ZIMSEC A Level Pure Mathematics Paper 1 November 2018

Question 101

[1 marks]Complex numbers

The equation

x^3 - 2x^2 + 4x - 8 = 0

has a root

x = 2i

. Find the other two roots.

A $x = -2i$ and $x = 2$
B $x = -2i$ and $x = -2$
C $x = 2i$ and $x = 2$
D $x = -2i$ and $x = 4$

Question 102

[2 marks]Complex numbers

Since the coefficients of

x^3-2x^2+4x-8=0

are real, the root

x=2i

implies

x=-2i

is also a root, so

(x-2i)(x+2i)=x^2+4

is a factor. Divide

x^3-2x^2+4x-8

x^2+4

to find the remaining linear factor, in the form

x+k

. State

k

Answer this when you sit the paper.

Question 201

[1 marks]Trigonometric equations

Solve

\cos(\theta - 75°) = \dfrac{\sqrt{6}-\sqrt{2}}{4}

for

0° \le \theta \le 360°

A $\theta = 0°,\ 150°,\ 360°$
B $\theta = 150°$ only
C $\theta = 15°$ and $\theta = 135°$
D $\theta = 75°$ and $\theta = 285°$

Question 202

[2 marks]Trigonometric equations

Recognise that

\dfrac{\sqrt6-\sqrt2}{4}=\cos75°

(using

\cos75°=\cos(45°+30°)

). Hence rewrite

\cos(\theta-75°)=\dfrac{\sqrt6-\sqrt2}{4}

in the form

\cos(\theta-75°)=\cos\phi°

. State

\phi

Answer this when you sit the paper.

Question 301

[1 marks]Simultaneous equations

Solve the simultaneous equations

e^{2x+y} = 1

and

4x - 3y = 10

A $x = 2$ , $y = -1$
B $x = \tfrac{5}{2}$ , $y = 0$
C $x = 1$ , $y = -2$
D $x = -1$ , $y = 2$

Question 302

[2 marks]Simultaneous equations

Taking natural logs of

e^{2x+y}=1

gives

2x+y=0

. Substitute

y=-2x

into

4x-3y=10

and simplify to find

x

Answer this when you sit the paper.

Question 401

[1 marks]Inequalities

Solve the inequality

\dfrac{21+13x}{2+3x^2} \ge 10

A $-\tfrac{1}{2} \le x \le \tfrac{1}{15}$
B $-\tfrac{1}{15} \le x \le \tfrac{1}{2}$
C $x \ge \tfrac{1}{2}$ only
D $x \le -\tfrac{1}{15}$ or $x \ge \tfrac{1}{2}$

Question 402

[2 marks]Inequalities

Since

2+3x^2>0

for all

x

, multiplying both sides of

\dfrac{21+13x}{2+3x^2}\ge10

2+3x^2

and simplifying gives an inequality of the form

30x^2+bx+c\le0

. State

b

and

c

Answer this when you sit the paper.

Question 403

[1 marks]Inequalities

Solve

30x^2-13x-1=0

using the quadratic formula. State the larger root.

Answer this when you sit the paper.

Question 501

[1 marks]Implicit differentiation

A curve has equation

y^2 = 3xy + x^2 - 3

. Find

\dfrac{dy}{dx}

A $\dfrac{3y + 2x}{2y - 3x}$
B $\dfrac{2y - 3x}{3y + 2x}$
C $\dfrac{3y - 2x}{2y + 3x}$
D $\dfrac{3x + 2y}{2x - 3y}$

Question 502

[1 marks]Implicit differentiation

Find the equation of the tangent to

y^2 = 3xy + x^2 - 3

at the point where

y = 1

and

x > 0

A $y = -5x - 6$
B $y = 5x - 4$
C $y = -5x + 6$
D $y = -\tfrac{1}{5}x + \tfrac{6}{5}$

Question 503

[2 marks]Implicit differentiation

Substitute

y=1

into

y^2=3xy+x^2-3

to find the value(s) of

x

, giving your answer(s) as solutions of a quadratic

x^2+bx+c=0

. State

b

and

c

Answer this when you sit the paper.

