WMA11 June 2022

• WMA11/01​​ Edexcel​​ IAL P1 June 2022, Q10 (Differentiation and Straight Line Graphs)

Figure 5 shows a sketch of the curve C with equation​​ 

y =27x3 +17x2 -52 x + k 

where k is a constant.​​ 

• Find​​ dydx​​ 

(2)

The line l, shown in Figure 5, is the normal to C at the point A with x coordinate​​ -72​​ . Given that l is also a tangent to C at the point B,​​ 

• show that the x coordinate of the point B is a solution of the equation​​ 

12x2​​ + 4x − 33 = 0

(4)

• Hence find the x coordinate of B, justifying your answer.​​ 

(2)

Given that the y intercept of l is −1​​ 

• find the value of k.​​ 

(4)

SOLUTION​​ 

a-

y =27x3 +17x2 -52 x + k

dydx=67x2+27x-52

b- To find the gradient of C at point A, substitute the value of x = -7/2.

dydx =67x2+27x-52

dydx=67-722+27-72-52

dydx=7

Since the line is perpendicular to the tangent of the curve at point A, the gradient of normal line, l will be​​ 

ml=-17

Where at point B, the gradient of curve would be the same as the gradient of line.​​ 

dydxpoint B = -17

67x2+27x-52= -17

7×67x2+27x-52=-1 

6x2+2x-352= -1

On further taking LCM and simplifying, we get

12x2+4x-33=0

c-​​ To find the x-cordinate of point B, solve the quadratic equation.​​ 

2x-36x+11=0

2x-3=0       6x+11=0

1      x=32          x= -116

2    x=-4±16-412-33212

x=32       ,       x=-116

B has a positive x-coordinate since the point B is on the positive x-axis side.​​ 

x=32

d- When x=3/2,​​ 

y =27x3 +17x2 -52 x + k

y=27323+17322-5232+k

y=-6928+k

At B line and curve intersects. So solving simultaneously.​​ 

-17x-1=-6928+k

Putting​​ x=3/2

-1732-1=-6928+k

-1714=-6928+k

k=54