ZIMSEC A Level Pure Mathematics Paper 1 June 2016

Question 101

[1 marks]functions / inverse functions

Find the inverse of

f(x) = ax + b

, where

a \ne 0

and

b

are constants.

A $f^{-1}(x) = ax - b$
B $f^{-1}(x) = \dfrac{x + b}{a}$
C $f^{-1}(x) = \dfrac{x - b}{a}$
D $f^{-1}(x) = \dfrac{1}{ax + b}$

Question 102

[1 marks]functions / inverse functions

Given

f(x)=ax+b

with inverse

f^{-1}(x)=\dfrac{x-b}a

, evaluate

f^{-1}(b)

Answer this when you sit the paper.

Question 201

[1 marks]partial fractions / comparing coefficients

Express

\dfrac{5x^3 - 3x^2 + 7x - 3}{(x^2+1)^2}

in the form

\dfrac{Ax+B}{x^2+1} + \dfrac{Cx+D}{(x^2+1)^2}

A $\dfrac{5x-3}{x^2+1} + \dfrac{2x}{(x^2+1)^2}$
B $\dfrac{5x+3}{x^2+1} + \dfrac{2x}{(x^2+1)^2}$
C $\dfrac{5x-3}{x^2+1} + \dfrac{7x-3}{(x^2+1)^2}$
D $\dfrac{5x-3}{x^2+1} + \dfrac{2x+3}{(x^2+1)^2}$

Question 202

[2 marks]partial fractions / comparing coefficients

Matching coefficients in

5x^3-3x^2+7x-3=(Ax+B)(x^2+1)+Cx+D

: the coefficient of

x^3

gives

A

directly, and the constant term gives

B+D=-3

. State

A

and

B+D

as 'A, B+D'.

Answer this when you sit the paper.

Question 301

[1 marks]coordinate geometry / inequalities

Find the set of values of

m

for which the gradient of the line through

(m, 4)

and

(1, 3-2m)

is less than 5.

A $m < 1$ or $m > 2$
B $m > 2$ only
C $m < -\tfrac{1}{2}$ or $m > 2$
D $1 < m < 2$

Question 302

[2 marks]coordinate geometry / inequalities

The gradient of the line through

(m,4)

and

(1,3-2m)

simplifies to

\dfrac{1+2m}{m-1}

. Setting this less than 5 and simplifying gives

\dfrac{2-m}{m-1}<0

. State the two critical values of

m

(where the numerator or denominator is zero), as 'a, b' in ascending order.

Answer this when you sit the paper.

Question 303

[2 marks]coordinate geometry / inequalities

For

\dfrac{2-m}{m-1}<0

(from the gradient condition), test

m=1.5

(between the critical values 1 and 2). Evaluate

\dfrac{2-m}{m-1}

m=1.5

Answer this when you sit the paper.

Question 401

[1 marks]sine and cosine rules

In a figure,

Q\hat{P}S = \tfrac{\pi}{6}

PS = 5

cm and

\tan P\hat{Q}S = \tfrac{3}{4}

. Use the sine rule in triangle

PQS

to find

QS

A $\tfrac{25}{6}$ cm
B $\tfrac{6}{25}$ cm
C $\tfrac{15}{4}$ cm
D $\tfrac{10}{3}$ cm

Question 402

[1 marks]sine and cosine rules

In triangle

RQS

QR = 3

cm,

QS = \tfrac{25}{6}

cm and

\cos R\hat{Q}S = \tfrac{3}{5}

. Find the exact length of

RS

A $\dfrac{\sqrt{409}}{36}$ cm
B $\dfrac{\sqrt{409}}{6}$ cm
C $\dfrac{\sqrt{625}}{6}$ cm
D $\dfrac{409}{36}$ cm

Question 403

[1 marks]sine and cosine rules

Given

\tan\angle PQS=\dfrac34

with

\angle PQS

acute, state

\sin\angle PQS

as a fraction.

Answer this when you sit the paper.

Question 404

[1 marks]sine and cosine rules

Given

\tan\angle PQS=\dfrac34

with

\angle PQS

acute, state

\cos\angle PQS

as a fraction.

Answer this when you sit the paper.