Question 504

[1 marks]Implicit differentiation

Factorise

x^2+3x-4

Answer this when you sit the paper.

Question 505

[1 marks]Implicit differentiation

Since

x>0

, state the

x

-coordinate of the point on the curve

y^2=3xy+x^2-3

where

y=1

Answer this when you sit the paper.

Question 506

[2 marks]Implicit differentiation

Using implicit differentiation on

y^2=3xy+x^2-3

, find

\dfrac{dy}{dx}

in the form

\dfrac{3y+2x}{2y-Kx}

. State

K

Answer this when you sit the paper.

Question 601

[1 marks]Functions and inverses

Express

f(x) = -x^2 + 8x - 12

in the form

a(x+b)^2 + c

A $-(x+4)^2 + 4$
B $(x-4)^2 - 4$
C $-(x-4)^2 - 4$
D $-(x-4)^2 + 4$

Question 602

[1 marks]Functions and inverses

The function

f(x) = -x^2 + 8x - 12

is defined for

4 \le x \le 6

. Find

f^{-1}(x)

and state its domain.

A $f^{-1}(x) = 4 + \sqrt{4-x}$ , domain $0 \le x \le 4$
B $f^{-1}(x) = 4 - \sqrt{4-x}$ , domain $0 \le x \le 4$
C $f^{-1}(x) = 4 + \sqrt{4-x}$ , domain $4 \le x \le 6$
D $f^{-1}(x) = 4 + \sqrt{x-4}$ , domain $x \ge 4$

Question 603

[2 marks]Functions and inverses

Which of the following explains why

f(x)=-x^2+8x-12

has an inverse on the domain

4\le x\le6

A $f$ is strictly increasing on $[4,6]$ because $f'(x)=8-2x\ge0$ there.
B $f$ is strictly decreasing on $[4,6]$ because $f'(x)=8-2x\le0$ there.
CThe vertex of $f$ lies outside the interval $[4,6]$ , so no repeated $y$ -values occur.
D $f(4)$ and $f(6)$ are both positive, so $f$ cannot repeat a value.

Question 604

[2 marks]Functions and inverses

The graphs of

y=f(x)

and

y=f^{-1}(x)

meet the line

y=x

where

f(x)=x

. Solve

-x^2+8x-12=x

to find both solutions of the resulting quadratic (only one lies in the domain

[4,6]

Answer this when you sit the paper.

Question 605

[1 marks]Functions and inverses

Which of the solutions of

-x^2+8x-12=x

lies within the domain

4\le x\le6

, and is therefore the point where the graphs of

y=f(x)

and

y=f^{-1}(x)

intersect?

Answer this when you sit the paper.

Question 606

[1 marks]Functions and inverses

State the range of

f(x)=-x^2+8x-12

4\le x\le6

(needed to know where

y=f^{-1}(x)

starts and ends on the sketch).

Answer this when you sit the paper.

Question 701

[1 marks]Complex numbers

The complex number

u

satisfies

\dfrac{-48+56i}{u} = 1 + 13i

. Express

u

in the form

a + ib

A $-4 + 4i$
B $8 + 8i$
C $4 - 4i$
D $4 + 4i$

Question 702

[1 marks]Complex numbers

Given

u = 4 + 4i

, state its modulus and argument.

A $|u| = 4\sqrt{2}$ , $\arg u = -\tfrac{\pi}{4}$
B $|u| = 4\sqrt{2}$ , $\arg u = \tfrac{\pi}{4}$
C $|u| = 4$ , $\arg u = \tfrac{\pi}{2}$
D $|u| = 8$ , $\arg u = \tfrac{\pi}{4}$

Question 703

[2 marks]Complex numbers

To find

u=\dfrac{-48+56i}{1+13i}

, multiply numerator and denominator by the conjugate

1-13i

. Find the denominator

(1+13i)(1-13i)

Answer this when you sit the paper.

