ZIMSEC A Level Pure Mathematics Paper 1 November 2009

Question 101

[1 marks]binomial expansion

In the expansion of

\left(x^2 - \dfrac{2}{x}\right)^9

, what is the coefficient of

x^{-3}

A $-10752$
B $-4608$
C $-2304$
D $4608$

Question 102

[3 marks]binomial expansion

In the expansion of

\left(x^2-\dfrac2x\right)^9

, the general term is

\binom{9}{r}(x^2)^{9-r}\left(-\dfrac2x\right)^r=\binom9r(-2)^rx^{18-3r}

. For the

x^{-3}

term, solve

18-3r=-3

for

r

Answer this when you sit the paper.

Question 201

[1 marks]trigonometry, cosine rule

In triangle

ABC

AC=7

cm,

CB=3

cm and

A\hat{B}C=60°

. Writing

AB=c

and applying the cosine rule at vertex

B

AC^2=AB^2+CB^2-2\cdot AB\cdot CB\cos B

, find the length of

AB

A $16$ cm
B $5$ cm
C $-5$ cm
D $8$ cm

Question 202

[2 marks]trigonometry, cosine rule

In triangle

ABC

with

AC=7

CB=3

, angle

B=60°

, the cosine rule

AC^2=AB^2+CB^2-2\cdot AB\cdot CB\cos B

gives a quadratic equation in

AB

(writing

c=AB

), in the form

c^2-3c-k=0

. State

k

Answer this when you sit the paper.

Question 203

[2 marks]trigonometry, cosine rule

Using

AB=8

(found from

AC=7

CB=3

, angle

B=60°

), evaluate the numerator

AC^2+CB^2-AB^2

used in

\cos C=\dfrac{AC^2+CB^2-AB^2}{2\cdot AC\cdot CB}

Answer this when you sit the paper.

Question 301

[1 marks]partial fractions

Writing

\dfrac{5x^2+7x+9}{(x+2)^2(3-x)}=\dfrac{A}{x+2}+\dfrac{B}{(x+2)^2}+\dfrac{C}{3-x}

, so that

5x^2+7x+9=A(x+2)(3-x)+B(3-x)+C(x+2)^2

, what is the value of

B

, found by substituting

x=-2

A $-3$
B $3$
C $5$
D $15$

Question 302

[1 marks]partial fractions

Express

\dfrac{5x^2+7x+9}{(x+2)^2(3-x)}

in partial fractions.

A $\dfrac{-2}{x+2}+\dfrac{3}{(x+2)^2}-\dfrac{3}{3-x}$
B $\dfrac{8}{x+2}+\dfrac{3}{(x+2)^2}+\dfrac{3}{3-x}$
C $\dfrac{-2}{x+2}+\dfrac{3}{(x+2)^2}+\dfrac{3}{3-x}$
D $\dfrac{2}{x+2}+\dfrac{3}{(x+2)^2}+\dfrac{3}{3-x}$

Question 303

[3 marks]partial fractions

Comparing coefficients of

x^2

5x^2+7x+9=A(x+2)(3-x)+B(3-x)+C(x+2)^2

gives

5=-A+C

. Given

C=3

(found by substituting

x=3

), state

A

Answer this when you sit the paper.

Question 401

[1 marks]functions

The function

f

is defined by

f:x\to\dfrac{1+x^2}{x}

x\in\mathbb{R}

x\neq k

. State the value of

k

Answer this when you sit the paper.

Question 402

[1 marks]functions

Given

f(x)=\dfrac{1+x^2}{x}

(

x\neq 0

), find

f\!\left(\dfrac{1}{x}\right)

in its simplest form.

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

Question 403

[1 marks]functions

Given

f(x)=\dfrac{1+x^2}{x}

(

x\neq0

), find the other element in the domain (besides

x=2

) that has the same image as

x=2

A $-2$
B $\dfrac12$
C $-\dfrac12$
D $1$

Question 404

[3 marks]functions

Setting

f(x)=\dfrac52

(the image of

x=2

under

f(x)=\dfrac{1+x^2}x

) gives the quadratic

2x^2-5x+2=0

, which factorises as

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

. State

c

Answer this when you sit the paper.

