Jan 2010 Edexcel Maths C1 solutions
1)
dy = 4x3 + 1x-2/3
dx
3
2)
a) 21 + 35 - 75 – 5 = 16 -45 = 4 (4 - 5)
b) Multiply the numerator and the denominator by (3 - 5). Remember that (x+y) *
(x-y) = x2 – y2 The said multiplication gives
4 (4 - 5)/ (9 – 1) = (4 - 5)//2
3)
a) In the line y = mx + c, the gradient is m. So rearranging the line gives
5y = -3x + 2 or y = -3x/5 + 2/5
Therefore, the gradients of l1 is –3/5
b) If two lines are perpendicular to each other than the product of their gradients
equal –1.
Therefore, the gradient of the line l2 is 5/3, giving the equation of l2 as y = 5x/3 + c
As the point (3,1) lies on the line l2, this co-ordinate must satisfy the equation of the
line. Substituting the co-ordinate gives
1 = 5 + c or c= -4
The equation of l2 becomes y = 5x/3 – 4
4)
The key to doing this question is to understand that xx is same as x3/2
y = 5x1/2 / (1/2) + x5/2 /(5/2) + c = 10x1/2 + 2 x5/2 /5 + c
Substituting the given values of x and y, we get
35 = 5 (4)1/2 * 2 + 2/5 (4)5/2 + c
35 = 20 + (2 * 32 /5 ) + c
35 = 20 + 64/5 + c
c = 11/5
Therefore, y = 10x + 2 x5/2 /5 + 11/5
5)
From equation 1, we have y = 3x – 2
Substituting, this value of y in equation 2 gives
(3x – 2)2 – x – 6x2 = 0
9x2 – 12x + 4 - x – 6x2 = 0
3x2 – 13x + 4 = 0
(3x – 1) (x – 4) = 0
Giving x = 1/3 or x = 4
For x = 1/3, by substituting in equation 1 we have
y –1+2 = 0 giving y = -1
and by substituting x = 4 we get
y – 12 + 2 = 0 giving y = 10
Therefore, the co-ordinates of the points of intersections are (1/3, -1) and (4, 10)
6)
a) We can re-write the equation as:
y = (x2 –5x -24) / x
Remember that (a + b + c)/d is same as a/d + b/d + c/d
Therefore, y = x – 5 - 24x-1
dy/dx = 1 + 24x--2 or dy/dx = (x2 + 24)/x2
b) At the point x=2, the dy/dx = 1 + 24/4 = 7
and y = (4 – 10 – 24)/2 = -15
Using y = mx = c, the equation of the tangent is y = 7x + c
Substituting, the co-ordinate (2, -15) in the above equation to find c gives
-15 = 14 + c
c = -29
Therefore, the equation of the tangent at the point (2, -15) is
y = 7x – 29
7)
a) The nth term of an arithmetic sequence is given by a + (n –1)d , where a is the first
term and d is the difference. In our case a=150, d=10 and n=10. By substitution,
we get the 10th term as 150 + 10(10 -1) =240
Therefore, in year 10 Jill gave £240
b) For an arithmetic sequence, the series (Sn) is (a+l)n/2, where a is the first term, l is
the last term and n is the number of terms. The last term in our case is 150 + 10(20
- 1) = 340
Therefore, Sn = (150 + 340) * 20 /2 = 4900
c) Sn for Kevin over twenty is 4900 * 2 = 9800, a = A and d=30
The last term, l, for Kevin is A + 30 (20 –1) = A + 570
Using Sn is (a+l)n/2, we get
9800 = (A + A + 570) * 20 /2
Simplifying, 980 = 2A + 570
2A = 980 – 570 = 410
Hence A = 205
8)
a)
The net effect of +2 on the curve y = f(x) is to translate it upwards by 2. The curve does not change
its shape at all but just shifts upwards, giving
b) The net effect of x4 is to expand the curve four times in y direction for each x, giving
c)
9)
The net effect of (x+1) is to shift the curve f(x) to the left by 1, ie the value which y had when x=0
now happens when x=-1
a) The equation can be written as x(x2-4)
The factors of (x2 – y2) are (x + y) and (x - y)., therefore the factor are x(x + 2)(x – 2).
b) From the factors it is clear that the curve C intercepts the x-axis at –2, 0 and 2. For a negative value
of x the x3 is also going to be negative, thus indicating that y is negative at the start of the curve. The
curve C would look like
(note: my graphic tool isn’t very good but the co-ordinates and the general shape are correct)
c)
For x= -1, y= -1+4 = 3 Therefore, the co-ordinate of point A is (-1, 3)
For x=3, y = 27 – 12 = 15 Therefore, the co-ordinate of point B is (3, 15)
The gradient of AB is (y2 –y1)/(x2 – x1) = (15 –3)/(3 +1) = 12/4 = 3
The equation of a straight line is y = mx + c, giving the equation as y = 3x + c
The point (-1,3) lies on the line so we can calculate for c by substitution
3 = -3 + c
c=6
The equation of line passing through points A and B is
y = 3x + 6
d) Applying Pythagoras’ Theorem we get
AB2 = (x2 – x1)2 + (y2 – y1)2
Substituting our values for co-ordinates of A and B, we get
AB2 = 42 + 122 = 16 + 144 = 160
AB = 160 = 410
Therefore k =4
10)
a)
x2 + 4kx = (x + 2k)2 – 4k2
Therefore, the equation becomes
(x + 2k)2 – 4k2 +3 + 11k
giving p = 2k and q = – 4k2 +3 + 11k
b) If a quadratic equation, ax2 + bx + c, doesn’t have real roots then the discriminant (b 2 – 4ac) has to
be less than zero. Applying this to our equation gives
(4k)2 – 4(3 + 11k) < 0
16k2 – 4(3 + 11k) < 0
4k2 –11k –3 < 0
(4k + 1) (k – 3) < 0
Therefore, the range satisfying the above inequality is
-1/4 < k < 3
c) We know from part b that the graph is not going to intercept the x axis for k=1 and by substituting in
the equation found in part a we get the equation of the curve as:
y = (x + 2)2 + 10
For x=0, the y=14
At the minimum point the dy/dx = 0. Differentiating we get
dy/dx= 2x + 4
2x + 4 = 0, at the minimum point,.
Therefore x = -2 and y = 10