Factor and Remainder Theorems

Edexcel IAL WMA12/01/P2/June/Oct 2020/Q06 (Long Division, Integration)

Figure 1 shows a sketch of part of the curves C1​​ and C2​​ with equations​​ 

C1 : y = x3 – 6x + 9 x≥ 0 

C2 : y = –2x2 + 7x – 1 x≥ 0 

The curves C1​​ and C2​​ intersect at the points A and B as shown in Figure 1.​​ 

The point A has coordinates (1, 4).​​ 

Using algebra and showing all steps of your working,​​ 

• find the coordinates of the point B.​​ 

(4)

The finite region R, shown shaded in Figure 1, is bounded by C1​​ and C2​​ 

• Use algebraic integration to find the exact area of R.​​ 

(5)

SOLUTION

a- ​​ Since point B is the point of intersection of both curves, solving the equations of the curves simultaneously.​​ 

y1 = x3 – 6x + 9

y2= –2x2 + 7x – 1

y1 =y2

x3-6x+9= -2x2+7x-1

x3+2x2-13x+10=0

Solving the cubic equation via long division method. As it is given that the point A has x-cordinate 1; this means​​ (x-1)​​ is a factor of​​ x3+2x2-13x+10=0.

      x2+3x-10  x-1⟌x3+2x2-13x+10-x3-x2                              3x2-13x                          -3x2-3x                                                  -10x+10                                                   -10x+10-            -             -

x-1x2+3x-10=0

x-1x+5x-2=0

x=1,        x=-5,           x=2

Substituting​​ x=2​​ because 1 is already been told to be the x-cordinate of A whereas​​ x=-5​​ lies in negative x-axis. Thus,​​ x=2​​ is left to be the x-coordinate of point B. ​​ 

y=x3-6x+9=23

y=62+9

y=5 

Hence, ​​ the coordinate of point B is​​ 2, 5. 

b- To find the area of the finite region R, shown shaded in Figure 1, which is bounded by C1​​ and C2​​ , use intaegration method.​​ 

With the help of integration of the​​ C2​​ within the limits 1 and 2, we will get the entire area bounded between the curve and x-axis. Now, to extract the area of the shaded region as shown in the question, we have to subtract the area of bounded between​​ C1​​ and x-axis (between the same limit)​​ ​​ from the answer/area we get from the integration of​​ C2.​​ 

Area of R=Area bounded between C2 and x-axis-Area bounded between C1 and x-axis

Area of R=∫12y2.dx-∫12y1.dx

Area of R=∫12(–2x2 + 7x – 1).dx-∫12(–2x2 + 7x – 1).dx

Area of R=∫12(–2x2 + 7x – 1).dx-∫12(–2x2 + 7x – 1).dx

Area of R=∫12-2x2+7x-1-x3-6x+9dx

Area of R=∫12-x3-2x2+13x-10dx

Applying the limits.​​ 

Area of R=-x44-2x33+13x22-10x12

Area of R= -164-283+1342-102--14-23+132-10

Area of R=-4-163+36-20+14+23-132+10

Area of R=12-143-254=1312 

Area of R=1112 Unit2