Question 501

[1 marks]complex numbers

Given

z=-3+2i

, find

|z|

A $\sqrt{13}$
B $5$
C $\sqrt5$
D $13$

Question 502

[1 marks]complex numbers

Given

z=-3+2i

, find

\arg z

(to 1 decimal place).

A $146.3°$
B $33.7°$
C $-146.3°$
D $213.7°$

Question 503

[1 marks]complex numbers

Given

z=-3+2i

and

w=5+4i

, find

\dfrac{z}{w}

in the form

a+ib

A $-\dfrac{7}{9}+\dfrac{22}{9}i$
B $-\dfrac{7}{41}+\dfrac{22}{41}i$
C $-\dfrac{7}{41}-\dfrac{22}{41}i$
D $\dfrac{7}{41}+\dfrac{22}{41}i$

Question 504

[3 marks]complex numbers

Multiplying

\dfrac{-3+2i}{5+4i}

\dfrac{5-4i}{5-4i}

, state the numerator

(-3+2i)(5-4i)

in the form

a+bi

(before dividing by the denominator 41).

Answer this when you sit the paper.

Question 601

[1 marks]polynomials

Given that

(3x-2)

is a factor of

6x^3+8x^2+kx+2

, find the value of

k

A $-11$
B $-9$
C $-8$
D $11$

Question 602

[1 marks]polynomials

Given that

k=-11

, so that

6x^3+8x^2-11x+2=6x^3+8x^2+kx+2

has

(3x-2)

as a factor and quotient

2x^2+4x-1

, what are the roots of

2x^2+4x-1=0

A $\dfrac{-2+\sqrt6}{2}$ and $\dfrac{-2-\sqrt6}{2}$
B $\dfrac{-4+\sqrt6}{4}$ and $\dfrac{-4-\sqrt6}{4}$
C $-2+\sqrt6$ and $-2-\sqrt6$
D $\dfrac{2+\sqrt6}{2}$ and $\dfrac{2-\sqrt6}{2}$

Question 603

[2 marks]polynomials

Dividing

6x^3+8x^2-11x+2

(3x-2)

gives a quadratic quotient

2x^2+bx+c

. State

b

and

c

as 'b, c'.

Answer this when you sit the paper.

Question 604

[2 marks]polynomials

Verifying

(3x-2)

is a factor: evaluate

6x^3+8x^2+2

x=\dfrac23

(ignoring the

kx

term for now), giving a fraction.

Answer this when you sit the paper.

Question 701

[1 marks]trigonometry, R-form

Expressing

3\cos x-5\sin x

in the form

R\sin(x+\alpha)

with

R>0

and

0°<\alpha<360°

, what are the values of

R

and

\alpha

(to 1 decimal place)?

A $R=\sqrt{34},\ \alpha\approx31.0°$
B $R=\sqrt{34},\ \alpha\approx239.0°$
C $R=\sqrt8,\ \alpha\approx149.0°$
D $R=\sqrt{34},\ \alpha\approx149.0°$

Question 702

[1 marks]trigonometry, R-form

Given that

3\cos x-5\sin x=\sqrt{34}\sin(x+149.0°)

, solve

3\cos x-5\sin x=2

for

0°<x<360°

, to the nearest

0.1°

A $x\approx190.9°$ or $51.0°$
B $x\approx10.9°$ or $231.0°$
C $x\approx20.1°$ or $159.9°$
D $x\approx128.9°$ or $349.1°$

Question 703

[2 marks]trigonometry, R-form

Matching

R\sin(x+\alpha)=R\sin x\cos\alpha+R\cos x\sin\alpha

with

3\cos x-5\sin x

, state

R\sin\alpha

and

R\cos\alpha

as 'a, b'.

Answer this when you sit the paper.

Question 704

[2 marks]trigonometry, R-form

Using

\sqrt{34}\sin(x+149.0°)=2

, evaluate

2/\sqrt{34}

, correct to 4 decimal places.

Answer this when you sit the paper.

