Transcript Questions & Answers

Harris Academy
Supported Study
Session 1
Paper 1 Questions and Answers
Non Calculator
Question 1
(Unit 1 LO3 Differentiation)
Find the stationary points of the function
f ( x)  x  12 x  4
3
and determine their nature.
marks (7)
Solution 1
ans:
Max TP at (-2,20)
Min TP at (2,-12)

Know to differentiate
1
f ( x )  ............

Differentiate
2
3 x 2 12

Derivative equal to zero
3
3 x 2 12  0

Factorise and solve
4
3( x  2)( x 2)0 x  2, 2

Find y coordinates
5
(2,20) (2,12)

Nature Table

State nature of TPs
6
x
f ( x )
7
 2  2 

0

0

shape
Max TP at (-2,20)
Min TP at (2,-12)
Question 2
(Unit 2 LO2 Integration)
Evaluate
x  7x  4
dx
2
1
x
2
3
2
marks (4)
Solution 2
2
ans:

3 12
1
x3
7x2
4
 2  2 dx
2
x
x
x
2

Prepare to integrate
2
x

7

4
x
dx
1 
1
2
x
1 
  7x  4x 
2
1
Integrate
2

Substitute
4
1
(
2

14

)

(
3
2
2  7  4)

Answer
4

2
3 12
Question 3
(Unit 2 LO3 Trigonometry)
If x is an acute angle such that
1
tan x 
3
find exact values for
(a)
sin 2 x
(b)
cos 2 x
marks (4,2)
Solution 3a
ans:
3
5
10
1
Construct a right angled
1
triangle and use Pythagoras
X

Find sin x and cos x
2
1
sin x 
10

Formula for sin2x
3 2sin xcos x
Substitute and answer
6
1
3
4 2 10  10  10


3
3
cos x 
10
Solution 3b
ans:

4
5
Formula for cos2x
 1 cos2 x  cos 2 x  sin 2 x
2

Substitute and answer
2
 3   1 

 

 10   10 

2
9 1 8
 
10 10 10
Note
cos2 x  2cos 2 x 1 or 12sin 2 x may be used
Question 4 (Unit 2 LO3 Trigonometry)
Solve
sin 2 x  3 cos x  0 ,
for 0  x  2
marks (4)
Solution 4

x

ans:
3



,
2
,
2
3
3
, 2
Use double angle formula  1
Factorise and form
equations
sin 2 x  2 sin x cos x
2 cos x ( 2 sin x  3 )  0
cos x  0 or sin x 

Solve
cos x  0
 3 x  2 or
3
2

Solve
sin x 
4 x  3 or
2
3
3
2
3
2
Harris Academy
Supported Study
Session 2
Paper 1 Questions and Answers
Non Calculator
Question 5
(Unit 1 LO1 Straight Line)
The line with equation
x  3 y 12
meets the
x and the y axes at the points A and B respectively
(a) Determine the coordinates of A and B
(b) Find the equation of the perpendicular bisector of AB.
marks (2,4)
Solution 5a
ans: A(12,0) B (0,4)

Coordinates of A
1
A(12,0)

Coordinates of B
2
B (0,4)
Solution 5b
ans:
y  3 x 16
Rearrange to y =……
and find gradient
1
y   13 x  4 m   13

Perpendicular gradient
2
m  3

Midpoint of AB
12 0
(
3
2 ,

Find equation of line
4 y  2  3( x  6)

0 4
2
)  ( 6, 2 )
Question 6
(Unit 1 LO2 Functions and graphs)
The diagram below shows part of the graph of
.
y  g( x)
The function has stationary points at
y
as shown
O
2
(0,3) and ( 2,0)
x
-3
Sketch the graph of the related function
y  3  g( x)
marks (3)
Solution 6
y
3



Reflection in x-axis
Move 3 units up
Annotate graph
1
2
3
O
y
2
x
6
(2,3)
O
x
Question 7
(Unit 1 Recurrence Relations)
A doctor administers 40ml of a drug to Mr Sick each week.
Over the same period 80% of the drug in the bloodstream
is removed.
If the level in the bloodstream rises above 55ml the drug
becomes toxic
(a) Write down a recurrence relation to model this situation.
(b) Find a limit and explain what it means in the context of
the question.
marks (2,4)
Solution 7a
ans:
un1  0.2un  40
 1 0.2