Question 704

[2 marks]Complex numbers

Find the numerator

(-48+56i)(1-13i)

, in the form

p+qi

Answer this when you sit the paper.

Question 705

[1 marks]Complex numbers

Given

u=4+4i

, write down the complex conjugate

u^*

Answer this when you sit the paper.

Question 706

[1 marks]Complex numbers

State the modulus of

u^*=4-4i

, in exact surd form.

Answer this when you sit the paper.

Question 801

[1 marks]Integration by parts and volumes of revolution

Find

\displaystyle\int xe^{-2x}\,dx

A $-\tfrac{1}{2}xe^{-2x} - \tfrac{1}{4}e^{-2x} + c$
B $-\tfrac{1}{2}xe^{-2x} + \tfrac{1}{4}e^{-2x} + c$
C $\tfrac{1}{2}xe^{-2x} - \tfrac{1}{4}e^{-2x} + c$
D $-\tfrac{1}{2}x^2e^{-2x} + c$

Question 802

[1 marks]Integration by parts and volumes of revolution

The region bounded by

y = 2x + e^{-2x}

, the

x

-axis and the lines

x = 0

and

x = 1

is rotated completely about the

x

-axis. Find the volume, correct to 3 significant figures.

A $2.17$
B $4.50$
C $4.96$
D $6.83$

Question 803

[2 marks]Integration by parts and volumes of revolution

Using integration by parts on

\int xe^{-2x}\,dx

with

u=x

dv=e^{-2x}dx

, state

v

(the antiderivative of

e^{-2x}

Answer this when you sit the paper.

Question 804

[2 marks]Integration by parts and volumes of revolution

Expand

(2x+e^{-2x})^2

(needed for the volume integral

\pi\int_0^1y^2\,dx

), giving the answer in the form

4x^2+Axe^{-2x}+e^{-4x}

. State

A

Answer this when you sit the paper.

Question 805

[1 marks]Integration by parts and volumes of revolution

Evaluate

\displaystyle\int_0^1 4x^2\,dx

, as an exact fraction.

Answer this when you sit the paper.

Question 806

[2 marks]Integration by parts and volumes of revolution

Evaluate

\displaystyle\int_0^1 e^{-4x}\,dx

, correct to 4 decimal places.

Answer this when you sit the paper.

Question 901

[1 marks]Partial fractions and integration

Express

f(x) = \dfrac{15x^2 + 40x + 2}{(4+x)(2+5x^2)}

in partial fractions.

A $\dfrac{1}{4+x} - \dfrac{10x}{2+5x^2}$
B $\dfrac{1}{4+x} + \dfrac{10x + 2}{2+5x^2}$
C $\dfrac{1}{4+x} + \dfrac{10x}{2+5x^2}$
D $\dfrac{10}{4+x} + \dfrac{x}{2+5x^2}$

Question 902

[1 marks]Partial fractions and integration

Evaluate

\displaystyle\int_0^2 \left(\dfrac{1}{4+x} + \dfrac{10x}{2+5x^2}\right)dx

, giving your answer as a single logarithm.

A $\ln 132$
B $\ln\tfrac{33}{2}$
C $\ln\tfrac{11}{2}$
D $\ln\tfrac{2}{33}$

Question 903

[2 marks]Partial fractions and integration

Using the cover-up rule, evaluate

A=\dfrac{15x^2+40x+2}{2+5x^2}

x=-4

(the root of

4+x

), to find the coefficient

A

\dfrac{A}{4+x}+\dfrac{Bx+C}{2+5x^2}

Answer this when you sit the paper.

Question 904

[2 marks]Partial fractions and integration

Comparing coefficients of

x^2

15x^2+40x+2\equiv A(2+5x^2)+(Bx+C)(4+x)

, with

A=1

5A+B=15

. Find

B

Answer this when you sit the paper.

