ZIMSEC A Level Pure Mathematics Paper 1 November 2010

Question 101

[1 marks]logarithms

Let

t = \log_a b

, where

a

and

b

are positive real numbers with

a \neq b

. If

t + \dfrac{2}{t} = 3

, which quadratic equation does

t

satisfy?

A $t^2 + 3t + 2 = 0$
B $t^2 - 3t - 2 = 0$
C $t^2 - 2t + 3 = 0$
D $t^2 - 3t + 2 = 0$

Question 102

[1 marks]logarithms

a

and

b

are positive real numbers with

a \neq b

, and

\log_a b + \dfrac{2}{\log_a b} = 3

, express

b

in terms of

a

A $b = a^2$
B $b = a^3$
C $b = \sqrt{a}$
D $b = a$

Question 103

[2 marks]logarithms

Why must

t=\log_a b=1

be rejected as a solution to

t+\frac{2}{t}=3

in this problem?

ABecause $t=1$ would make $\log_a b$ negative, which is not permitted for real logarithms.
BBecause $t=1$ makes the equation $t+\frac2t=3$ false, so it is not a real solution.
CBecause $t=1$ gives $\log_a b=1$ , i.e. $b=a$ , which contradicts the condition $a\neq b$ given in the problem.
DBecause $t$ must be a positive integer greater than $1$ for logarithms to be defined.

Question 201

[1 marks]algebra

Find the product of

x^2 - 3x - 5

and

x^2 + 3x - 2

A $x^4 - 16x^2 + 9x + 10$
B $x^4 - 16x^2 - 9x + 10$
C $x^4 - 14x^2 - 9x + 10$
D $x^4 + 16x^2 - 9x + 10$

Question 202

[1 marks]algebra

Solve the inequality

2x^2 - 3x < 5

A $-\dfrac{5}{2} < x < 1$
B $-1 < x < 5$
C $-1 < x < \dfrac{5}{2}$
D $x < -1$ or $x > \dfrac{5}{2}$

Question 203

[2 marks]algebra

Rearrange

2x^2-3x<5

2x^2-3x-5<0

and factorise the left-hand side.

Answer this when you sit the paper.

Question 301

[1 marks]complex numbers

Express

z = \dfrac{1+i}{3+4i}

in the form

a+bi

, where

a

and

b

are real.

A $-\dfrac{1}{25} - \dfrac{1}{25}i$
B $\dfrac{1}{5} - \dfrac{1}{5}i$
C $\dfrac{7}{25} - \dfrac{1}{25}i$
D $\dfrac{7}{25} + \dfrac{1}{25}i$

Question 302

[1 marks]complex numbers

Given

z = \dfrac{7}{25} - \dfrac{1}{25}i

, find

|z|

in the form

c\sqrt{d}

, where

d

is a prime number.

A $\dfrac{7\sqrt2}{25}$
B $\dfrac{\sqrt2}{5}$
C $\dfrac{\sqrt2}{25}$
D $\dfrac{2\sqrt2}{5}$

Question 303

[1 marks]complex numbers

When rationalising

z=\dfrac{1+i}{3+4i}

by multiplying by

3-4i

, what does the denominator become?

Answer this when you sit the paper.

Question 304

[2 marks]complex numbers

Expand

(1+i)(3-4i)

to find the numerator (before dividing by 25).

Answer this when you sit the paper.

Question 401

[1 marks]arithmetic progression

An arithmetic progression has first term

-10

. For a certain number of terms

n

, the

n

th (last) term of the progression is

25

. Using

S_n = \dfrac{n}{2}(\text{first term}+\text{last term})

, find the smallest number of terms

n

for which the sum of the progression first exceeds

300

A $n = 41$
B $n = 42$
C $n = 40$
D $n = 21$

Question 402

[1 marks]arithmetic progression

An arithmetic progression has first term

-10

and

41

terms, with the

41

st term equal to

25

. Find the common difference

d

A $\dfrac{7}{8}$
B $\dfrac{35}{41}$
C $-\dfrac{7}{8}$
D $\dfrac{7}{40}$

Question 403

[1 marks]arithmetic progression

Write

S_n

for this progression in terms of

n

, using first term

-10

and last term

25

Answer this when you sit the paper.

