Sec 4 A-Math Linear law worksheet with answers

A free Sec 4 A-Math worksheet on linear law: 7 questions, 40 marks, short questions first and then structured ones, in the school exam style. The answers and full worked solutions are on the last page.

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Sec 4 A-Math worksheet: Linear law

Total: 40 marksSuggested time: 1 hour

Name: ______________________________Class: ____________ Date: ____________

Answer all questions. Show all your working. Give non-exact answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless the question says otherwise.

Section A

The variables

x

and

y

are related by the equation

y = ae^{bx}

, where

a

and

b

are constants. When

\ln y

is plotted against

x

, a straight line passing through the points

(2, -2.35)

and

(6, -8.35)

is obtained. Find the value of

a

, correct to 3 significant figures.

Answer[3]

The variables

x

and

y

are related by the equation

y = ae^{bx}

, where

a

and

b

are constants. When

\ln y

is plotted against

x

, a straight line passing through the points

(1, 1.45)

and

(4, 3.85)

is obtained. Find the value of

b

Answer[3]

The variables

x

and

y

are related by the equation

y = \dfrac{a}{x} + b

, where

a

and

b

are constants. When

y

is plotted against

\dfrac{1}{x}

, a straight line passing through the points

(2, -2)

and

(5, 1)

is obtained. Find the value of

b

Answer[3]

(a)The variables

x

and

y

are related by the equation

y = ax + \dfrac{b}{x}

, where

a

and

b

are constants. When

xy

is plotted against

x^{2}

, a straight line passing through the points

(2, 7)

and

(5, 13)

is obtained. Find the value of

b

Answer (a)[3]

(b)The variables

x

and

y

are related by the equation

y = a\sqrt{x} + \dfrac{b}{\sqrt{x}}

, where

a

and

b

are constants. When

y\sqrt{x}

is plotted against

x

, a straight line passing through the points

(1, 5)

and

(3, 1)

is obtained. Find the value of

a

Answer (b)[2]

(a)The variables

x

and

y

are related by the equation

y = kx^{n}

, where

k

and

n

are constants. When

\lg y

is plotted against

\lg x

, a straight line passing through the points

(0.1, 1.42)

and

(0.9, 3.82)

is obtained. Find the value of

k

, correct to 3 significant figures.

Answer (a)[3]

(b)The variables

x

and

y

are related by the equation

y = kx^{n}

, where

k

and

n

are constants. When

\lg y

is plotted against

\lg x

, a straight line passing through the points

(0.5, 1.6)

and

(1.5, 3.1)

is obtained. Find the value of

k

, correct to 3 significant figures.

Answer (b)[3]

Section B

The table shows experimental values of two variables,

x

and

y

. It is known that

x

and

y

are related by the equation

y = ax^{2} + bx

, where

a

and

b

are constants.

$x$	1	2	3	4	5
$y$	26	56	90	128	170

(a)A straight line graph is to be drawn to find the values of

a

and

b

. State the variables that should be plotted.

Answer (a)[1]

(b)Using the values in the table, the graph of

Y

against

X

is drawn, as shown above, where

Y

and

X

are the variables in part (a). Calculate the value of

Y

for the point plotted when

x = 2

Answer (b)[1]

(c)Use the graph to find the value of

a

Answer (c)[2]

(d)Use the graph to find the value of

b

Answer (d)[1]

(e)Use the graph to estimate the value of

y

when

x = 3.5

Answer (e)[2]

(a)The variables

x

and

y

are related by the equation

y = a\sqrt{x} + \dfrac{b}{\sqrt{x}}

, where

a

and

b

are constants. When

y\sqrt{x}

is plotted against

x

, a straight line passing through the points

(2, 7)

and

(5, 13)

is obtained. Find the value of

b

Answer (a)[4]

(b)The variables

x

and

y

are related by the equation

y = kx^{n}

, where

k

and

n

are constants. When

\lg y

is plotted against

\lg x

, a straight line passing through the points

(0.3, 1.72)

and

(1.1, 3.32)

is obtained. Find the value of

n

Answer (b)[4]

(c)The variables

x

and

y

are related by the equation

y = ae^{bx}

, where

a

and

b

are constants. When

\ln y

is plotted against

x

, a straight line passing through the points

(2, 2.25)

and

(5, 4.65)

is obtained. Find the value of

a

, correct to 3 significant figures.