Question 405

[2 marks]sine and cosine rules

Since

\angle PQR=\dfrac{\pi}2

and

\angle RQS=\dfrac{\pi}2-\angle PQS

, and

\sin\angle PQS=\dfrac35

, state

\cos\angle RQS

(used in triangle RQS to find RS).

Answer this when you sit the paper.

Question 501

[1 marks]binomial expansion / approximation

Expand

(9-4x)^{-1/2}

in ascending powers of

x

up to and including the term in

x^2

A $3 + \tfrac{2}{3}x + \tfrac{2}{9}x^2$
B $\tfrac{1}{3} + \tfrac{2}{9}x + \tfrac{2}{27}x^2$
C $\tfrac{1}{3} - \tfrac{2}{27}x + \tfrac{2}{81}x^2$
D $\tfrac{1}{3} + \tfrac{2}{27}x + \tfrac{2}{81}x^2$

Question 502

[1 marks]binomial expansion / approximation

State the range of values of

x

for which the expansion of

(9-4x)^{-1/2}

is valid.

A $-\tfrac{4}{9} < x < \tfrac{4}{9}$
B $|x| < 1$
C $|x| < 9$
D $-\tfrac{9}{4} < x < \tfrac{9}{4}$

Question 503

[1 marks]binomial expansion / approximation

By putting

x = \tfrac{1}{9}

in the expansion of

(9-4x)^{-1/2}

, find an approximation for

\sqrt{77}

A $\dfrac{6561}{2243}$
B $\dfrac{19683}{2243}$
C $\dfrac{2243}{19683}$
D $\dfrac{2243}{6561}$

Question 504

[1 marks]binomial expansion / approximation

Factoring

(9-4x)^{-1/2}=9^{-1/2}\left(1-\dfrac{4x}9\right)^{-1/2}

, state

9^{-1/2}

as a fraction.

Answer this when you sit the paper.

Question 505

[2 marks]binomial expansion / approximation

In the binomial expansion

(1+u)^n\approx1+nu+\ldots

with

n=-\dfrac12

u=-\dfrac{4x}9

(used for

\left(1-\frac{4x}9\right)^{-1/2}

, before multiplying by

9^{-1/2}

), state the coefficient of

x

Answer this when you sit the paper.

Question 506

[1 marks]binomial expansion / approximation

Substituting

x=\dfrac19

into

9-4x

(used to approximate

\sqrt{77}

) gives

9-\dfrac49=\dfrac p9

for some integer

p

. State

p

Answer this when you sit the paper.

Question 601

[1 marks]geometric series / convergence

A geometric series has first term

x

and second term

x^2 - x

, with all terms positive. Find the set of values of

x

for which the series converges.

A $x > 1$
B $1 < x < 2$
C $-1 < x < 1$
D $0 < x < 2$

Question 602

[1 marks]geometric series / convergence

A geometric series has first term

\tfrac{5}{3}

and common ratio

\tfrac{2}{3}

. Find the smallest

n

for which the sum of the first

n

terms differs from the sum to infinity by less than

0.001

A $n = 20$
B $n = 23$
C $n = 22$
D $n = 21$

Question 603

[2 marks]geometric series / convergence

A geometric series has first term

x

and second term

x^2-x

. State the common ratio

r

in terms of

x

Answer this when you sit the paper.

Question 604

[2 marks]geometric series / convergence

For the geometric series with

x=\dfrac53

(so the common ratio is

r=x-1=\dfrac23

), state the sum to infinity

S_\infty

Answer this when you sit the paper.

Question 605

[1 marks]geometric series / convergence

With

S_\infty=5

for this series, the condition

S_\infty-S_n<0.001

becomes

5(2/3)^n<0.001

, i.e.

(2/3)^n<k

. State

k

Answer this when you sit the paper.

