Mathematics · Oct / Nov 2025 · Paper 31 · Question 2

2

+

−

2(1)

2

+

3(1)

−

4

1

2(2)

3(2)

4

1

=

−

=

and

f(2)

=

−

=

...

f(1)

...

1

2

2

2

e

1

e

and

There is a sign change and f(x) is continuous over the

interval hence root (in the interval [1,2] )

2

+

−

2

x

3

x

4

1

−

=

0

2

x

4

+

3

x

3

=

e

x

+

4

x

2

e

x

x

2

3 e

x

+

4

x

2

x

3

(2

x

+

3)

=

e

x

+

4

x

2

=

*

x

2

x

+

3

1

+

2

e

4(1)

x

=

=

awrt1.1035

x

=

awrt1.1484

(i)

3

2

2(1)

+

3

3

(ii)

1.1813

(a)

M1:

Attempts f (1) and f (2) . Values embedded in the expression is sufficient

(accept slips if the intention is clear), or if no substitution is seen accept

one correct value (truncated or rounded to 1sf) with an attempt at both

made.

A1:

Both f (1) and f (2) correct (truncated or rounded to 1sf), reason (e.g. “sign

change” or < 0, > 0 shown) and conclusion. Must refer to the function

being continuous (over the interval) in some way, though be tolerant with

language used.

Note: a narrower interval may be used but if so must contain the root 1.183… to

score the M.

(b)

M1:

Sets f(x) = 0 (may be implied), attempts to multiply both sides by

and isolates the quartic and cubic terms on one side.

A1*:

Factorises the lefthand side and proceeds to the given answer via an

intermediate step with no errors seen.

Note: Working backwards is possible. For the M look for cubing and cross

multiplying to reach a similar intermediate stage, and all correct for the A

mark.

(c)

(i)

M1:

Substitutes

x = into the iterative formula. May be implied by awrt 1.10 or

1

awrt 1.15 if substitution not seen. Condone miscopies if the substitution is

shown. NB Correct value for x3 stated implies this mark.

A1:

(x3 =) awrt 1.1484. Must be x3 and not another term (if labelled).

(ii)

A1:

cao 1.1813 as the final answer (cannot be scored without M1 being scored

in (c)(i))

−