WMA11 October 2022

5.​​ WMA11/01 Edexcel IAL P1 October 2022, Q5 (Integration & Differentaition:​​ Equation of Normal)

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

The curve C has equation y = f(x), x > 0​​ 

Given that​​ 

• f'x=12x +x3 – 4 

• the point P(9, 8) lies on C​​ 

(a) find, in simplest form, f(x)​​ 

(5)

The line l is the normal to C at P​​ 

(b) Find the coordinates of the point at which l crosses the y‑axis.​​ 

(4)

SOLUTION

a-​​ 

To find the f(x) from f’(x) expression, we need to integrate the f’(x).​​ 

f'x=12x-12 +13x-4

fx=∫(12x-12 +13x-4).dx

When integrating a polynomial function, you can integrate the terms one at a time.

fx=∫12x-12 .dx+∫13x.dx-∫4.dx

fx=12x1212+13x22-4x+c

fx=24x12+16x2-4x+c

Since​​ P(9, 8) lies on C, it must satisfy the equation of C. hence, putting the values of x and y coordinates in f(x) expression.​​ 

8=24 912+1692-49+c

8=243+816-36+c

8=992+c

c=8-992

c= -832

fx=24x12+16x2-4x-832

b-​​ 

Since the line is normal to C at P, so the slope/gradient of the tangent to the curve at pont P is​​ 

​​ 

mT=f'9=129+93-4

mT=123+3-4

mT=3

So, the gradient/slope of the normal is​​ 

mT×mN=-1

mN=-1mT

mN=-13

Now, since we have slope of normal and point through which it passes that is​​ point P(9, 8), let us use point-slope formula to get the equation of normal.​​ 

y-y1=mN x-x1 

y-8= -13x-9

y-8= -13 -9

y-8=-13-9

y-8=3

y=11

Hence, the coordinates of the point at which​​ I​​ crosses the y‑axis​​ (0, 11).​​