Question 701

[1 marks]optimisation

A closed cylindrical tin of radius

r

has fixed volume

V

. Express its total surface area

S

in terms of

\pi

r

and

V

A $S = 2\pi r^2 + \dfrac{V}{r}$
B $S = 2\pi r^2 + \dfrac{2V}{r}$
C $S = \pi r^2 + \dfrac{2V}{r}$
D $S = 2\pi r^2 + \dfrac{2V}{\pi r}$

Question 702

[1 marks]optimisation

For a closed cylinder of fixed volume, the total surface area

S = 2\pi r^2 + \dfrac{2V}{r}

is least when

Athe height equals the diameter
Bthe height equals the radius
Cthe radius is twice the height
Dthe volume equals $\pi r^3$

Question 703

[2 marks]optimisation

For a closed cylinder of fixed volume

V

S=2\pi r^2+\dfrac{2V}r

, so

\dfrac{dS}{dr}=4\pi r-\dfrac{2V}{r^2}

. Setting

\dfrac{dS}{dr}=0

and rearranging for

V

gives

V=k\pi r^3

. State

k

Answer this when you sit the paper.

Question 704

[2 marks]optimisation

For

S=2\pi r^2+\dfrac{2V}r

, find

\dfrac{d^2S}{dr^2}

, in terms of

\pi

V

and

r

Answer this when you sit the paper.

Question 705

[1 marks]optimisation

For a closed cylinder of fixed volume,

\dfrac{d^2S}{dr^2}=4\pi+\dfrac{4V}{r^3}

is positive for every

r>0

. What does this confirm about the stationary point found by setting

\dfrac{dS}{dr}=0

AIt is a minimum
BIt is a maximum
CIt is a point of inflection
DIt does not exist

Question 801

[1 marks]Newton-Raphson method

Let

f(x) = e^x - \cos x - 2

. Which calculation shows that

e^x - \cos x = 2

has a root between

x = 0

and

x = 1

A $f(0) = 0$ and $f(1) = 1$
B $f(0) = -2$ and $f(1) = -0.178$
C $f(0) = -2$ and $f(1) = 0.178$
D $f(0) = 2$ and $f(1) = -0.178$

Question 802

[1 marks]Newton-Raphson method

Starting with

x_0 = 1

, apply the Newton-Raphson method twice to

e^x - \cos x - 2 = 0

and give the root correct to 3 decimal places.

A $0.945$
B $0.949$
C $0.950$
D $1.000$

Question 803

[1 marks]Newton-Raphson method

For

f(x)=e^x-\cos x-2

, evaluate

f(0)

exactly.

Answer this when you sit the paper.

Question 804

[1 marks]Newton-Raphson method

For

f(x)=e^x-\cos x-2

, evaluate

f(1)

, correct to 3 decimal places.

Answer this when you sit the paper.

Question 805

[2 marks]Newton-Raphson method

For the Newton-Raphson method on

f(x)=e^x-\cos x-2

, with

f'(x)=e^x+\sin x

, evaluate

f'(1)

, correct to 3 decimal places.

Answer this when you sit the paper.

Question 806

[1 marks]Newton-Raphson method

Using

x_0=1

f(1)\approx0.178

and

f'(1)\approx3.560

, evaluate

x_1=x_0-\dfrac{f(x_0)}{f'(x_0)}

(the first Newton-Raphson iterate), correct to 3 decimal places.

Answer this when you sit the paper.

Question 901

[1 marks]logarithmic and exponential equations

Solve exactly the equation

\tfrac{1}{2}\log_5(x-2) = 3\log_5 2 - \tfrac{3}{2}\log_5(x-2)

A $x = 10$
B $x = 2 - \sqrt{8}$
C $x = 8$
D $x = 2 + \sqrt{8}$

Question 902

[1 marks]logarithmic and exponential equations

Solve exactly the equation

2e^{2x} = 7e^x - 3

A $x = \ln 3$ only
B $x = \ln\tfrac{1}{2}$ or $x = \ln 3$
C $x = \tfrac{1}{2}$ or $x = 3$
D $x = \ln 2$ or $x = \ln 3$

Question 903

[2 marks]logarithmic and exponential equations

Combining

\dfrac12\log_5(x-2)+\dfrac32\log_5(x-2)

(from rearranging

\dfrac12\log_5(x-2)=3\log_5 2-\dfrac32\log_5(x-2)

) gives

c\log_5(x-2)

. State

c

Answer this when you sit the paper.

Question 904

[2 marks]logarithmic and exponential equations

The equation

2\log_5(x-2)=3\log_5 2

rearranges to

\log_5\left[(x-2)^2\right]=\log_5(2^3)

, so

(x-2)^2=2^k

. State

k

Answer this when you sit the paper.