Question 404

[1 marks]arithmetic progression

Using

n=41

terms, write the equation for

d

from

u_n=a+(n-1)d=25

(substitute the known values, before solving).

Answer this when you sit the paper.

Question 405

[1 marks]arithmetic progression

Using

u_n=a+(n-1)d

with

n=41

, what is the value of

(n-1)

substituted into the formula?

Answer this when you sit the paper.

Question 501

[1 marks]vectors

The position vectors of points

A

and

B

relative to the origin

O

are

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

and

\overrightarrow{OB}=-4\mathbf{i}+5\mathbf{j}+3\mathbf{k}

. Find

\cos(A\hat{O}B)

A $-\dfrac{4}{\sqrt{38}}$
B $\dfrac{2}{\sqrt{38}}$
C $\dfrac{4}{\sqrt{50}}$
D $\dfrac{4}{\sqrt{38}}$

Question 502

[1 marks]vectors

With

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

and

\overrightarrow{OB}=-4\mathbf{i}+5\mathbf{j}+3\mathbf{k}

, point

P

lies on

OB

such that

AP

is perpendicular to

OB

. Find the position vector of

P

A $-\dfrac{8}{5}\mathbf{i}+2\mathbf{j}+\dfrac{6}{5}\mathbf{k}$
B $-2\mathbf{i}+2.5\mathbf{j}+1.5\mathbf{k}$
C $-\dfrac{4}{5}\mathbf{i}+\mathbf{j}+\dfrac{3}{5}\mathbf{k}$
D $\dfrac{8}{5}\mathbf{i}-2\mathbf{j}-\dfrac{6}{5}\mathbf{k}$

Question 503

[1 marks]vectors

Calculate

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

for

\overrightarrow{OA}=i+3j+3k

\overrightarrow{OB}=-4i+5j+3k

Answer this when you sit the paper.

Question 504

[1 marks]vectors

Find

|\overrightarrow{OA}|

for

\overrightarrow{OA}=i+3j+3k

, exact form.

Answer this when you sit the paper.

Question 505

[2 marks]vectors

With

\overrightarrow{OP}=\lambda\overrightarrow{OB}

, set up

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

and solve for

\lambda

Answer this when you sit the paper.

Question 601

[1 marks]logarithms

Putting

10^x = m

, the equation

\dfrac{10^x+10^{-x}}{10^x-10^{-x}}=k

can be rearranged to give

m^2 =

A $\dfrac{k+1}{k-1}$
B $\dfrac{k+1}{1-k}$
C $\dfrac{k-1}{k+1}$
D $\dfrac{1-k}{1+k}$

Question 602

[1 marks]logarithms

Given that

x = \lg 2 + \dfrac12\lg 3 - \dfrac12

(the solution of

\dfrac{10^x+10^{-x}}{10^x-10^{-x}}=k

when

k=11

), find the value of

x

, correct to 3 decimal places.

A $-0.040$
B $0.040$
C $0.079$
D $0.540$

Question 603

[2 marks]logarithms

For

k=11

, evaluate

\dfrac{k+1}{k-1}

, simplified to lowest terms.

Answer this when you sit the paper.

Question 604

[2 marks]logarithms

Using

12=4\times3

and

\lg4=2\lg2

, write

\frac12\lg12

in terms of

\lg2

and

\lg3

(before subtracting the

\lg10

term).

Answer this when you sit the paper.