Question 801

[1 marks]parametric equations, coordinate geometry

Find, in the form

y=mx+c

, the equation of a line with gradient

m

that passes through the point

(m,m)

A $y=mx-m-m^2$
B $y=mx$
C $y=mx+m^2-m$
D $y=mx+m-m^2$

Question 802

[1 marks]parametric equations, coordinate geometry

A curve is given parametrically by

x=2+\sin\theta

y=6\cos\theta

. Find

\dfrac{dy}{dx}

in terms of

\theta

A $6\tan\theta$
B $-6\cot\theta$
C $-6\tan\theta$
D $-\dfrac16\tan\theta$

Question 803

[1 marks]parametric equations, coordinate geometry

A curve is given parametrically by

x=2+\sin\theta

y=6\cos\theta

. Which of the following is the correct Cartesian equation relating

x

and

y

A $y^2-36x^2+144x-108=0$
B $y^2+36x^2-144x+108=0$
C $y^2+36x^2-144x+144=0$
D $y^2+36x^2-72x+108=0$

Question 804

[2 marks]parametric equations, coordinate geometry

For

x=2+\sin\theta

y=6\cos\theta

, express

\sin\theta

in terms of

x

Answer this when you sit the paper.

Question 805

[2 marks]parametric equations, coordinate geometry

For

x=2+\sin\theta

y=6\cos\theta

, express

\cos\theta

in terms of

y

Answer this when you sit the paper.

Question 806

[1 marks]parametric equations, coordinate geometry

Using

\sin^2\theta+\cos^2\theta=1

with

\sin\theta=x-2

\cos\theta=y/6

, state the equation before simplifying, in the form

(x-2)^2+y^2/36=k

. State

k

Answer this when you sit the paper.

Question 901

[1 marks]trigonometric equations

Solve

|2\sin x-3|=1

for

0°\leq x\leq360°

ANo solution
B $x=90°$ only
C $x=90°$ and $x=270°$
D $x=30°$ and $x=150°$

Question 902

[1 marks]trigonometric equations

Solve

\dfrac{3}{2}\sin2\theta=2\sin^2\theta

for

0°\leq\theta\leq360°

, to the nearest

0.1°

A $\theta=0°,180°,360°$ only
B $\theta=0°,180°,360°,33.7°,213.7°$
C $\theta=0°,180°,360°,56.3°,236.3°$
D $\theta=56.3°,236.3°$ only

Question 903

[2 marks]trigonometric equations

Solving

|2\sin x-3|=1

splits into

2\sin x-3=1

(giving

\sin x=2

, no solution) and

2\sin x-3=-1

. State the value of

\sin x

from the second case.

Answer this when you sit the paper.

Question 904

[2 marks]trigonometric equations

Rewriting

\dfrac32\sin2\theta=2\sin^2\theta

using

\sin2\theta=2\sin\theta\cos\theta

gives

3\sin\theta\cos\theta=2\sin^2\theta

, i.e.

\sin\theta(3\cos\theta-2\sin\theta)=0

. From the second factor, state

\tan\theta

as a fraction.

Answer this when you sit the paper.

Question 905

[2 marks]trigonometric equations

From the factor

\sin\theta=0

\sin\theta(3\cos\theta-2\sin\theta)=0

, state all solutions for

0°\leq\theta\leq360°

, as 'a, b, c'.

Answer this when you sit the paper.

Question 1001

[1 marks]differentiation, rates of change

Given

y=\dfrac{3x-6}{x}

, where

x

and

y

are functions of

t

with

\dfrac{dy}{dt}=0.6

, find

\dfrac{dx}{dt}

when

y=1

A $0.225$
B $0.4$
C $0.9$
D $1.11$

Question 1002

[1 marks]differentiation, rates of change

Given

y=5-\dfrac{3}{x}

, and that

y

increases from

4

by a small amount

\dfrac{p}{25}

, find the approximate change in

x

in terms of

p

A $\dfrac{3p}{25}$
B $\dfrac{p}{25}$
C $\dfrac{p}{75}$
D $\dfrac{16p}{75}$

Question 1003

[1 marks]differentiation, rates of change

Given

y=5-\dfrac{3}{x}

, and that

y

increases from

4

by a small amount

\dfrac{p}{25}

, giving an approximate change in

x

\dfrac{3p}{25}

(at

x=3

), find the corresponding percentage change in

x

, in terms of

p

A $4p\%$
B $\dfrac{4p}{3}\%$
C $12p\%$
D $\dfrac{4p}{25}\%$

Question 1004

[2 marks]differentiation, rates of change

For

y=\dfrac{3x-6}{x}=3-\dfrac6x

, find

\dfrac{dy}{dx}

in terms of

x

Answer this when you sit the paper.

