ordered list of stationary values straddles $0$ , together with the end behaviour. A turning value $> 0$ and the next one $< 0$ force a crossing between them; consecutive stationary values of the same sign give no crossing there. For a cubic this collapses to the clean rule: three distinct real roots exactly when local max $> 0$ and local min $< 0$ .

Trigger: a polynomial (or polynomial-after-rearranging) and a question about the number of real roots or the values of a constant giving a prescribed number.

Instances:
(i) evaluate $f$ at every stationary point and count sign changes across the list to read off the crossings;
(ii) for three distinct roots impose (local max value) $\times$ (local min value) $< 0$ , which turns into a band on the parameter;
(iii) fix where the roots sit (e.g. how many are positive) by also testing the sign of $f$ at $x = 0$ or another reference point.

Linked questions (5)

TMUA 2016 Paper 1 — Q13

How many real roots does the equation

3 x^{5} - 10 x^{3} - 120 x + 30 = 0

have?

(A)

1

(B)

2

(C)

3

(D)

4

(E)

5

Voir la correction

Let

f (x) = 3 x^{5} - 10 x^{3} - 120 x + 30

. Differentiate:

f^{'} (x) = 15 x^{4} - 30 x^{2} - 120 = 15 (x^{4} - 2 x^{2} - 8) = 15 (x^{2} - 4) (x^{2} + 2)

. The factor

x^{2} + 2 > 0

always, so stationary points occur at

x = \pm 2

. At

x = - 2

f (- 2) = 3 (- 32) - 10 (- 8) - 120 (- 2) + 30 = - 96 + 80 + 240 + 30 = 254 > 0

(local maximum). At

x = 2

f (2) = 3 (32) - 10 (8) - 120 (2) + 30 = 96 - 80 - 240 + 30 = - 194 < 0

(local minimum). Since

f (x) \to - \infty

x \to - \infty

f (x) \to + \infty

x \to + \infty

, and the function has a local max above zero and a local min below zero, the graph crosses the

x

-axis exactly 3 times. Hence the answer is C.

TMUA 2018 Paper 1 — Q9

Find the complete set of values of the constant

c

for which the cubic equation

2 x^{3} - 3 x^{2} - 12 x + c = 0

has three distinct real solutions.

(A)

- 20 < c < 7

(B)

- 7 < c < 20

(C)

c > 7

(D)

c > - 7

(E)

c < 20

(F)

c < - 20

Voir la correction

Let

g (x) = 2 x^{3} - 3 x^{2} - 12 x + c

. Differentiating:

g^{'} (x) = 6 x^{2} - 6 x - 12 = 6 (x - 2) (x + 1)

, giving stationary points at

x = - 1

and

x = 2

. Evaluating:

g (- 1) = - 2 - 3 + 12 + c = 7 + c

(local maximum) and

g (2) = 16 - 12 - 24 + c = - 20 + c

(local minimum). For three distinct real roots, the local maximum must be positive and the local minimum must be negative:

7 + c > 0 ⟹ c > - 7 and - 20 + c < 0 ⟹ c < 20.

Combining:

- 7 < c < 20

. Hence the answer is B.

TMUA 2018 Paper 2 — Q15

It is given that

f (x) = x^{3} + 3 q x^{2} + 2

, where

q

is a real constant.

The equation

f (x) = 0

has 3 distinct real roots.

Which of the following statements is/are necessarily true?

I. The equation

f (x) + 1 = 0

has 3 distinct real roots.

II. The equation

f (x + 1) = 0

has 3 distinct real roots.

III. The equation

f (- x) - 1 = 0

has 3 distinct real roots.

(A) none of them
(B) I only
(C) II only
(D) III only
(E) I and II only
(F) I and III only
(G) II and III only
(H) I, II and III

Voir la correction

The stationary points of

f (x) = x^{3} + 3 q x^{2} + 2

satisfy

f^{'} (x) = 3 x^{2} + 6 q x = 0

, giving

x = 0

x = - 2 q

. The stationary values are

f (0) = 2

and

f (- 2 q) = 2 + 4 q^{3}

.

