(MPC1) January 2010 - Chatterton Tuition

Mathematics AQA Advanced Subsidiary GCE
Core 1 (MPC1) January 2010
Link to past paper on AQA website:
http://store.aqa.org.uk/qual/gce/pdf/AQA-MPC1-W-QP-JAN10.PDF
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Question 1
a) We set (x + 3) = 0 to get x = -3 and then use the remainder theorem
p(-3) = (-3)3 - (13 x -3) - 12
p(-3) = -27 - - 39 – 12 = -27 + 39 – 12 = 0
as p(-3) = 0 this means that (x + 3) is a factor of p(x)
b) now we know that (x + 3) is a factor we can work out how many times this goes into the cubic
equation
we can do this by inspection, by algebra or by polynomial division
algebraically
the quadratic equation that we will get will be of the form Ax2 + Bx + C
we can easily see that A must be 1 as the leading term is x3
(x + 3)(x2 +Bx + C) = x3 - 13x - 12
Just looking at the x2 terms
(x + 3)(x2 +Bx + C) = x3 - 13x - 12
Bx2 + 3x2 = 0x2
B+3 =0
subtract 3 from both sides
B = -3
just looking at the x terms
(x + 3)(x2 +Bx + C) = x3 - 13x - 12
3Bx +Cx = -13x
3B + C = -13
we know that B = -3
-9 + C = -13
add 9 to both sides
C = -4
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Mathematics AQA Advanced Subsidiary GCE
Core 1 (MPC1) January 2010
check with the units
(x + 3)(x2 +Bx + C) = x3 - 13x - 12
3C = - 12
divide both sides by 3
C = -4 A = 1, B = -3, C = -4
(x + 3)(x2 - 3x - 4)
now we can factorise the quadratic
Two numbers multiply to give -4 and add to give -3. The two numbers are -4 and +1.
(x + 3)(x - 4)(x + 1)
By Polynomial division
We must make sure that each power of x in the polynomial is included (so we need 0x2)
x2 - 3x - 4
x+3
x3 + 0x2 – 13x - 12
x3 + 3x2
-3x2 - 13x
-3x2 –9x
-4x - 12
-4x - 12
0
(x + 3)(x2 - 3x - 4)
now we can factorise the quadratic
Two numbers multiply to give -4 and add to give -3. The two numbers are -4 and +1.
(x + 3)(x - 4)(x + 1)
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Mathematics AQA Advanced Subsidiary GCE
Core 1 (MPC1) January 2010
Question 2
a) the gradient of a line is calculated by dividing the difference in the y values by the difference in
the x values
gradient = i) gradient AB = = ! = 2
ii) we will know that angle ABC is a right angle if lines AB and BC are perpendicular to each other
"
!
gradient BC = = =
!
if the product of the two gradients is -1 then the lines are perpendicular
this is the same as saying the gradients are the negative reciprocal of each other
2x
=
!
-1 so angle ABC is a right angle
b) i) to find the midpoint of two points you add the two x values and divide by 2 and add the two y
values and divide by 2
# # "
,
)
!
!
M=(
= (0, 6)
ii) length AB2 = (x2 – x1)2 + (y2 – y1)2
AB2 = (3 – 1)2 + (7 – 3)2 = 22 + 42 = 4 + 16 = 20, length AB = √20
BC2 = (3 - -1)2 + (7 – 9)2 = 42 + (-2)2 = 16 + 4 = 20, length BC = √20
length AB = length BC
iii) the line of symmetry through triangle ABC will be the line that goes through B and M
we have the two coordinates for B and M (3,7) and (0, 6) so we can calculate the gradient
'
gradient BM = ( = we have the gradient and a point that the line goes through so we can use
y – y1 = m(x – x1) to get the equation
y – 6 = (x – 0)
add 6 to both sides
y= +6
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Mathematics AQA Advanced Subsidiary GCE
Core 1 (MPC1) January 2010
Question 3
to differentiate we multiply by the power and then reduce the power by 1
)
a) i) * = +t3 – 4t + 4 = !t3 – 4t + 4
, )
ii) * , = !t2 – 4
)
b) stationary values are minimum or maximum, they happen when * = 0
substituting t = 2 into
)
*
)
*
we should get 0
!
= -2)3 – (4 x 2) + 4 = 4 – 8 + 4 = 0
, )
To determine whether the point is a maximum or a minimum we use ,
*
If this comes out positive then the point is a minimum
If this comes out negative then the point is a maximum
When t = 2
, )
* ,
= ! -2)2 – 4 = 6 – 4 = 2 which is positive
The stationary point is a minimum
c) i) we want
)
*
!
)
*
when t = 1
!
= (1)3 – (4 x 1) + 4 = – 4 + 4 =
!
)
ii) * is positive which means that y is increasing
)
( * is a measure of the gradient of the slope and if this is positive then y (the depth) is increasing)
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Mathematics AQA Advanced Subsidiary GCE
Core 1 (MPC1) January 2010
Question 4
i) √50 = √25 x 2 = √25 x √2 = 5√2
and
√18 = √9 x 2 = √9 x √2 = 3√2
and
√8 = √4 x 2 = √4 x √2 = 2√2
so we have
√4(#√+
√+
=
4√!#√! +√!
=
=
!√!
!√!