Question 701

[1 marks]functions and inverses

The graphs of

y=f(x)

y=g(x)=x

and

y=f^{-1}(x)

are sketched on the same axes;

f^{-1}(x)

is the reflection of

f(x)

in the line

g(x)=x

. In relation to

f

and

f^{-1}

, the line

g(x)=x

is best described as

Athe common tangent line to $f$ and $f^{-1}$
Bthe line of symmetry that maps $f$ onto $f^{-1}$
Cthe horizontal asymptote shared by $f$ and $f^{-1}$
Dthe axis about which $f$ is symmetric to itself

Question 702

[1 marks]functions and inverses

The graphs of

f(x)

g(x)=x

and

f^{-1}(x)

pass through the points

(0,3)

and

(0,-1\frac12)

. The graph of

f^{-1}(x)

meets the

x

-axis at

A

and the graph of

f(x)

meets the

x

-axis at

B

. What are the coordinates of

A

and

B

A $A(0,3)$ and $B(0,-1\frac12)$
B $A(1\frac12,0)$ and $B(-3,0)$
C $A(-1\frac12,0)$ and $B(3,0)$
D $A(3,0)$ and $B(-1\frac12,0)$

Question 703

[1 marks]functions and inverses

The line

f(x)

passes through

(0,3)

and

(-1\frac12,0)

, so

f(x)=2x+3

and

f^{-1}(x)=0.5x-1.5

. Find the coordinates of

C

, the point where

f(x)

f^{-1}(x)

and

g(x)=x

all intersect.

A $C(3,3)$
B $C(-1\frac12,-1\frac12)$
C $C(-3,-3)$
D $C(0,3)$

Question 704

[1 marks]functions and inverses

Find the gradient of the line

f^{-1}(x)

using its points

A(3,0)

and

(0,-1\frac12)

Answer this when you sit the paper.

Question 705

[2 marks]functions and inverses

Hence write

f^{-1}(x)

in the form

y=mx+c

Answer this when you sit the paper.

Question 706

[1 marks]functions and inverses

Since

f

is the reflection of

f^{-1}

y=x

, write

f(x)

in the form

y=mx+c

Answer this when you sit the paper.

Question 801

[1 marks]integration and trigonometry

By means of the substitution

u=\sin x

, find the exact value of

\displaystyle\int_0^{\frac{\pi}{2}}\cos x\sqrt{\sin x}\,dx

A $\dfrac{\sqrt2}{6}$
B $\dfrac{2}{3}$
C $\dfrac{1}{3}$
D $\dfrac{2^{1/4}}{3}$

Question 802

[1 marks]integration and trigonometry

Solve the equation

\cos 3y = -\dfrac{1}{\sqrt{2}}

for

0^\circ < y < 270^\circ

A $y = 15°, 25°, 55°, 65°$
B $y = 45°, 75°, 165°, 195°$
C $y = 135°, 225°$
D $y = 45°, 75°, 165°$

Question 803

[1 marks]integration and trigonometry

After substituting

u=\sin x

, find the new upper limit for

u

(when

x=\pi/2

Answer this when you sit the paper.

Question 804

[2 marks]integration and trigonometry

Evaluate

\int u^{1/2}\,du

(the indefinite integral in terms of

u

, ignoring the constant).

Answer this when you sit the paper.

Question 805

[2 marks]integration and trigonometry

For

0°<y<270°

, find the range of values of

3y

(in degrees).

Answer this when you sit the paper.

Question 901

[1 marks]numerical methods

Let

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

, so that

f'(x)=3x^2+3

. Since

f'(x)>0

for every real

x

f

is strictly increasing everywhere. How many real roots does the equation

f(x)=0

have?

Azero
Bexactly three
Cexactly one
Dexactly two

Question 902

[1 marks]numerical methods

Let

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

. Evaluate

f(0.8)

and

f(1)

, and state what this shows about the root of

f(x)=0

A $f(0.8)=-0.088$ , $f(1)=1$ ; the change of sign shows a root lies between $-0.8$ and $-1$
B $f(0.8)=0.088$ , $f(1)=1$ ; since both are positive, this interval contains no root
C $f(0.8)=-0.088$ , $f(1)=-1$ ; since both are negative, this interval contains no root
D $f(0.8)=-0.088$ , $f(1)=1$ ; the change of sign shows a root lies between $0.8$ and $1$

Question 903

[1 marks]numerical methods

Let

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

, with

f'(x)=3x^2+3

. Using one application of the Newton-Raphson formula

x_2=x_1-\dfrac{f(x_1)}{f'(x_1)}

, starting from

x_1=0.8

, find

x_2

correct to 3 significant figures.