Answer (c)[5]

END OF PAPER

Answers and worked solutions

Question 1

Answer:

= 1.92

(3 s.f.)[3]

\ln y = \ln a + bx

\ln y = bx + \ln a

b

is the gradient and

\ln a

is the

Y

-intercept.

b = \dfrac{Y_2 - Y_1}{X_2 - X_1}

b = \dfrac{-8.35 - (-2.35)}{6 - 2}

= -1.5

-2.35 = -1.5(2) + \ln a

\ln a = 0.65

a = e^{0.65}

= 1.92

(3 s.f.)

Question 2

Answer:

= 0.8

[3]

\ln y = \ln a + bx

\ln y = bx + \ln a

b

is the gradient and

\ln a

is the

Y

-intercept.

b = \dfrac{Y_2 - Y_1}{X_2 - X_1}

b = \dfrac{3.85 - 1.45}{4 - 1}

= 0.8

Question 3

Answer:

b = -4

[3]

Write a ÷ x as a times (1 ÷ x).

y = a\left(\dfrac{1}{x}\right) + b

a

is the gradient and

b

is the

Y

-intercept.

a = \dfrac{Y_2 - Y_1}{X_2 - X_1}

a = \dfrac{1 - (-2)}{5 - 2}

= 1

-2 = 1(2) + b

b = -4

Question 4

(a)Answer:

b = 3

[3]

Multiply both sides by x.

xy = ax^{2} + b

a

is the gradient and

b

is the

Y

-intercept.

a = \dfrac{Y_2 - Y_1}{X_2 - X_1}

a = \dfrac{13 - 7}{5 - 2}

= 2

7 = 2(2) + b

b = 3

(b)Answer:

= -2

[2]

Multiply both sides by √x.

y\sqrt{x} = ax + b

a

is the gradient and

b

is the

Y

-intercept.

a = \dfrac{Y_2 - Y_1}{X_2 - X_1}

a = \dfrac{1 - 5}{3 - 1}

= -2

Question 5

(a)Answer:

= 13.2

(3 s.f.)[3]

\lg y = \lg k + n\lg x

\lg y = n\lg x + \lg k

n

is the gradient and

\lg k

is the

Y

-intercept.

n = \dfrac{Y_2 - Y_1}{X_2 - X_1}

n = \dfrac{3.82 - 1.42}{0.9 - 0.1}

= 3

1.42 = 3(0.1) + \lg k

\lg k = 1.12

k = 10^{1.12}

= 13.2

(3 s.f.)

(b)Answer:

= 7.08

(3 s.f.)[3]

\lg y = \lg k + n\lg x

\lg y = n\lg x + \lg k

n

is the gradient and

\lg k

is the

Y

-intercept.

n = \dfrac{Y_2 - Y_1}{X_2 - X_1}

n = \dfrac{3.1 - 1.6}{1.5 - 0.5}

= 1.5

1.6 = 1.5(0.5) + \lg k

\lg k = 0.85

k = 10^{0.85}

= 7.08

(3 s.f.)

