x 2  x  2   x  2  x  1  0

3  12



2

 3  12  2 36  27

2

x  3 or x  1 c)

dV  27  2 x  4  54 x  108 dx 2 d 2V d ) i) 2  x  3  27  2  3  4   27  2  54  0 dx d 2V  x  1  27  2 1  4   27  2  54  0 dx2 ii) There is a maximum for x  1 iii) For x  1, V  81  54  9  36

b) i)

dV  9  3x2  54  2 x  81  27 x 2  108x  81 dx dV  27  x2  4 x  3 dx dV ii)  0 when x 2  4 x  3  0 dx  x  3 x  1  0

V  9x3  54 x2  81x

V  x  81  9x 2  54x 

a) y  9  3x so V  xy 2  x(9  3x)2

V  xy 2

Question 3: 3x  y  9 x  0, y  0



y 12 2 3   2 x 3 3

c) ( x  y)2 

b)

a ) xy  3  12  36  6

x  3 and y  12

Question 2:

x  2 or x  1 and y  3x  1 y  5 or y  4 The line and the curve cross at (2,5) and (  1,  4)

c)

x2  x  2  0

 y  3x  1 b)Solve simultaneously  by identification 2  y  x  2x  3  y   x 2  2 x  3  3x  1

1 a) The line crosses the y-axis at (0,  1) and the x-axis at ( ,0) 3 The curve crosses the y-axis at (0,  3) and the x-axis at (  3,0) and (1,0)

CurveC : y  ( x  3)( x  1)  x 2  2 x  3

Question 1: Line L : y  3x  1

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AQA – Core 1 - Jan 2007 – Answers

2

2

2

2

translation

1

triangle AHC ) 1 41 25   3  15  41 52  22 12  18 190 2

 area beneath the curve   (area of

4

1  1  32    b) i)  16  x dx  16 x  x 5   16     32   2 5  2  5  5   33  48   48  6 53  41 25 5 ii ) Call the point H (1, 0) the area of the shaded region is

1

y  16  x A(2, 0), B(2, 0) and C (1,15) 15  0 a) Gradient of AC  mAC  5 1 2 Equation of AC : y  0  5( x  2) y  5 x  10

4

Question 5:

 3   by the translation vector  2  7  4 

y  x 2 is mapped into y  x 2  3 x  4

translation 1.5 unitsin x  dir

2

7 unitsin y  dir 3 f  3 3 7  4 c) x  x     x      x    2 2 2 4  

2

3 3 7 7   b) For all x,  x    0 so  x     2 2 4 4   7 The minimum value is 4

2

3 3 3 7   a) x 2  3x  4   x       4   x    2 2 2 4  

Question 4:

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2

c) i ) By substituting x by 0, we have p(0)  12 The curve passes through the point (0,  12) ii) The curve crosses the x  axis at (3,0) and it is tangent to it at (  2,0).

  x  2   x  3

2

  x  2  x  3 x  2 

ii) x 3  x 2  8 x  12   x  2   x 2  x  6 

 8  4  16  12  0  2 is a root of p, so ( x  2) is a factor of p.

b) i) p (2)  (2)3  (2)2  8   2   12

p(1)  1  1  8  12  18

a) The remainder of the division by  x  1 is p (1)

p( x)  x  x  8 x  12

3

Question 6:

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2

 x  8   y  13

2

 132

Component title Core 1 – Unit PC1

Max mark 75

3 or k  3 4

(4)

(4k  3)(k  3)  0 for k  

b) 4k 2  9k  9  0 (4k  3)(k  3)  0 3 critical values  and 3 4

4k 2  9k  9  0

16k 2  36k  36  0

a) (4k )2  4  (k  1)  9  0

(k  1) x  4kx  9  0 has real roots which means that the discriminant  0.

2

Question 8:

GRADE BOUNDARIES A B 59 51

CI  144  12

 132  52  169  25  144

Using pythagoras' theorem: CI2  CP 2  PI 2

The equation of the tangent is : y  1  

5  x  3 12 12 y  12  5 x  15 5 x  12 y  27 iii ) Call I the midpoint of PQ. The triangle PIC is a right-angled triangle and the shortest distance from C to the chord PQ is the distance CI.

i ) gradient  mPC 

13  1 12  83 5 ii ) The tangent at P is perpendicular to the radius PC 1 5 the gradient of the tangent is   mPC 12

b) P(3,1) lies on the circle

a) Equation of S:

Circle S has centre C (8,13) and touches the x-axis The radius is 13 ( y-coordinate of C )

Question 7:

C 43

D 35

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E 28