Question 905

[1 marks]Partial fractions and integration

Comparing the constant terms in

15x^2+40x+2\equiv A(2+5x^2)+(Bx+C)(4+x)

2A+4C=2

. With

A=1

, find

C

Answer this when you sit the paper.

Question 906

[2 marks]Partial fractions and integration

The antiderivative

\ln(4+x)+\ln(2+5x^2)

combines into a single log,

\ln[(4+x)(2+5x^2)]

. State the argument of this logarithm (the numeric value inside) at

x=2

Answer this when you sit the paper.

Question 1001

[1 marks]Vectors

\overrightarrow{OA} = -4\lambda\mathbf{i} - 3\lambda\mathbf{j} + 4\lambda\mathbf{k}

and

\overrightarrow{OB} = 3\mathbf{i} + 9\mathbf{j} - 6\mathbf{k}

. Given that

\overrightarrow{AB}

is perpendicular to

\overrightarrow{OB}

, find

\lambda

A $\lambda = -2$
B $\lambda = -\tfrac{1}{2}$
C $\lambda = 63$
D $\lambda = 2$

Question 1002

[1 marks]Vectors

With

\overrightarrow{OA} = (8, 6, -8)

and

\overrightarrow{OB} = (3, 9, -6)

, find angle

A\hat{O}B

correct to the nearest

0.1°

A $61.3°$
B $90.0°$
C $151.3°$
D $28.7°$

Question 1003

[1 marks]Vectors

O

A(8, 6, -8)

and

B

lie on a circle in which angle

OBA = 90°

. Find the centre and radius of the circle.

Acentre $(4, 3, -4)$ , radius $2\sqrt{41}$
Bcentre $(4, 3, -4)$ , radius $41$
Ccentre $(4, 3, -4)$ , radius $\sqrt{41}$
Dcentre $(8, 6, -8)$ , radius $2\sqrt{41}$

Question 1004

[2 marks]Vectors

Form

\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=(3+4\lambda)\mathbf{i}+(9+3\lambda)\mathbf{j}+(-6-4\lambda)\mathbf{k}

. Using

\overrightarrow{AB}\cdot\overrightarrow{OB}=0

, simplify to a linear equation in

\lambda

, in the form

63\lambda=k

. State

k

Answer this when you sit the paper.

Question 1005

[2 marks]Vectors

Find

\overrightarrow{OA}\cdot\overrightarrow{OB}

for

\overrightarrow{OA}=8\mathbf{i}+6\mathbf{j}-8\mathbf{k}

and

\overrightarrow{OB}=3\mathbf{i}+9\mathbf{j}-6\mathbf{k}

Answer this when you sit the paper.

Question 1006

[2 marks]Vectors

Since

\angle OBA=90°

OA

is a diameter of the circle through

O

A(8,6,-8)

and

B

. State the coordinates of the midpoint of

OA

(the centre of the circle).

Answer this when you sit the paper.

Question 1101

[1 marks]Binomial expansion

In the expansion of

(1+ax)^n

with

n < 0

, the coefficients of

x

and

x^3

are

-1

and

-\tfrac{14}{3}

. Find

n

A $n = -3$
B $n = -\tfrac{1}{3}$
C $n = -\tfrac{2}{9}$
D $n = \tfrac{2}{9}$

Question 1102

[1 marks]Binomial expansion

In the expansion of

(1+ax)^n

with

n = -\tfrac{1}{3}

and the coefficient of

x

equal to

-1

, find

a

and the set of values of

x

for which the expansion is valid.

A $a = -3$ , valid for $|x| < \tfrac{1}{3}$
B $a = 3$ , valid for $|x| < 3$
C $a = \tfrac{1}{3}$ , valid for $|x| < 3$
D $a = 3$ , valid for $|x| < \tfrac{1}{3}$

Question 1103

[2 marks]Binomial expansion

In the expansion of

(1+ax)^n

, the coefficient of

x

na=-1

, so

a=-\dfrac1n

. The coefficient of

x^3

\dfrac{n(n-1)(n-2)}{6}a^3=-\dfrac{14}{3}

, becomes

-\dfrac{(n-1)(n-2)}{6n^2}=-\dfrac{14}{3}

after substituting

a=-\tfrac1n

. Simplify to

(n-1)(n-2)=kn^2

. Find

k

Answer this when you sit the paper.