Question 905

[2 marks]logarithmic and exponential equations

Substituting

u=e^x

into

2e^{2x}=7e^x-3

gives

2u^2-7u+3=0

, which factorises as

(2u-1)(u-c)=0

. State

c

Answer this when you sit the paper.

Question 1001

[1 marks]trigonometric identities and equations

Simplify

\tan^2\theta - \sin^2\theta

A $\sin^2\theta\cos^2\theta$
B $\tan^2\theta\sin^2\theta$
C $1 - \cos^2\theta$
D $\tan^2\theta\cos^2\theta$

Question 1002

[1 marks]trigonometric identities and equations

Solve the equation

4\sin^2\theta\tan\theta - \tan\theta = 0

for

0 \le \theta \le 2\pi

A $\theta = 0,\ \tfrac{\pi}{3},\ \tfrac{2\pi}{3},\ \pi,\ \tfrac{4\pi}{3},\ \tfrac{5\pi}{3},\ 2\pi$
B $\theta = 0,\ \tfrac{\pi}{6},\ \tfrac{5\pi}{6},\ \pi,\ \tfrac{7\pi}{6},\ \tfrac{11\pi}{6},\ 2\pi$
C $\theta = \tfrac{\pi}{6},\ \tfrac{5\pi}{6},\ \tfrac{7\pi}{6},\ \tfrac{11\pi}{6}$ only
D $\theta = 0,\ \pi,\ 2\pi$ only

Question 1003

[2 marks]trigonometric identities and equations

Rewriting

\tan^2\theta-\sin^2\theta

\sin^2\theta\left(\dfrac1{\cos^2\theta}-1\right)

, simplify

\dfrac1{\cos^2\theta}-1

as a single fraction with denominator

\cos^2\theta

. State the numerator, in terms of

\cos\theta

Answer this when you sit the paper.

Question 1004

[2 marks]trigonometric identities and equations

Factoring

4\sin^2\theta\tan\theta-\tan\theta=0

\tan\theta(4\sin^2\theta-c)=0

, state

c

Answer this when you sit the paper.

Question 1005

[2 marks]trigonometric identities and equations

From the factor

4\sin^2\theta-1=0

(in

4\sin^2\theta\tan\theta-\tan\theta=0

), state

\sin^2\theta

as a fraction.

Answer this when you sit the paper.

Question 1006

[1 marks]trigonometric identities and equations

From the factor

\tan\theta=0

(in

4\sin^2\theta\tan\theta-\tan\theta=0

), state the solution strictly between

0

and

2\pi

(excluding both endpoints), in terms of pi.

Answer this when you sit the paper.

Question 1101

[1 marks]complex numbers

Express

w = 8 + \dfrac{4-i}{1+2i}

in the form

x + iy

A $\tfrac{12}{5} - \tfrac{9}{5}i$
B $12 + 15i$
C $\tfrac{42}{5} + \tfrac{9}{5}i$
D $\tfrac{42}{5} - \tfrac{9}{5}i$

Question 1102

[1 marks]complex numbers

Given

w = \tfrac{42}{5} - \tfrac{9}{5}i

, find

|w|

in the form

a\sqrt{b}

A $\tfrac{1}{5}\sqrt{205}$
B $\tfrac{9}{5}\sqrt{205}$
C $\tfrac{3}{5}\sqrt{205}$
D $3\sqrt{205}$

Question 1103

[1 marks]complex numbers

Find the argument of

w = \tfrac{42}{5} - \tfrac{9}{5}i

, in radians correct to 2 decimal places.

A $-1.36$
B $-0.21$
C $0.21$
D $2.93$

Question 1104

[2 marks]complex numbers

Multiplying

\dfrac{4-i}{1+2i}

\dfrac{1-2i}{1-2i}

, state the numerator

(4-i)(1-2i)

in the form

a+bi

(before dividing by the denominator).

Answer this when you sit the paper.

Question 1105

[1 marks]complex numbers

State the denominator obtained from

(1+2i)(1-2i)

(used to simplify

\dfrac{4-i}{1+2i}

Answer this when you sit the paper.

Question 1106

[2 marks]complex numbers

For

w=\dfrac{42}5-\dfrac95i

, computing

|w|^2=\left(\dfrac{42}5\right)^2+\left(\dfrac95\right)^2

, state

|w|^2

as a single fraction with denominator 25.