4
ii) we need to rationalise the denominator, so multiply the numerator and the
denominator by (2√7- 5)
-!√6
-!√#46
-!√46
-!√ 6-!√ 46
=
√ 4 6 7!√# 48-!√ 46
x -!
when we expand the numerator we get
-2√7 9 16-2√7 9 56 = (2 x 2 x √7 x √7) - 2√7 + 5 - 10√7 = 28 - 2√7 + 5 - 10√7 = 33 - 12√7
expanding the denominator we get
-2√7 : 56-2√7 9 56 = (2 x 2 x √7 x √7) + 10√7 – 25 - 10√7 = 28 – 25 = 3
so we have
!√ - √6
=
=
11 - 4√7
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Mathematics AQA Advanced Subsidiary GCE
Core 1 (MPC1) January 2010
Question 5
a) we first need to expand the brackets
(x – 5)(x – 3) + 2 = x2 – 5x – 3x + 15 + 2 = x2 - 8x + 17
now we can complete the square
(x – 4)2 – 42 + 17 = (x – 4)2 + 1
p = 4, q = 1
b) i) the minimum point will be when x = 4 and when y = 1 (4, 1)
this is because the smallest (x – 4)2 can ever be is 0 (since it is a square of something)
and (x – 4)2 is 0 when x = 4
the graph crosses the y axis when x = 0
this happens when y = 02 – (8 x 0) + 17 = 17
crosses y axis at (0, 17)
ii) the tangent to the curve at the minimum point is just the straight line y = 1
c) we know from part a) that y = (x – 4)2 + 1
the changes from y = x2 to (x – 4)2 + 1 are two translations (since we are adding or subtracting
values)
one of the changes is inside the brackets so affects the x values. When the x values are affected the
change is in the opposite direction to what you expect. This is a translation of 4 units in the positive
x direction.
The other change is outside of the brackets so affects the y values. This is a translation of 1 unit in
the positive y direction
Overall we have vector translation of 78
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Mathematics AQA Advanced Subsidiary GCE
Core 1 (MPC1) January 2010
Question 6
a) i) to find the gradient we need to differentiate
to differentiate we multiply by the power and then reduce the power by 1
)
= 24x – 19 – 6x2
)
To find the gradient at the point (2, -6) substitute x = 2 into Gradient = (24 x 2) – 19 – (6 x 22) = 48 – 19 – 24 = 5
ii) the normal is perpendicular to the tangent
the gradient of the tangent at A is 5
the gradient of the normal at A will be the negative reciprocal of 5 which is 4
We now have the gradient and a point (2, -6) that the line goes through so we can work out the
equation of the line using the formula y – y1 = m(x – x1)
y - -6 = - 4(x – 2)
multiply by 5
5(y + 6) = -(x – 2)
expand the brackets
5y + 30 = -x + 2
add x to both sides
x + 5y + 30 = 2
subtract 2 from both sides
x + 5y + 28 = 0
b) i) to integrate you increase the power by 1 and then divide by the new power
! <
;
9
" ,
!
9
!=
!
> = (4(23) –
(
" ? !,
!
–
!=
)–
!
(0) = 32 – 38 – 8 = -14
ii) what we have done in part i) is work out the whole area enclosed by the curve and the x axis
between the lines x = 0 and x = 2
the area came out as negative because it is below the x axis
we need to work out the area of the triangle and subtract this from 14
area of triangle = ½ x base x height = ½ x 2 x 6 = 6
shaded area = 14 – 6 = 8 square units
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Mathematics AQA Advanced Subsidiary GCE
Core 1 (MPC1) January 2010
Question 7
a) we need to complete the square
then when the equation is in the form
(x – x1)2 + (y – y1)2 = r2 we will be able to work out the centre (x1, y1) and radius r
x2 – 4x + y2 + 12y + 15 = 0
complete the square on both x and y
(x – 2)2 – 22 + (y + 6)2 – 62 + 15 = 0
group the units
(x – 2)2 + (y + 6)2 -25 = 0
add 25 to both sides
(x – 2)2 + (y + 6)2 = 25 (= 52)
i) centre will be (2, -6)
ii) radius will be 5
b) the centre is below the x axis (as the y coordinate of centre is -6)
the radius is 5
the highest the circle will reach will be -6 + 5 = -1 and as this is negative it is still below the x axis
c)i) length PC2 = (x2 – x1)2 + (y2 – y1)2
PC2 = (5 – 2)2 + (k – -6)2 = 32 + (k + 6)2 = 9 + (k + 6)(k + 6)
expand the brackets
PC2 = 9 + k2 + 6k + 6k + 36
group terms
PC2 = k2 + 12k + 45
ii) we know that P is outside of the circle so PC must be more than the radius which is 5
PC @ 5
so also PC2 @ 25
k2 + 12k + 45 @ 25
subtract 25 from both sides
k2 + 12k + 20 @ 0
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Mathematics AQA Advanced Subsidiary GCE
Core 1 (MPC1) January 2010
iii) we can factorise if we can find two numbers that multiply to make 20 and add to make 12
these two numbers are +2 and +10
(k + 2)(k + 10) @ 0
If we set this equal to 0 then the critical values will be k = -2 and k = -10
Now we need to decide if we want k to be between these two values or either side of them
Try k = -3 (which is in between), when k = -3: we get 9 – 36 + 20 = -7 which is less than 0 (not what
we want) so we don’t want k to be between the two values
Alternatively if we know the way that a quadratic graph (with a leading positive k2 term lies (like a
big A) then we can also see that this curve is above 0 when k is either side of the two critical values
k B -10, k @ -2
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