Question 1104

[2 marks]Binomial expansion

Expand

(n-1)(n-2)

, in the form

n^2+bn+c

Answer this when you sit the paper.

Question 1105

[1 marks]Binomial expansion

Solve

27n^2+3n-2=0

using the quadratic formula. Find the discriminant

3^2-4(27)(-2)

Answer this when you sit the paper.

Question 1106

[2 marks]Binomial expansion

Using the discriminant

225

(so

\sqrt{225}=15

), find the positive root

n=\dfrac{-3+15}{54}

27n^2+3n-2=0

(not the required negative root,

n<0

Answer this when you sit the paper.

Question 1201

[1 marks]Numerical methods (Newton-Raphson)

For

f(x) = x^3 - 4e^{x/2}

, the Newton-Raphson iterative formula for a root of

f(x) = 0

A $x_{n+1} = x_n - \dfrac{x_n^3 - 4e^{x_n/2}}{3x_n^2 - 4e^{x_n/2}}$
B $x_{n+1} = x_n - \dfrac{x_n^3 - 4e^{x_n/2}}{3x_n^2 - 2e^{x_n/2}}$
C $x_{n+1} = x_n - \dfrac{3x_n^2 - 2e^{x_n/2}}{x_n^3 - 4e^{x_n/2}}$
D $x_{n+1} = x_n + \dfrac{x_n^3 - 4e^{x_n/2}}{3x_n^2 - 2e^{x_n/2}}$

Question 1202

[1 marks]Numerical methods (Newton-Raphson)

Taking

x_1 = 2

, use the Newton-Raphson formula for

x^3 - 4e^{x/2} = 0

to find

x_2

correct to 3 decimal places.

A $1.900$
B $2.348$
C $2.438$
D $2.500$

Question 1203

[2 marks]Numerical methods (Newton-Raphson)

Evaluate

f(1{,}9)=1{,}9^3-4e^{0{,}95}

, correct to 3 decimal places (used to verify the root lies between 1.9 and 2.5).

Answer this when you sit the paper.

Question 1204

[2 marks]Numerical methods (Newton-Raphson)

Evaluate

f(2{,}5)=2{,}5^3-4e^{1{,}25}

, correct to 3 decimal places.

Answer this when you sit the paper.

Question 1205

[2 marks]Numerical methods (Newton-Raphson)

Find

f'(x)=3x^2-2e^{x/2}

evaluated at

x=2

, correct to 3 decimal places (the denominator of the Newton-Raphson formula at

x_1=2

Answer this when you sit the paper.

Question 1206

[1 marks]Numerical methods (Newton-Raphson)

Find

f(2)=2^3-4e^1

, correct to 3 decimal places (the numerator of the Newton-Raphson formula at

x_1=2

Answer this when you sit the paper.

Question 1301

[1 marks]Arithmetic and geometric progressions

In an arithmetic progression the sum of the 5th and 10th terms is 27 and the 3rd term is

-9

. Find the first term and common difference.

A $a = 19$ , $d = -5$
B $a = -9$ , $d = 5$
C $a = -19$ , $d = -5$
D $a = -19$ , $d = 5$

Question 1302

[1 marks]Arithmetic and geometric progressions

An arithmetic progression has

a = -19

and

d = 5

. Find the smallest value of

n

for which the sum of the first

n

terms is positive.

A $n = 8$
B $n = 10$
C $n = 9$
D $n = 43$

Question 1303

[1 marks]Arithmetic and geometric progressions

A geometric progression with

r < 0

has 4th term

2.56

and 6th term

1.6348

. Find its sum to infinity, correct to 2 decimal places.