Answer this when you sit the paper.

Question 1107

[2 marks]complex numbers

For

w=\dfrac{42}5-\dfrac95i

, state

\dfrac yx

(used to find the argument of

w

) as a fraction in lowest terms.

Answer this when you sit the paper.

Question 1201

[1 marks]differential equations / partial fractions

The mass

x

grammes of an animal increases at a rate directly proportional to

\left(10\,000 - x^2\right)

, where

0 < x < 100

. Write down the differential equation connecting

x

and

t

A $\dfrac{dx}{dt} = k\left(10\,000 - x^2\right)$
B $\dfrac{dx}{dt} = \dfrac{k}{10\,000 - x^2}$
C $\dfrac{dx}{dt} = kx\left(10\,000 - x^2\right)$
D $\dfrac{dx}{dt} = k\left(100 - x\right)$

Question 1202

[1 marks]differential equations / partial fractions

Separating

\dfrac{dx}{dt} = k\left(10\,000 - x^2\right)

with partial fractions gives which general solution?

A $\ln\left(10\,000 - x^2\right) = kt + c$
B $\tfrac{1}{200}\ln\left(\dfrac{100-x}{100+x}\right) = kt + c$
C $\tfrac{1}{100}\ln\left(\dfrac{100+x}{100-x}\right) = kt + c$
D $\tfrac{1}{200}\ln\left(\dfrac{100+x}{100-x}\right) = kt + c$

Question 1203

[2 marks]differential equations / partial fractions

Separating

\dfrac{dx}{dt}=k(10\,000-x^2)

gives

\dfrac1{200}\ln\left(\dfrac{100+x}{100-x}\right)=kt+c

. Using

x=10

t=0

, evaluate

c

in the form

\dfrac1{200}\ln(p)

. State

p

as a fraction.

Answer this when you sit the paper.

Question 1204

[2 marks]differential equations / partial fractions

Using

x=50

t=10

\dfrac1{200}\ln\left(\dfrac{100+x}{100-x}\right)=kt+c

, evaluate the left-hand side at that point, in the form

\dfrac1{200}\ln(q)

. State

q

Answer this when you sit the paper.

Question 1205

[2 marks]differential equations / partial fractions

Using

c=\dfrac1{200}\ln(11/9)

and

10k+c=\dfrac1{200}\ln3

, state

10k

in the form

\dfrac1{200}\ln(r)

(i.e.

r=3\div\tfrac{11}9

Answer this when you sit the paper.

Question 1206

[2 marks]differential equations / partial fractions

From

10k=\dfrac1{200}\ln(27/11)

, state

k

in the form

\dfrac1{d}\ln(27/11)

. State

d

Answer this when you sit the paper.

Question 1301

[1 marks]graph transformations

The function

f

is defined by

f(x) = (x-2)(x+3)

. State the coordinates of the turning point of

y = f(x)

A $\left(-\tfrac{1}{2},\ -6\right)$
B $\left(-\tfrac{1}{2},\ \tfrac{25}{4}\right)$
C $\left(-\tfrac{1}{2},\ -\tfrac{25}{4}\right)$
D $\left(\tfrac{1}{2},\ -\tfrac{25}{4}\right)$

Question 1302

[1 marks]graph transformations

For

f(x) = (x-2)(x+3)

, state the

x

-intercepts and turning point of

y = f(2x)

Aintercepts $-\tfrac{3}{2}$ and $1$ ; turning point $\left(-\tfrac{1}{4}, -\tfrac{25}{4}\right)$
Bintercepts $-\tfrac{3}{2}$ and $1$ ; turning point $\left(-\tfrac{1}{2}, -\tfrac{25}{4}\right)$
Cintercepts $-6$ and $4$ ; turning point $\left(-1, -\tfrac{25}{4}\right)$
Dintercepts $-3$ and $2$ ; turning point $\left(-\tfrac{1}{4}, -\tfrac{25}{2}\right)$

Question 1303

[1 marks]graph transformations

For

f(x) = (x-2)(x+3)

, state the turning point and

y

-intercept of

y = -2f(x)