Question 1005

[2 marks]differentiation, rates of change

When

y=1

y=3-\dfrac6x

, solve for

x

Answer this when you sit the paper.

Question 1006

[2 marks]differentiation, rates of change

For

y=5-\dfrac3x

, find

\dfrac{dy}{dx}

in terms of

x

Answer this when you sit the paper.

Question 1101

[1 marks]vectors

Points

A(1,6,2)

and

B(-1,11,-1)

are given, along with a point

C(-2,-5,11)

and a point

P

with position vector

(2p+1,\,-5p+6,\,3p+2)

. Find

\overrightarrow{CP}

in terms of

p

A $(2p+3,\ -5p+1,\ 3p-9)$
B $(2p-1,\ -5p+1,\ 3p+13)$
C $(-2p-3,\ 5p-11,\ -3p+9)$
D $(2p+3,\ -5p+11,\ 3p-9)$

Question 1102

[1 marks]vectors

Given

\overrightarrow{AB}=(-2,5,-3)

and

\overrightarrow{CP}=(2p+3,\,-5p+11,\,3p-9)

, find

\overrightarrow{CP}\cdot\overrightarrow{AB}

in terms of

p

A $-38p+22$
B $-38p+76$
C $-22p+76$
D $38p-76$

Question 1103

[1 marks]vectors

Given that

\overrightarrow{CP}\cdot\overrightarrow{AB}=-38p+76

and that

CP

is perpendicular to

AB

, find the value of

p

A $-2$
B $0$
C $0.5$
D $2$

Question 1104

[1 marks]vectors

Points

A(1,6,2)

B(-1,11,-1)

and

P=(2p+1,-5p+6,3p+2)

are collinear for all values of

p

. Point

C

has position vector

(-2,-5,11)

. Given that

\overrightarrow{CP}

is perpendicular to

\overrightarrow{AB}

when

p=2

, find the shortest distance from

C

to the line

AB

A $59$
B $\sqrt{58}$
C $\sqrt{38}$
D $\sqrt{59}$

Question 1105

[2 marks]vectors

State

\overrightarrow{AB}=B-A

for

A=(1,6,2)

B=(-1,11,-1)

, as 'x,y,z'.

Answer this when you sit the paper.

Question 1106

[2 marks]vectors

State

\overrightarrow{AP}=P-A

in terms of

p

, for

P=(2p+1,-5p+6,3p+2)

A=(1,6,2)

, as 'x,y,z'.

Answer this when you sit the paper.

Question 1107

[2 marks]vectors

Given

\overrightarrow{AP}=p(2,-5,3)

and

\overrightarrow{AB}=(-2,5,-3)

, state the scalar

m

such that

\overrightarrow{AB}=m(2,-5,3)

(used to confirm

A

B

P

are collinear).

Answer this when you sit the paper.

Question 1201

[1 marks]calculus, integration

Given that

\dfrac{d}{dx}\left(\dfrac{x}{\sqrt{4-x^2}}\right)=\dfrac{4}{(4-x^2)^{3/2}}

, evaluate

\displaystyle\int_0^1\dfrac{8}{(4-x^2)^{3/2}}\,dx

A $\dfrac{2\sqrt3}{3}$
B $\dfrac{4\sqrt3}{3}$
C $\dfrac23$
D $\dfrac{\sqrt3}{3}$

Question 1202

[1 marks]calculus, integration

The region enclosed by

y=\dfrac{x}{\sqrt{4-x^2}}

, the

x

-axis, and the lines

x=-1

and

x=1

is rotated about the

x

-axis through

360°

. Which of the following gives the volume generated?