A $0.782$
B $0.800$
C $0.818$
D $0.846$

Question 904

[1 marks]numerical methods

The equation

x^3+3x-3=0

has a root near

x=0.8

. Using the iteration

x_{n+1}=\dfrac{3-x_n^3}{3}

, starting from

x_0=0.8

, find the value obtained after two iterations, correct to 3 significant figures.

A $0.771$
B $0.800$
C $0.810$
D $0.829$

Question 905

[1 marks]numerical methods

For

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

, find

f'(x)

Answer this when you sit the paper.

Question 906

[2 marks]numerical methods

Using

x_{n+1}=\dfrac{3-x_n^3}{3}

with

x_0=0.8

, calculate

x_1

, to 4 decimal places.

Answer this when you sit the paper.

Question 1001

[1 marks]coordinate geometry - circles

Write down the equation of the circle with centre

(4,-3)

and radius

5

A $(x-4)^2+(y+3)^2=5$
B $(x-4)^2+(y-3)^2=25$
C $(x+4)^2+(y-3)^2=25$
D $(x-4)^2+(y+3)^2=25$

Question 1002

[1 marks]coordinate geometry - circles

For the circle

(x-4)^2+(y+3)^2=25

, state the condition satisfied by points

(x,y)

that lie strictly inside the circle.

A $(x-4)^2+(y+3)^2 \le 25$
B $(x-4)^2+(y+3)^2 < 25$
C $(x+4)^2+(y-3)^2 < 25$
D $(x-4)^2+(y+3)^2 > 25$

Question 1003

[1 marks]coordinate geometry - circles

The circle

(x-4)^2+(y+3)^2=25

intersects the line

2x+y=3

at two points. Find the range of values of

x

for which the point

(x,y)

on this line lies inside the circle.

A $-\dfrac{27}{5} < x < 1$
B $1 < x < 5$
C $-1 < x < \dfrac{27}{5}$
D $1 < x < \dfrac{27}{5}$

Question 1004

[2 marks]coordinate geometry - circles

Substitute

y=3-2x

into

(x-4)^2+(y+3)^2<25

and simplify to a quadratic inequality

ax^2+bx+c<0

. State the quadratic expression (just the expression).

Answer this when you sit the paper.

Question 1005

[2 marks]coordinate geometry - circles

Factorise

5x^2-32x+27

Answer this when you sit the paper.

Question 1101

[1 marks]parametric differentiation

A curve is defined parametrically by

x=\ln(3+2t)

and

y=e^{3t^2}

. Find

\dfrac{dy}{dx}

in terms of

t

A $3t(3+2t)e^{3t^2}$
B $6t(3+2t)e^{3t^2}$
C $\dfrac{3te^{3t^2}}{3+2t}$
D $3t(3-2t)e^{3t^2}$

Question 1102

[1 marks]parametric differentiation

For the curve

x=\ln(3+2t)

y=e^{3t^2}

, given that

\dfrac{dy}{dx}=3t(3+2t)e^{3t^2}

, find the coordinates of the curve's only turning point.

A $(\ln 3, 1)$
B $(-1.5, e^{6.75})$
C $(0, 1)$
D $(\ln 3, 0)$

Question 1103

[1 marks]parametric differentiation

Find

\dfrac{dx}{dt}

for

x=\ln(3+2t)

Answer this when you sit the paper.

Question 1104

[1 marks]parametric differentiation

Find

\dfrac{dy}{dt}

for

y=e^{3t^2}

Answer this when you sit the paper.

Question 1105

[3 marks]parametric differentiation

Why is

t=-\frac32

rejected as a solution to

\frac{dy}{dx}=0

, even though it satisfies

3t(3+2t)=0

ABecause $t=-\frac32$ makes $x=\ln(3+2t)$ undefined, since the argument $3+2t$ becomes zero.
BBecause $t=-\frac32$ gives a negative value of $x$ , which is not allowed on this curve.
CBecause $t=-\frac32$ makes $\frac{dy}{dx}$ positive instead of zero, so it is not really a root.
DBecause $t=-\frac32$ makes $y=e^{3t^2}$ undefined, since the exponent becomes negative.