Amaximum $\left(-\tfrac{1}{2}, \tfrac{25}{2}\right)$ ; $y$ -intercept $12$
Bminimum $\left(-\tfrac{1}{2}, \tfrac{25}{2}\right)$ ; $y$ -intercept $12$
Cmaximum $\left(-\tfrac{1}{2}, \tfrac{25}{4}\right)$ ; $y$ -intercept $6$
Dmaximum $\left(\tfrac{1}{2}, \tfrac{25}{2}\right)$ ; $y$ -intercept $-12$

Question 1304

[2 marks]graph transformations

For

f(x)=(x-2)(x+3)

, state the

x

-intercepts of

y=f(x)

, as 'a, b' in ascending order.

Answer this when you sit the paper.

Question 1305

[1 marks]graph transformations

For

f(x)=(x-2)(x+3)

, state the

y

-intercept of

y=f(x)

Answer this when you sit the paper.

Question 1306

[2 marks]graph transformations

For

f(x)=(x-2)(x+3)

, with

y=f(x)

having minimum point

\left(-\tfrac12,-\tfrac{25}4\right)

, state the coordinates of the turning point of

y=|f(x)|

arising from reflecting this minimum above the

x

-axis, as 'x,y'.

Answer this when you sit the paper.

Question 1307

[1 marks]graph transformations

For

f(x)=(x-2)(x+3)

, state the

y

-intercept of

y=|f(x)|

Answer this when you sit the paper.

Question 1308

[1 marks]graph transformations

For

f(x)=(x-2)(x+3)

, state the

y

-intercept of

y=f(2x)

Answer this when you sit the paper.

Question 1309

[2 marks]graph transformations

For

f(x)=(x-2)(x+3)

with

x

-intercepts at

x=-3,2

, state the

x

-intercepts of

y=f(x-1)

, as 'a, b' in ascending order.

Answer this when you sit the paper.

Question 1401

[1 marks]integration / trapezium rule / volumes of revolution

The region

R

is bounded by

f(x) = x + x^{-1}

, the

x

-axis and the lines

x = 2

and

x = 5

. Calculate the exact area of

R

A $21 + \ln\tfrac{5}{2}$
B $10\tfrac{1}{2} + \ln\tfrac{2}{5}$
C $\tfrac{25}{2} + \ln 5$
D $10\tfrac{1}{2} + \ln\tfrac{5}{2}$

Question 1402

[1 marks]integration / trapezium rule / volumes of revolution

Use the trapezium rule with 4 ordinates to find the approximate area of the region bounded by

f(x) = x + x^{-1}

, the

x

-axis and the lines

x = 2

and

x = 5

, correct to 3 decimal places.

A $10.916$
B $11.416$
C $11.433$
D $22.867$

Question 1403

[1 marks]integration / trapezium rule / volumes of revolution

The region bounded by

f(x) = x + x^{-1}

, the

x

-axis and the lines

x = 2

and

x = 5

is rotated completely about the

x

-axis. Find the exact volume generated.

A $45.3$ cubic units
B $\dfrac{453\pi}{10}$ cubic units
C $\dfrac{453\pi}{5}$ cubic units
D $\dfrac{125\pi}{3}$ cubic units

Question 1404

[2 marks]integration / trapezium rule / volumes of revolution

For

f(x)=x+x^{-1}

, evaluate

\displaystyle\int_2^5 x\,dx

(one part of the area of region R).

Answer this when you sit the paper.

Question 1405

[2 marks]integration / trapezium rule / volumes of revolution

For

f(x)=x+x^{-1}

, evaluate

\displaystyle\int_2^5\dfrac1x\,dx

exactly (the other part of the area of region R).

Answer this when you sit the paper.

Question 1406

[2 marks]integration / trapezium rule / volumes of revolution

Using the trapezium rule with 4 ordinates for

f(x)=x+x^{-1}

[2,5]

(strip width

h=1

, ordinates at

x=2,3,4,5

), state

f(3)

, correct to 4 decimal places.

Answer this when you sit the paper.

Question 1407

[2 marks]integration / trapezium rule / volumes of revolution

For the same trapezium rule calculation on

f(x)=x+x^{-1}

, state

f(4)

, correct to 2 decimal places.

Answer this when you sit the paper.