Question 6

(a)Answer:

\dfrac{y}{x}

against

x

[1]

Divide both sides by

x

\dfrac{y}{x} = ax + b

Plot

\dfrac{y}{x}

against

x

: the gradient is

a

and the vertical intercept is

b

(b)Answer:

Y = 28

[1]

When

x = 2

y = 56

\dfrac{y}{x} = \dfrac{56}{2}

\phantom{\dfrac{y}{x}} = 28

(c)Answer:

a = 2

[2]

a

is the gradient of the line.Gradient

= \dfrac{34 - 26}{5 - 1}

= 2

(d)Answer:

b = 24

[1]

b

is the vertical intercept: the line meets the

Y

-axis at

24

b = 24

(e)Answer:

y \approx 108.5

[2]

When

x = 3.5

X = 3.5

From the line,

Y = 31

y = 31 \times 3.5

= 108.5

Question 7

(a)Answer:

b = 3

[4]

Multiply both sides by √x.

y\sqrt{x} = ax + b

a

is the gradient and

b

is the

Y

-intercept.

a = \dfrac{Y_2 - Y_1}{X_2 - X_1}

a = \dfrac{13 - 7}{5 - 2}

= 2

7 = 2(2) + b

b = 3

(b)Answer:

= 2

[4]

\lg y = \lg k + n\lg x

\lg y = n\lg x + \lg k

n

is the gradient and

\lg k

is the

Y

-intercept.

n = \dfrac{Y_2 - Y_1}{X_2 - X_1}

n = \dfrac{3.32 - 1.72}{1.1 - 0.3}

= 2

(c)Answer:

= 1.92

(3 s.f.)[5]

\ln y = \ln a + bx

\ln y = bx + \ln a

b

is the gradient and

\ln a

is the

Y

-intercept.

b = \dfrac{Y_2 - Y_1}{X_2 - X_1}

b = \dfrac{4.65 - 2.25}{5 - 2}

= 0.8

2.25 = 0.8(2) + \ln a

\ln a = 0.65

a = e^{0.65}

= 1.92

(3 s.f.)

Practice questions by skill

Sec 4 A-Math Linear law worksheet with answers

In the print window, choose Save as PDF to download it.

Need the formulas first? Read the Linear law notes. Or practise it online with step-by-step help when you get one wrong.

Section A

The variables

x

and

y

are related by the equation

y = ae^{bx}

, where

a

and

b

are constants. When

\ln y

is plotted against

x

, a straight line passing through the points

(2, -2.35)

and

(6, -8.35)

is obtained. Find the value of

a

, correct to 3 significant figures.

Answer[3]

Section B

The table shows experimental values of two variables,

x

and

y

. It is known that

x

and

y

are related by the equation

y = ax^{2} + bx

, where

a

and

b

are constants.

$x$	1	2	3	4	5
$y$	26	56	90	128	170

(a)A straight line graph is to be drawn to find the values of

a

and

b

. State the variables that should be plotted.

Answer (a)[1]

(b)Using the values in the table, the graph of

Y

against

X

is drawn, as shown above, where

Y

and

X

are the variables in part (a). Calculate the value of

Y

for the point plotted when

x = 2

Answer (b)[1]

(c)Use the graph to find the value of

a

Answer (c)[2]

(d)Use the graph to find the value of

b

Answer (d)[1]

(e)Use the graph to estimate the value of

y

when

x = 3.5

Answer (e)[2]

Answers and worked solutions

Question 1

Answer:

= 1.92

(3 s.f.)[3]

\ln y = \ln a + bx

\ln y = bx + \ln a

b

is the gradient and

\ln a

is the

Y

-intercept.

b = \dfrac{Y_2 - Y_1}{X_2 - X_1}

b = \dfrac{-8.35 - (-2.35)}{6 - 2}

= -1.5

-2.35 = -1.5(2) + \ln a

\ln a = 0.65

a = e^{0.65}

= 1.92

(3 s.f.)

