WMA11 Jan 2022

7. WMA11/01 Edexcel IAL P1 January 2022, Q7 (Graphs and Transformations, Differentiation, &​​ Quadratics)​​ 

Figure 3 shows a sketch of part of the curve with equation y = f(x), where​​ 

f(x) = (x + 4)(x – 2)(2x – 9)

Given that the curve with equation y = f(x) – p passes through the point with coordinates (0, 50)​​ 

(a) find the value of the constant p.​​ 

(2)

Given that the curve​​ with equation y = f(x + q) passes through the origin,​​ 

(b) write down the possible values of the constant q.​​ 

(2)​​ 

(c) Find f′(x).​​ 

(4)​​ 

(d) Hence find the range of values of x for which the gradient of the curve with equation y = f(x) is less than –18​​ 

(3)

SOLUTION

a- ​​ Since the curve​​ y = f(x) – p passes through the point (0, 50), it must satisfy the equation.​​ 

y = f(x) – p

y = (x + 4)(x – 2)(2x – 9)–p

On substituting​​ (0, 50),​​ 

50= (0+4)(0–2)(2×0–9)–p

72-p=50

p=72-50

P=22

b-​​ 

Remember, adding or subtracting a constant inside the function translates the graph horizontally. In case of​​ this part, the graph has translated only horizontally by q units that now it passes through the origin point. This means that one of the solution of y = f(x + q) is zero. So there are 3 possibilities as shown below.​​ 

• fx-4

• fx+2

Hence, the possible values of​​ q are​​ 

q=-4

q=2

c-

fx=2x-9x2+2x-8

fx=2x3+4x2-16x-9x2 18x+72

fx=2x3-5x2-34x+72

Now, differentiating f(x).

f'x=6x2-10x-34

d-​​ 

f'x=6x2-10x-34

6x2-10x-34<-18

6x2-10x-34+18<0

6x2-10x-16<0

3x2-5x-8<0

Finding critical values.​​ 

3x2-5x-8=0

3x-8x+1=0

x=83     x=-1

The yellow portion of line represents the value of x as the inequality sign in was ‘’less than’’.​​ 

Hence, the range of x is​​ 

-1<x<83