Question 1201

[1 marks]polynomial equations

One root of

x^4+ax^3+bx^2+16x-12=0

2

and another root is

-2

. Find the values of

a

and

b

A $a=4,\ b=-1$
B $a=-4,\ b=1$
C $a=-4,\ b=-1$
D $a=-4,\ b=-7$

Question 1202

[1 marks]polynomial equations

Given that

x^4-4x^3-x^2+16x-12=0

has roots

x=2

and

x=-2

, find the other two roots.

A $x=3$ and $x=4$
B $x=1$ and $x=3$
C $x=-1$ and $x=-3$
D $x=1$ and $x=-3$

Question 1203

[1 marks]polynomial equations

Substitute

x=2

into

x^4+ax^3+bx^2+16x-12=0

and simplify to an equation in

a

and

b

Answer this when you sit the paper.

Question 1204

[1 marks]polynomial equations

Substitute

x=-2

into

x^4+ax^3+bx^2+16x-12=0

and simplify to an equation in

a

and

b

Answer this when you sit the paper.

Question 1205

[2 marks]polynomial equations

Divide

x^4-4x^3-x^2+16x-12

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

to find the quotient quadratic.

Answer this when you sit the paper.

Question 1206

[2 marks]polynomial equations

Factorise

x^2-4x+3

Answer this when you sit the paper.

Question 1301

[1 marks]trigonometry - triangles

In triangle

ABC

AB=23.2

cm,

BC=15.3

cm and angle

ABC=121°

. Find angle

BAC

, giving your answer to the nearest

0.1°

A $36.1°$
B $27.0°$
C $67.1°$
D $22.9°$

Question 1302

[1 marks]trigonometry - triangles

In triangle

PQR

PQ=2x

units,

PR=(x-2)

units, and angle

P=\left(\dfrac{\pi}{2}-x\right)

radians, where

x

is small so that

\cos P\approx\sin x\approx x

. Using the cosine rule

QR^2=PQ^2+PR^2-2\cdot PQ\cdot PR\cdot\cos P

, show that

QR^2

is approximately equal to which cubic expression in

x

A $-4x^3+13x^2-4x+4$
B $-4x^3+9x^2-4x+4$
C $-4x^3+13x^2+4x+4$
D $4x^3+13x^2-4x+4$

Question 1303

[2 marks]trigonometry - triangles

Using the cosine rule, calculate

AC

(to 1 d.p.) for triangle

ABC

with

AB=23.2

cm,

BC=15.3

cm, angle

ABC=121°

Answer this when you sit the paper.

Question 1304

[1 marks]trigonometry - triangles

What does

\cos\left(\frac{\pi}{2}-x\right)

equal exactly (before applying the small-angle approximation)?

Answer this when you sit the paper.

Question 1305

[3 marks]trigonometry - triangles

Using the cosine rule

QR^2=PQ^2+PR^2-2\cdot PQ\cdot PR\cdot\cos P

with

PQ=2x

and

PR=x-2

, expand and simplify

PQ^2+PR^2

only (before subtracting the cosine term).

Answer this when you sit the paper.

Question 1401

[1 marks]partial fractions and integration

Show that

2x^3-x^2+8x-4=(2x-1)(x^2+4)

by expanding the right-hand side. What is the correct expansion of

(2x-1)(x^2+4)

A $2x^3+x^2+8x-4$
B $2x^3-x^2+4x-4$
C $2x^3-x^2+8x-4$
D $2x^3-x^2+8x+4$

Question 1402

[1 marks]partial fractions and integration

Given that

\dfrac{x^2+2x+20}{2x^3-x^2+8x-4}=\dfrac{A}{2x-1}+\dfrac{Bx+C}{x^2+4}

, using

2x^3-x^2+8x-4=(2x-1)(x^2+4)