Question 2

Answer:

= 0.8

[3]

\ln y = \ln a + bx

\ln y = bx + \ln a

b

is the gradient and

\ln a

is the

Y

-intercept.

b = \dfrac{Y_2 - Y_1}{X_2 - X_1}

b = \dfrac{3.85 - 1.45}{4 - 1}

= 0.8

Question 3

Answer:

b = -4

[3]

Write a ÷ x as a times (1 ÷ x).

y = a\left(\dfrac{1}{x}\right) + b

a

is the gradient and

b

is the

Y

-intercept.

a = \dfrac{Y_2 - Y_1}{X_2 - X_1}

a = \dfrac{1 - (-2)}{5 - 2}

= 1

-2 = 1(2) + b

b = -4

Question 4

(a)Answer:

b = 3

[3]

Multiply both sides by x.

xy = ax^{2} + b

a

is the gradient and

b

is the

Y

-intercept.

a = \dfrac{Y_2 - Y_1}{X_2 - X_1}

a = \dfrac{13 - 7}{5 - 2}

= 2

7 = 2(2) + b

b = 3

(b)Answer:

= -2

[2]

Multiply both sides by √x.

y\sqrt{x} = ax + b

a

is the gradient and

b

is the

Y

-intercept.

a = \dfrac{Y_2 - Y_1}{X_2 - X_1}

a = \dfrac{1 - 5}{3 - 1}

= -2

Question 5

(a)Answer:

= 13.2

(3 s.f.)[3]

\lg y = \lg k + n\lg x

\lg y = n\lg x + \lg k

n

is the gradient and

\lg k

is the

Y

-intercept.

n = \dfrac{Y_2 - Y_1}{X_2 - X_1}

n = \dfrac{3.82 - 1.42}{0.9 - 0.1}

= 3

1.42 = 3(0.1) + \lg k

\lg k = 1.12

k = 10^{1.12}

= 13.2

(3 s.f.)

(b)Answer:

= 7.08

(3 s.f.)[3]

\lg y = \lg k + n\lg x

\lg y = n\lg x + \lg k

n

is the gradient and

\lg k

is the

Y

-intercept.

n = \dfrac{Y_2 - Y_1}{X_2 - X_1}

n = \dfrac{3.1 - 1.6}{1.5 - 0.5}

= 1.5

1.6 = 1.5(0.5) + \lg k

\lg k = 0.85

k = 10^{0.85}

= 7.08

(3 s.f.)

Question 6

(a)Answer:

\dfrac{y}{x}

against

x

[1]

Divide both sides by

x

\dfrac{y}{x} = ax + b

Plot

\dfrac{y}{x}

against

x

: the gradient is

a

and the vertical intercept is

b

(b)Answer:

Y = 28

[1]

When

x = 2

y = 56

\dfrac{y}{x} = \dfrac{56}{2}

\phantom{\dfrac{y}{x}} = 28

(c)Answer:

a = 2

[2]

a

is the gradient of the line.Gradient

= \dfrac{34 - 26}{5 - 1}

= 2

(d)Answer:

b = 24

[1]

b

is the vertical intercept: the line meets the

Y

-axis at

24

b = 24

(e)Answer:

y \approx 108.5

[2]

When

x = 3.5

X = 3.5

From the line,

Y = 31

y = 31 \times 3.5

= 108.5

Question 7

(a)Answer:

b = 3

[4]

Multiply both sides by √x.

y\sqrt{x} = ax + b

a

is the gradient and

b

is the

Y

-intercept.

a = \dfrac{Y_2 - Y_1}{X_2 - X_1}

a = \dfrac{13 - 7}{5 - 2}

= 2

7 = 2(2) + b

b = 3

(b)Answer:

= 2

[4]

\lg y = \lg k + n\lg x

\lg y = n\lg x + \lg k

n

is the gradient and

\lg k

is the

Y

-intercept.

n = \dfrac{Y_2 - Y_1}{X_2 - X_1}

n = \dfrac{3.32 - 1.72}{1.1 - 0.3}

= 2

(c)Answer:

= 1.92

(3 s.f.)[5]

\ln y = \ln a + bx

\ln y = bx + \ln a

b

is the gradient and

\ln a

is the

Y

-intercept.

b = \dfrac{Y_2 - Y_1}{X_2 - X_1}

b = \dfrac{4.65 - 2.25}{5 - 2}

= 0.8

2.25 = 0.8(2) + \ln a

\ln a = 0.65

a = e^{0.65}

= 1.92

(3 s.f.)