A $-27.78$
B $-5.00$
C $-2.78$
D $2.78$

Question 1304

[3 marks]Arithmetic and geometric progressions

Using the first term

a=-19

and common difference

d=5

of the arithmetic progression, find

S_9

(the sum of the first 9 terms), using

S_n=\tfrac n2(2a+(n-1)d)

Answer this when you sit the paper.

Question 1305

[1 marks]Arithmetic and geometric progressions

Find

S_8

(the sum of the first 8 terms), and confirm it is not yet positive.

Answer this when you sit the paper.

Question 1306

[1 marks]Arithmetic and geometric progressions

Using

r^2=\dfrac{1.6348}{2.56}\approx0{,}6386

, giving

r\approx-0{,}8

, and

ar^3=2{,}56

, find

r^3

Answer this when you sit the paper.

Question 1307

[1 marks]Arithmetic and geometric progressions

Hence find

a=\dfrac{2.56}{r^3}

using

r^3=-0{,}512

Answer this when you sit the paper.

Question 1308

[1 marks]Arithmetic and geometric progressions

State

|r|

for

r=-0{,}8

, used to confirm

|r|<1

so the sum to infinity exists.

Answer this when you sit the paper.

Question 1309

[2 marks]Arithmetic and geometric progressions

Find

T_5=a+4d

and

T_{10}=a+9d

using

a=-19

d=5

Answer this when you sit the paper.

Question 1401

[1 marks]Coordinate geometry of the circle

A circle has centre

A(-3, k)

and the tangent at

B(3, 3k)

has gradient

-\tfrac{3}{4}

. Find

k

A $k = -4$
B $k = 8$
C $k = 4$
D $k = 3$

Question 1402

[1 marks]Coordinate geometry of the circle

A circle has centre

A(-3, 4)

and passes through

B(3, 12)

. Find its equation.

A $(x+3)^2 + (y-4)^2 = 100$
B $(x+3)^2 + (y+4)^2 = 100$
C $(x+3)^2 + (y-4)^2 = 10$
D $(x-3)^2 + (y-4)^2 = 100$

Question 1403

[1 marks]Coordinate geometry of the circle

The circle

(x+3)^2 + (y-4)^2 = 100

crosses the

x

-axis. Find the

x

-coordinates of the crossing points, correct to 2 decimal places.

A $x = -9.17$ and $x = 3.17$
B $x = -10.00$ and $x = 10.00$
C $x = -12.17$ and $x = 6.17$
D $x = -6.17$ and $x = 12.17$

Question 1404

[2 marks]Coordinate geometry of the circle

The perpendicular distance from the centre

A(-3,4)

to the

x

-axis is

d=4

. With radius

r=10

, find

\cos\theta=\dfrac{d}{r}

, where

2\theta

is the central angle subtended by the chord (the

x

-axis intercepts).

Answer this when you sit the paper.

Question 1405

[2 marks]Coordinate geometry of the circle

Find

\theta=\arccos(0{,}4)

, in radians, correct to 4 decimal places.

Answer this when you sit the paper.

Question 1406

[2 marks]Coordinate geometry of the circle

Hence find the central angle

\alpha=2\theta

subtended by the chord, in radians, correct to 4 decimal places.

Answer this when you sit the paper.

Question 1407

[2 marks]Coordinate geometry of the circle

Find

\sin\alpha

for

\alpha=2{,}3186

radians, correct to 4 decimal places (needed for the segment area formula).

Answer this when you sit the paper.

Question 1408

[2 marks]Coordinate geometry of the circle

Using the segment area formula

\text{Area}=\tfrac12r^2(\alpha-\sin\alpha)

with

r=10

\alpha=2{,}3186

\sin\alpha=0{,}7332

, find the area of the minor segment cut off by the

x

-axis, correct to 3 significant figures.

Answer this when you sit the paper.

Question 1501

[1 marks]Differentiation and curve sketching

The function

f(x) = \tfrac{1}{32}x + \tfrac{1}{2}(x-3)^{-1}

is defined for

x \ne k

. State the value of

k

and find the

x

-coordinates of the turning points.

