Download C1 2009 June solutions

C1 January 09 Worked answer
1.
2 −3 =
2.
State the value of each of the following.
[1] (ii) 90
(i) 2−3
1 1
=
23 8
90 = 1
Find the equation of the line passing through (−1, −9) and (3, 11).
Give your answer in the form y = mx + c.
Gradient
y − y1 11 − (−9) 20
m= 2
=
=
=5
x 2 − x1
3 − (−1)
4
3.
[1]
Equation of a straight line is:
y − y1 = m( x − x1 )
y − 11 = 5( x − 3)
y − 11 = 5 x − 15
y = 5 x − 15 + 11
y = 5x − 4
Solve the inequality 7 − x < 5x − 2.
Add x to both sides.
Add 2 to both sides.
[3]
7 − x < 5x − 2
+x +x
7 < 6x − 2
+2
+2
9 < 6x
6x > 9
4.
x>
9
6
x>
3
2
You are given that f(x) = x4 + ax − 6 and that x − 2 is a factor of f(x). Find the value of a.
x – 2 is a factor of f(x) ∴ f(2) = 0
5.
[3]
f(2) = 24 + a × 2 – 6
= 16 + 2a – 6
= 10 + 2a
10 + 2a = 0
∴a=–5
(i) Find the coefficient of x3 in the expansion of (x2 − 3)(x3 + 7x + 1).
(ii) Find the coefficient of x2 in the binomial expansion of (1 + 2x)7.
(x2 − 3)(x3 + 7x + 1)
= x2 ×(x3 + 7x + 1) − 3 × (x3 + 7x + 1)
= x5 + 7x3 + x2 − 3x3 – 21x − 3
= x5 + 4x3 + x2 – 21x − 3
∴the coefficient of x3 is 4
[3]
The term in x2 is 7C5 × (1) 5 × (2 x) 2
21 × 1 × 4x5
84x5
The coefficient of x2 is 84.
[2]
[3]
6.
Solve the equation
3x + 1
= 4.
2x
[3]
3x + 1
=4
2x
3x + 1 = 8 x Takeaway 3x from both sides
– 3x – 3x
1
∴ x=
1 = 5x
5
7.
(i) Express 125 5 in the form 5k.
[2]
(ii) Simplify ( 4a b
[2]
)
3 5 2
( 4a b )
3 5 2
125 5
1
42 × a 3×2 × b5×2
= 53 × 5 2
3+
=5
1
2
= 53.5
16a 6b10
8. Find the range of values of k for which the equation 2x2 + kx + 18 = 0 does not have real roots. [4]
No real roots means the discriminant b2 – 4ac is
negative.
k 2 − 4 × 2 × 18 < 0
k 2 − 144 < 0
k 2 − 122 < 0
(k − 12)(k + 12) < 0
−12 < k < 12
9. Rearrange y + 5 = x(y + 2) to make y the subject of the formula.
y (1 − x) = 2 x − 5
y + 5 = x( y + 2)
2x − 5
y + 5 = xy + 2 x
y=
1− x
y − xy = 2 x − 5
75 − 48 in the form a 3 .
14
in the form b + c d .
(ii) Express
3− 2
10. (i) Express
75 + 48
25 × 3 + 16 × 3
5 3+4 3
9 3
[4]
[2]
[3]
=
14
3+ 2
×
3− 2 3+ 2
14(3 + 2)
9−2
14(3 + 2)
=
7
=
= 2(3 + 2) = 6 + 2 2
11.
Figure shows the points A and B, which have
coordinates (−1, 0) and (11, 4) respectively.
(i) Show that the equation of the circle with
AB as diameter may be written as (x − 5)2 +
(y − 2)2 = 40.
4
(ii) Find the coordinates of the points of
intersection of this circle with the y-axis. Give
your answer in the form a ± b .
4
(iii) Find the equation of the tangent to the
circle at B. Hence find the coordinates of the
points of intersection of this tangent with the
axes.
6
AB is diameter
 −1 + 11 0 + 4 
∴ centre 
,
 = (5 , 2)
2 
 2
Radius
equation of the circle is
(5 − −1)2 + (2 − 0) 2 = 62 + 22 = 40
intersection of this circle with the y-axis
means x = 0
(0 − 5)2 + ( y − 2)2 = 40
( x − 5)2 + ( y − 2) 2 =
(
40
)
2
( x − 5)2 + ( y − 2) 2 = 40
( y − 2) 2 = 40 − 25
( y − 2) 2 = 15
4−0
4 1
=
=
11 − (−1) 12 3
Gradient of the tangent at B = – 3
Gradient of AB =
y − 2 = ± 15
y = 2 ± 15
y – 4 = – 3(x – 11)
y – 4 = – 3x + 33
3x + y =37
12. (i) Find algebraically the coordinates of the points of intersection of the curve y = 3x2 + 6x + 10
and the line y = 2 − 4x.
[5]
(ii) Write 3x2 + 6x + 10 in the form a(x + b)2 + c.
[4]
2
(iii) Hence or otherwise, show that the graph of y = 3x + 6x + 10 is always above the x-axis. [2]
Part-ii
Part-i
3x2 + 6x + 10 = 3(x2 +2x +10/3)
y = 3x2 + 6x + 10 and y = 2 − 4x
= 3{(x +1)2 – 1 +10/3}
∴ 3x2 + 6x + 10 = 2 − 4x
= 3{(x +1)2 +7/3}
3x2 + 10x + 8 = 0
= 3(x +1)2 + 7
(3x + 4)(x + 2)=0
x = – 2 or – 4/3
y = 10 or 22/3
Part-iii
2
Minimum point is (-1 , 7)
2
3x + 6x + 10 = 3(x +1) + 7
(x +1)2 is always positive and other term is
positive as well
∴the graph of y = 3x2 + 6x + 10 is always above
the x-axis
13. Answer part (i) of this question on the insert provided. The insert shows the graph of y =
(i) On the insert, on the same axes, plot the graph of y = x2 − 5x + 5 for 0 ≤ x ≤ 5 .
1
x
[4]
1
(ii) Show algebraically that the x-coordinates of the points of intersection of the curves y =
x
2
3
2
and y = x − 5x + 5 satisfy the equation x − 5x + 5x − 1 = 0.
[2]
(iii) Given that x = 1 at one of the points of intersection of the curves, factorise x3 − 5x2 + 5x − 1
into a linear and a quadratic factor.
Show that only one of the three roots of x3 − 5x2 + 5x − 1 = 0 is rational.
[5]
Part -i
Part-ii
1
= x 2 − 5x + 5
x
1 = x × ( x 2 − 5 x + 5)
To solve x2 – 4x + 1 = 0
1 = x3 − 5 x 2 + 5 x
x=
x3 − 5 x 2 + 5 x − 1 = 0
Part-iii
x = 1 at one of the points of intersection of the
curves ∴ (x – 1) is a factor of x3 − 5x2 + 5x − 1.
x − 4x +1
x − 1 x3 − 5x 2 + 5x − 1
2
x3 − x 2
− 4x2 + 5x − 1
− 4x2 + 4x
x −1
x −1
−( −4) ± 42 − 4 ×1×1
2 ×1
4 ± 12
4±2 3
=
=
2
2
= 2± 3
∴Only rational root is 1; other roots are (real
irrational
and)