, find the values of

A

B

and

C

A $A=4,\ B=-2,\ C=1$
B $A=5,\ B=-2,\ C=4$
C $A=5,\ B=-2,\ C=0$
D $A=5,\ B=2,\ C=0$

Question 1403

[1 marks]partial fractions and integration

Given that

\dfrac{x^2+2x+20}{2x^3-x^2+8x-4}=\dfrac{5}{2x-1}-\dfrac{2x}{x^2+4}

, evaluate

\displaystyle\int_1^3\left(\dfrac{5}{2x-1}-\dfrac{2x}{x^2+4}\right)dx

, correct to 3 significant figures.

A $2.56$
B $3.06$
C $3.07$
D $5.63$

Question 1404

[1 marks]partial fractions and integration

Substituting

x=\frac12

into

x^2+2x+20=A(x^2+4)+(Bx+C)(2x-1)

, evaluate the left side,

x^2+2x+20

, at

x=\frac12

Answer this when you sit the paper.

Question 1405

[2 marks]partial fractions and integration

Hence solve for

A

, dividing

21.25

by the coefficient of

A

x=\frac12

(i.e.

\frac14+4

Answer this when you sit the paper.

Question 1406

[2 marks]partial fractions and integration

Write the antiderivative of

\dfrac{5}{2x-1}-\dfrac{2x}{x^2+4}

in terms of natural logs (ignore the constant of integration).

Answer this when you sit the paper.

Question 1407

[2 marks]partial fractions and integration

Evaluate

\dfrac{7}{2}\ln5

alone, to 3 decimal places.

Answer this when you sit the paper.

Question 1501

[1 marks]calculus - differentiation and integration

Express

12x-4x^2

in the form

c-(ax+b)^2

, where

a

b

and

c

are constants.

A $4-(2x-3)^2$
B $9-(x-3)^2$
C $9-(2x+3)^2$
D $9-(2x-3)^2$

Question 1502

[1 marks]calculus - differentiation and integration

The curve

y=12x-4x^2

passes through the point where

y=8

and

x<\dfrac32

. Find the equation of the tangent to the curve at this point.

A $y=8x$
B $y=4x+4$
C $y=4x-4$
D $y=-4x+16$

Question 1503

[1 marks]calculus - differentiation and integration

The region

R

is bounded by the

y

-axis, the tangent line

y=4x+4

, and the curve

y=12x-4x^2

, between

x=0

and

x=1

(the tangent line lies above the curve throughout this interval, touching it at

x=1

). Find, in terms of

\pi

, the exact volume generated when

R

is rotated through

360°

about the

x

-axis.

A $\dfrac{152}{15}$
B $\dfrac{16\pi}{5}$
C $\dfrac{136\pi}{5}$
D $\dfrac{152\pi}{15}$

Question 1504

[2 marks]calculus - differentiation and integration

Rearrange

12x-4x^2=8

4x^2-12x+8=0

and factorise.

Answer this when you sit the paper.

Question 1505

[1 marks]calculus - differentiation and integration

Find

\dfrac{dy}{dx}

for

y=12x-4x^2

Answer this when you sit the paper.

Question 1506

[1 marks]calculus - differentiation and integration

Evaluate the gradient of the curve at

x=1

, using

y'=12-8x

Answer this when you sit the paper.

Question 1507

[3 marks]calculus - differentiation and integration

Which integral correctly gives the volume generated when region R (bounded by the y-axis, the tangent

y=4x+4

, and the curve

y=12x-4x^2

, between

x=0

and

x=1

) is rotated

360°

about the x-axis?

A $\pi\displaystyle\int_0^2[(4x+4)^2-(12x-4x^2)^2]\,dx$
B $\displaystyle\int_0^1[(4x+4)^2-(12x-4x^2)^2]\,dx$
C $\pi\displaystyle\int_0^1[(4x+4)^2-(12x-4x^2)^2]\,dx$
D $\pi\displaystyle\int_0^1[(12x-4x^2)^2-(4x+4)^2]\,dx$

Question 1508

[3 marks]calculus - differentiation and integration

Evaluate

\displaystyle\int_0^1(4x+4)^2\,dx

Answer this when you sit the paper.