For

f (x) = 0

to have 3 distinct roots, the stationary values must have opposite signs. Since

f (0) = 2 > 0

, we need

2 + 4 q^{3} < 0

, i.e.

q^{3} < - \frac{1}{2}

.

Statement I:

y = f (x) + 1

shifts the graph up by 1. The new stationary values are

3

and

3 + 4 q^{3}

. Since

2 + 4 q^{3} < 0

, we have

3 + 4 q^{3} < 1

, which could be positive or negative. If

- 1 < 4 q^{3} + 2 < 0

, then

3 + 4 q^{3}

is between 2 and 3, which is positive, meaning

y = f (x) + 1

has only one root. So I is not necessarily true.

Statement II:

y = f (x + 1)

is a horizontal shift, which does not change the number of real roots. So II is necessarily true.

Statement III:

y = f (- x) - 1

is a reflection in the

y

-axis followed by a downward shift of 1. Reflection does not change root count. The new stationary values are

2 - 1 = 1 > 0

(at the local max after reflection) and

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

. Since

q^{3} < - \frac{1}{2}

, we have

4 q^{3} < - 2

, so

1 + 4 q^{3} < - 1 < 0

. The two stationary values have opposite signs, guaranteeing 3 distinct roots. So III is necessarily true.

Statements II and III are necessarily true. Hence the answer is G.

TMUA 2019 Paper 1 — Q20

What is the complete range of values of

k

for which the curves with equations

y = x^{3} - 12 x

and

y = k - (x - 2)^{2}

intersect at three distinct points, of which exactly two have positive

x

-coordinates?

(A)

- 4 < k < 0

(B)

- 4 < k < 4

(C)

- 4 < k < 16

(D)

- 16 < k < 0

(E)

- 16 < k < 4

(F)

- 16 < k < 16

Voir la correction

Setting the two expressions equal:

x^{3} - 12 x = k - (x - 2)^{2}

, so

x^{3} + x^{2} - 16 x + 4 - k = 0

. Let

f (x) = x^{3} + x^{2} - 16 x + (4 - k)

. Find stationary points:

f^{'} (x) = 3 x^{2} + 2 x - 16 = 0

, giving

x = \frac{- 2 \pm 14}{6}

, so

x = 2

x = - \frac{8}{3}

. We have

f (0) = 4 - k

and

f (2) = 8 + 4 - 32 + 4 - k = - 16 - k

. For exactly three distinct intersection points with exactly two having positive

x

-coordinates, the cubic must have one negative root and two positive roots. This requires the local minimum at

x = 2

to lie below the

x

-axis and

f (0) > 0

f (2) < 0 \Rightarrow - 16 - k < 0 \Rightarrow k > - 16

f (0) > 0 \Rightarrow 4 - k > 0 \Rightarrow k < 4

. These two conditions together give

- 16 < k < 4

. Hence the answer is E.

TMUA Specimen Paper 1 — Q13

How many real roots does the equation

x^{4} - 4 x^{3} + 4 x^{2} - 10 = 0

have?

(A)

0

(B)

1

(C)

2

(D)

3

(E)

4

Voir la correction

Let

f (x) = x^{4} - 4 x^{3} + 4 x^{2} - 10

. Differentiating:

f^{'} (x) = 4 x^{3} - 12 x^{2} + 8 x = 4 x (x^{2} - 3 x + 2) = 4 x (x - 1) (x - 2)

. The stationary points are at

x = 0, 1, 2

. Evaluating:

f (0) = - 10

f (1) = 1 - 4 + 4 - 10 = - 9

f (2) = 16 - 32 + 16 - 10 = - 10

. All three stationary values are negative, and

f (x) \to + \infty

x \to \pm \infty

. The function rises from

- \infty

, passes through local min

f (0) = - 10

, local max

f (1) = - 9

(still negative), local min

f (2) = - 10

, then rises to

+ \infty

. Crossing zero occurs exactly twice: once for

x < 0

and once for

x > 2

. Hence the answer is C.