A $2\pi(\ln3-1)$
B $2\pi(1-\ln3)$
C $\pi(\ln3-1)$
D $2\pi(\ln3+1)$

Question 1203

[2 marks]calculus, integration

Differentiating

\dfrac{x}{\sqrt{4-x^2}}=x(4-x^2)^{-1/2}

by the product rule needs

\dfrac{d}{dx}(4-x^2)^{-1/2}

. State this derivative, in terms of

x

Answer this when you sit the paper.

Question 1204

[2 marks]calculus, integration

Evaluate

\dfrac{1}{\sqrt{4-1^2}}=\dfrac1{\sqrt3}

as a decimal, correct to 3 decimal places (used in

\left[\dfrac{x}{\sqrt{4-x^2}}\right]_0^1

Answer this when you sit the paper.

Question 1205

[2 marks]calculus, integration

The integrand

\dfrac{x^2}{4-x^2}

can be rewritten as

-1+\dfrac4{4-x^2}

. Confirm by expanding

-1+\dfrac4{4-x^2}

over the common denominator

4-x^2

: state the resulting numerator (it should equal

x^2

Answer this when you sit the paper.

Question 1206

[2 marks]calculus, integration

Evaluate

[-x]_0^1

(one part of

\left[-x+\ln\dfrac{2+x}{2-x}\right]_0^1

used in the volume integral).

Answer this when you sit the paper.

Question 1301

[1 marks]completing the square, Newton-Raphson

Expressing

12x-3x^2+15

in the form

a+b(x+c)^2

gives

27-3(x-2)^2

. What are the coordinates of the turning point of

y=12x-3x^2+15

, and is it a maximum or minimum?

A $(2,27)$ , minimum
B $(-2,27)$ , maximum
C $(2,-27)$ , maximum
D $(2,27)$ , maximum

Question 1302

[1 marks]completing the square, Newton-Raphson

The graphs of

y=e^{-x}

and

y=12x-3x^2+15

are sketched on the same axes. How many real roots does the equation

e^{-x}+3x^2-12x-15=0

have?

A $0$
B $1$
C $2$
D $3$

Question 1303

[1 marks]completing the square, Newton-Raphson

Using

x_1=-0.8

and the Newton-Raphson formula

x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}

, where

f(x)=e^{-x}+3x^2-12x-15

, find the root of

f(x)=0

correct to 4 decimal places.

A $-0.9000$
B $-0.8650$
C $-0.8000$
D $-0.7900$

Question 1304

[2 marks]completing the square, Newton-Raphson

Complete the square: write

12x-3x^2+15

-3(x-2)^2+d

for some constant

d

. State

d

Answer this when you sit the paper.

Question 1305

[2 marks]completing the square, Newton-Raphson

For

f(x)=e^{-x}+3x^2-12x-15

, state

f'(x)

in terms of

x

Answer this when you sit the paper.

Question 1306

[2 marks]completing the square, Newton-Raphson

Using

x_1=-0.8

, evaluate

f(-0.8)

for

f(x)=e^{-x}+3x^2-12x-15

, correct to 4 decimal places.

Answer this when you sit the paper.

Question 1307

[2 marks]completing the square, Newton-Raphson

Evaluate

f'(-0.8)

for

f'(x)=-e^{-x}+6x-12

, correct to 4 decimal places.

Answer this when you sit the paper.

Question 1401

[1 marks]arithmetic and geometric progressions

In an arithmetic progression, the 6th term is half the 4th term, and the 3rd term is 15. Find the first term

a

and common difference

d

A $a=21,\ d=3$
B $a=-15,\ d=15$
C $a=21,\ d=-3$
D $a=15,\ d=0$

Question 1402

[1 marks]arithmetic and geometric progressions

For the arithmetic progression with first term

a=21

and common difference

d=-3

, find the least number of terms

n

for which the sum

S_n

becomes, and remains, less than

65

A $n=3$
B $n=4$
C $n=12$
D $n=11$

Question 1403

[1 marks]arithmetic and geometric progressions

Evaluate

\displaystyle\sum_{r=1}^{10}3\left(\dfrac34\right)^r

, correct to 4 significant figures.

A $2.250$
B $8.493$
C $9.000$
D $11.49$

Question 1404

[2 marks]arithmetic and geometric progressions

In the AP,

T_3=a+2d=15

and

T_6=\frac12T_4

gives

a+7d=0

, so

a=-7d

. Substituting into

a+2d=15

gives an equation in

d

, in the form

md=15

. State

m

Answer this when you sit the paper.