Question 1601

[1 marks]differential equations

Juice is poured into a cylindrical container of radius 5 cm at a rate of

100\text{ cm}^3

per minute and drains out through a hole at a rate of

2.5h\text{ cm}^3

per minute, where

h

cm is the depth of juice at time

t

minutes. Using

V=\pi r^2 h

, which differential equation does

h

satisfy?

A $\dfrac{dh}{dt}=\dfrac{40-h}{10\pi}$
B $\dfrac{dh}{dt}=\dfrac{40-h}{2\pi}$
C $\dfrac{dh}{dt}=\dfrac{h-40}{10\pi}$
D $\dfrac{dh}{dt}=\dfrac{100-2.5h}{25}$

Question 1602

[1 marks]differential equations

Solving

\dfrac{dh}{dt}=\dfrac{40-h}{10\pi}

with

h=0

t=0

gives

h

in terms of

t

A $h=40e^{-t/(10\pi)}$
B $h=40\left(1-e^{-t/(10\pi)}\right)$
C $h=40-e^{-t/(10\pi)}$
D $h=10\pi\left(1-e^{-t/40}\right)$

Question 1603

[1 marks]differential equations

When the tap in the previous part is closed, the juice drains according to

\dfrac{dh}{dt}=-\dfrac{h}{10\pi}

, with

h=40

t=0

. Solving gives

t=10\pi\ln\!\left(\dfrac{40}{h}\right)

. Find the time taken, to the nearest minute, for the height to fall to

0.88

cm.

A $52$ minutes
B $120$ minutes
C $240$ minutes
D $12$ minutes

Question 1604

[1 marks]differential equations

Write the volume of juice in the cylinder,

V

, in terms of

h

(radius

5

cm).

Answer this when you sit the paper.

Question 1605

[2 marks]differential equations

Write the rate balance

\dfrac{dV}{dt}=\text{rate in}-\text{rate out}

in terms of

h

(before dividing by

25\pi

Answer this when you sit the paper.

Question 1606

[2 marks]differential equations

Separate variables in

\dfrac{dh}{dt}=\dfrac{40-h}{10\pi}

to write

\int\frac{dh}{40-h}=\int\frac{dt}{10\pi}

, then integrate the left side (state the antiderivative, ignoring the constant).

Answer this when you sit the paper.

Question 1607

[1 marks]differential equations

Using

h=0

t=0

, find the constant of integration

C

-\ln|40-h|=\frac{t}{10\pi}+C

Answer this when you sit the paper.

Question 1608

[2 marks]differential equations

Why can the height

h

never actually reach

40

cm during filling, even though it is the value

h

approaches?

ABecause the outflow rate $2.5h$ becomes negative once $h$ exceeds $40$ , which is impossible.
BBecause at $h=40$ the rate of inflow becomes exactly $0$ , stopping the tap automatically.
CBecause $h=40(1-e^{-t/(10\pi)})$ equals $40$ only in the limit as $t\to\infty$ , never at any finite time $t$ .
DBecause the cylinder would overflow physically before $h$ reaches $40$ cm, due to its fixed height.

Question 1609

[2 marks]differential equations

For the drainage equation

\dfrac{dh}{dt}=-\dfrac{h}{10\pi}

, separate variables and integrate the left side

\int\frac{dh}{h}

(state the antiderivative, ignoring the constant).

Answer this when you sit the paper.

Question 1610

[1 marks]differential equations

Using

h=40

t=0

(start of drainage), find the constant of integration

C

t=-10\pi\ln h+C

Answer this when you sit the paper.

Question 1611

[2 marks]differential equations

Evaluate

\ln\left(\dfrac{40}{0.88}\right)

to 3 decimal places.

Answer this when you sit the paper.

Pure Mathematics Paper 1 November 2010

Sit this paper online

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