Question 1408

[2 marks]integration / trapezium rule / volumes of revolution

Finding the volume generated when the region bounded by

f(x)=x+x^{-1}

is rotated about the

x

-axis needs

(x+x^{-1})^2=x^2+c+x^{-2}

. State

c

Answer this when you sit the paper.

Question 1501

[1 marks]vectors

Points

P

Q

R

have position vectors

\begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix}

\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}

and

\begin{pmatrix} -3 \\ 1 \\ -2 \end{pmatrix}

. Find

P\hat{Q}R

correct to the nearest

0.1°

A $13.8°$
B $76.2°$
C $103.8°$
D $166.2°$

Question 1502

[1 marks]vectors

The vector

\mathbf{W} = \begin{pmatrix} m - \tfrac{1}{2} \\ m + \tfrac{1}{2} \\ m\sqrt{6} \end{pmatrix}

is a unit vector. Find the possible values of

m

A $m = \pm\tfrac{1}{8}$
B $m = \pm\tfrac{1}{16}$
C $m = \pm\tfrac{1}{4}$
D $m = \pm\tfrac{1}{2}$

Question 1503

[1 marks]vectors

\mathbf{W} = \begin{pmatrix} m - \tfrac{1}{2} \\ m + \tfrac{1}{2} \\ m\sqrt{6} \end{pmatrix}

is normal to

\mathbf{V} = \begin{pmatrix} m+3 \\ 2m \\ -2\sqrt{6} \end{pmatrix}

. Find the values of

m

A $m = \tfrac{1}{6}$ or $m = -3$
B $m = \pm\tfrac{1}{4}$
C $m = 2$ only
D $m = -\tfrac{1}{6}$ or $m = 3$

Question 1504

[2 marks]vectors

Points

P

Q

R

have position vectors

\begin{pmatrix}2\\3\\4\end{pmatrix}

\begin{pmatrix}1\\0\\-1\end{pmatrix}

and

\begin{pmatrix}-3\\1\\-2\end{pmatrix}

. State

\overrightarrow{QP}=P-Q

as 'x,y,z'.

Answer this when you sit the paper.

Question 1505

[2 marks]vectors

For

P

Q

R

with position vectors

\begin{pmatrix}2\\3\\4\end{pmatrix}

\begin{pmatrix}1\\0\\-1\end{pmatrix}

and

\begin{pmatrix}-3\\1\\-2\end{pmatrix}

, state

\overrightarrow{QR}=R-Q

as 'x,y,z'.

Answer this when you sit the paper.

Question 1506

[2 marks]vectors

Given

\overrightarrow{QP}=(1,3,5)

and

\overrightarrow{QR}=(-4,1,-1)

(used to find

\angle PQR

), evaluate

\overrightarrow{QP}\cdot\overrightarrow{QR}

Answer this when you sit the paper.

Question 1507

[2 marks]vectors

\mathbf W=\begin{pmatrix}m-\frac12\\m+\frac12\\m\sqrt6\end{pmatrix}

is a unit vector, so

\left(m-\frac12\right)^2+\left(m+\frac12\right)^2+6m^2=1

. Expanding and simplifying (the cross terms cancel) gives

km^2=\dfrac12

. State

k

Answer this when you sit the paper.

Question 1508

[2 marks]vectors

\mathbf W

is normal to

\mathbf V=\begin{pmatrix}m+3\\2m\\-2\sqrt6\end{pmatrix}

, giving (after doubling to clear fractions) the quadratic

6m^2-17m-3=0

. Evaluate the discriminant

b^2-4ac

used to solve this quadratic.

Answer this when you sit the paper.

Question 1509

[1 marks]vectors

\mathbf U=\begin{pmatrix}m-\frac12\\n\\2\sqrt6\end{pmatrix}

and

\mathbf W=\begin{pmatrix}m-\frac12\\m+\frac12\\m\sqrt6\end{pmatrix}

are parallel (

\mathbf W=\lambda\mathbf U

). Since both share the same

x

-component

m-\frac12

(nonzero in general), this forces

\lambda=1

. Using the

z

-components,

m\sqrt6=1\times2\sqrt6

. State

m

Answer this when you sit the paper.

Pure Mathematics Paper 1 June 2016

Sit this paper online

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