WMA11 Jan 2023

• WMA11/01 (Edexcel) IAL P1 January 2023, Q10 (Graphs & Transformation: Cubic, Differentiation)

Figure 4 shows a sketch of part of the curve C with equation y = f(x), where

fx=3x+20x+62x-3

(a) Use the given information to state the values of x for which

f(x) > 0

(2)

(b) Expand​​ 3x+20x+62x-3, writing your answer as a polynomial in simplest form.

(3)

The straight line / is the tangent to C at the point where C cuts the y-axis.

Given that / cuts C at the point P, as shown in Figure 4,

(c) find, using algebra, the x coordinate of P

(Solutions based on calculator technology are not acceptable.)

(5)

SOLUTION

a-​​ 

fx=0

3x+20x+62x-3=0 

3x+20=0        x+6=0        2x -3=0

x=-203        x=-6         x=1

Since the inequality sign is greater than, so we need to consider only that portion of cubic graph that is above x-axis (yellow line portion).​​ 

Hence, the range of x is​​ 

-623<x<6​​ &​​ x>1.5

​​ b-​​ 

fx=3x+20x+62x-3

=3x+20 2x2-3x+12x -18

=3x+202x2+9x-18=6x3+27x2-54x +40x2+180x-360

f(x)=6x3+67x2+126x-360 

c- ​​ Since the tangent to curve is at the point where the curve intersects y-axis. Thus, we need to find the grdient of ltangent at x=0.​​ 

y=6x3+67 x2+126x-360

dydx=18x2+134 x+126

Gradient at x=0

dydx= 18(0)2+134 (0)+ 126

dydx=  126

The line y-intercept is equal to the y-intercept of curve. So the equation of a line is given as​​ 

y=mx+c

y=126x-360

For coordinate of P, taking use of the given information that curve an dline intersects at P. so, equating both equations to solve them simultaneously.​​ 

6x3+67x2+126x-360=126x-360

6x3+67x2=0

x2  6x+67=0

6x+67=0

x=-676