A $k = 0$ ; turning points at $x = -1$ and $x = 7$
B $k = 3$ ; turning points at $x = 1$ and $x = -7$
C $k = 3$ ; turning point at $x = 3$ only
D $k = 3$ ; turning points at $x = -1$ and $x = 7$

Question 1502

[1 marks]Differentiation and curve sketching

For

f(x) = \tfrac{1}{32}x + \tfrac{1}{2}(x-3)^{-1}

, state the turning points and their nature.

A $\left(-1, -\tfrac{5}{32}\right)$ minimum and $\left(7, \tfrac{11}{32}\right)$ maximum
B $\left(-1, -\tfrac{5}{32}\right)$ maximum and $\left(7, \tfrac{11}{32}\right)$ minimum
C $\left(-1, \tfrac{5}{32}\right)$ maximum and $\left(7, -\tfrac{11}{32}\right)$ minimum
Dboth are points of inflexion

Question 1503

[1 marks]Differentiation and curve sketching

Show that the graph of

f(x) = \tfrac{1}{32}x + \tfrac{1}{2}(x-3)^{-1}

does not cross the

x

-axis. Which reason is correct?

A $f(x) = 0$ leads to $x^2 + 3x - 16 = 0$ , which has no integer roots
BThe function is always positive
CThe function has a vertical asymptote at $x = 3$
D $f(x) = 0$ leads to $x^2 - 3x + 16 = 0$ , whose discriminant is $-55 < 0$

Question 1504

[2 marks]Differentiation and curve sketching

Differentiate

f(x)=\dfrac1{32}x+\dfrac12(x-3)^{-1}

to find

f'(x)

, in the form

\dfrac1{32}-\dfrac12(x-3)^{-K}

. State

K

Answer this when you sit the paper.

Question 1505

[2 marks]Differentiation and curve sketching

Setting

f'(x)=0

\dfrac1{32}=\dfrac1{2(x-3)^2}

gives

(x-3)^2=16

. Solve for

x

, stating both values.

Answer this when you sit the paper.

Question 1506

[1 marks]Differentiation and curve sketching

Find

f(7)

, as an exact fraction (the

y

-coordinate of one turning point).

Answer this when you sit the paper.

Question 1507

[1 marks]Differentiation and curve sketching

Find

f(-1)

, as an exact fraction (the

y

-coordinate of the other turning point).

Answer this when you sit the paper.

Question 1508

[2 marks]Differentiation and curve sketching

Find

f''(x)

, in the form

(x-3)^{-K}

. State

K

Answer this when you sit the paper.

Question 1509

[2 marks]Differentiation and curve sketching

Evaluate

f''(7)

, as an exact fraction (used to determine the nature of the turning point at

x=7

Answer this when you sit the paper.

Pure Mathematics Paper 1 November 2018

Sit this paper online

The questions

Question 101

Question 102

Question 201

Question 202

Question 301

Question 302

Question 401

Question 402

Question 403

Question 501

Question 502

Question 503

Question 504

Question 505

Question 506

Question 601

Question 602

Question 603

Question 604

Question 605

Question 606

Question 701

Question 702

Question 703

Question 704

Question 705

Question 706

Question 801

Question 802

Question 803

Question 804

Question 805

Question 806

Question 901

Question 902

Question 903

Question 904

Question 905

Question 906

Question 1001

Question 1002

Question 1003

Question 1004

Question 1005

Question 1006

Question 1101

Question 1102

Question 1103

Question 1104

Question 1105

Question 1106

Question 1201

Question 1202

Question 1203

Question 1204

Question 1205

Question 1206

Question 1301

Question 1302

Question 1303

Question 1304

Question 1305

Question 1306

Question 1307

Question 1308

Question 1309

Question 1401

Question 1402

Question 1403

Question 1404

Question 1405

Question 1406

Question 1407

Question 1408

Question 1501

Question 1502

Question 1503