For correct multiplier

For recurrence relation  2
un1  0.2un  40
Solution 7b
ans: 50 55 so not toxic

justify limit
1
limit exists as  1 0.2 1

use limit formula
2
b
L
1 a

calculate limit
3
40
L
1 0.2

explanation
4
 50
50 55 so not toxic
Question 8 (Unit 2 LO1 Polynomials)
(a) For what value of k is
x2
a factor of
x3  x 2  2kx  8 ?
(b) Hence fully factorise the expression x 3  x 2  2kx  8
when k takes this value
marks (3,2)
Solution 8a
ans: k  5
-2

Use synthetic division
1
-1
2k
-8
1
-2

Complete division
2

Calculate k
3
1
1
-1
2k
-8
-2
6
-4k-12
-3
2k+6
-4k-20
 4k  20  0
k  5
Solution 8b
ans: ( x  2)( x 1)( x  4)
-2

Find quotient
1
1
1
-1
-10
-8
-2
6
8
-3
-4
0
( x 2  3 x  4)

Factorise fully
2
( x  2)( x 2  3 x  4)
( x  2)( x 1)( x  4)
Harris Academy
Supported Study
Session 3
Paper 1 Questions and Answers
Non Calculator
Question 9
(Unit 2 LO1 Polynomials)
For what value of p does the equation
( p 1) x  px  1  0
2
have equal roots?
marks (4)
Solution 9
ans:
p2

Use the discriminant
 1 b2  4ac  0

Find values of a, b and c
 2 a  p  1 b   p c 1
Substitute and simplify
3

p 2  4( p  1)  0
p2  4 p  4  0

Calculate value of p
4
( p  2)( p  2)  0
p2
Question 10 (Unit 2 LO4 Circle)
A block of wood of thickness t has to pass between
two rollers
The equations of the two circles are
y
x2  y2  4 x  6 y  4  0
and
x  y  22 x  30 y  246  0
2
2
x
Find the maximum possible value of t
t
marks (4)
Solution 10
ans: t  2




Find centre and radius
of small circle
 1 centre  (2,3) r  3
Find centre and radius
of large circle
2 centre  (11,15) r 10
Calculate distance
between centres
3 d  (112)2 (153)2
Calculate distance, t
4 t 15  (10  3) t  2
d 15
Question 11
(Unit 1 LO3 Differentiation)
Part of the graph of
is shown in the diagram
y  x ( x  5 x 6)
2
The tangent to the curve at the point where x = 1
is also shown
y
o
1
x
Find the equation of the tangent at the point where x = 1
marks (5)
Solution 11
ans:
yx  3

knowing to differentiate  1

differentiate

gradient at x = 1
2
dy
dx
 .......
dy
dx
 3 x 2 10 x  6
dy
m

at x 1
3
dx
m  3 10  6   1

y coordinate
4 at x 1, y 1 5  6  2

equation
5 y  2  1( x 1)
Question 12 (Unit 1 LO2 Functions and graphs)
The functions
f ( x)  x  9
2
and
h( x )  3  2 x
are defined on the set of real numbers.
(a) Evaluate
h ( f (3))
(b) Find an expression for
f (h( x ))
(c) For what value(s) of x does
f (h( x ))  f ( x )
marks (1,2,3)
Solution 12a
ans: 3

Evaluate h(f (3))
 1 h( f (3))  h(0)  3
Solution 12b
2
(
3

2
x
)
9
ans:

Apply h
1
f (3  2 x )

Apply f
2
(3  2 x)  9
2
Solution 12c
ans:
x  1 or 3

Equation
 1 (3  2 x)2  9  x 2  9.

Rearrange to ..... = 0
2 3 x 2 12 x  9  0

Factorise and solve
3 3( x  3)( x 1)  0
x  1 or 3