Question 1405

[2 marks]arithmetic and geometric progressions

Using

S_n=\dfrac{3n(15-n)}2

(for the AP with

a=21

d=-3

), evaluate

S_{10}

Answer this when you sit the paper.

Question 1406

[2 marks]arithmetic and geometric progressions

Using

S_n=\dfrac{3n(15-n)}2

, evaluate

S_{11}

Answer this when you sit the paper.

Question 1407

[2 marks]arithmetic and geometric progressions

For

\sum_{r=1}^{10}3\left(\frac34\right)^r

, the series has first term

3\times\frac34

. State this first term as a decimal.

Answer this when you sit the paper.

Question 1501

[1 marks]calculus, partial fractions

The radius of a circle depends on time according to

r=t^3+1

, and its area is

A=\pi r^2

. Find

\dfrac{dA}{dt}

t=2

, in terms of

\pi

A $108\pi$
B $192\pi$
C $72\pi$
D $216\pi$

Question 1502

[1 marks]calculus, partial fractions

The curve

y=x^2\ln x

has

\dfrac{dy}{dx}=x(2\ln x+1)

, which is zero at

x=e^{-1/2}

, giving the point

\left(e^{-1/2},\,-\tfrac12e^{-1}\right)

. Using

\dfrac{d^2y}{dx^2}=2\ln x+3

, determine the nature of this turning point.

AMaximum, since $\dfrac{d^2y}{dx^2}=2>0$ at that point
BPoint of inflection, since $\dfrac{d^2y}{dx^2}=0$ at that point
CMinimum, since $\dfrac{d^2y}{dx^2}=2>0$ at that point
DMaximum, since $\dfrac{d^2y}{dx^2}=-1<0$ at that point

Question 1503

[1 marks]calculus, partial fractions

Express

\dfrac{6x^2+7x-2}{(x-1)(x-2)(x+2)}

in partial fractions.

A $\dfrac{-11/3}{x-1}+\dfrac{9}{x-2}-\dfrac{2/3}{x+2}$
B $\dfrac{-11/3}{x-1}+\dfrac{9}{x-2}+\dfrac{2/3}{x+2}$
C $\dfrac{11/3}{x-1}+\dfrac{9}{x-2}+\dfrac{2/3}{x+2}$
D $\dfrac{-11/3}{x-1}+\dfrac{2/3}{x-2}+\dfrac{9}{x+2}$

Question 1504

[2 marks]calculus, partial fractions

For

r=t^3+1

and

A=\pi r^2

, state

\dfrac{dA}{dt}

in terms of

t

(general expression, before substituting

t=2

Answer this when you sit the paper.

Question 1505

[2 marks]calculus, partial fractions

For

y=x^2\ln x

, find

\dfrac{dy}{dx}

in terms of

x

Answer this when you sit the paper.

Question 1506

[2 marks]calculus, partial fractions

Setting

\dfrac{dy}{dx}=x(2\ln x+1)=0

for

x>0

, the turning point requires

2\ln x+1=0

. State

\ln x

at the turning point.

Answer this when you sit the paper.

Question 1507

[2 marks]calculus, partial fractions

For

y=x^2\ln x

, state

\dfrac{d^2y}{dx^2}

in terms of

x

Answer this when you sit the paper.

Question 1508

[2 marks]calculus, partial fractions

For

\dfrac{6x^2+7x-2}{(x-1)(x-2)(x+2)}=\dfrac A{x-1}+\dfrac B{x-2}+\dfrac C{x+2}

, substituting

x=2

gives

36=B(1)(4)

. State

B

Answer this when you sit the paper.

Question 1509

[2 marks]calculus, partial fractions

Verifying the partial fractions

A=-\frac{11}3

B=9

C=\frac23

: evaluate

6x^2+7x-2

x=0

(this should match

-4A-2B+2C

from the identity).

Answer this when you sit the paper.

Pure Mathematics Paper 1 November 2009

Sit this paper online

The questions

Question 101

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Question 1501

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Question 1508

Question 1509