WMA11 Jan 2022

6. WMA11/01 Edexcel​​ IAL P1 January 2022,​​ Q6​​ (Differention:​​ Gradients,​​ Tangents, & Normals)

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

Given that​​ 

• f'x=x+32xx

• the point P(4, 20) lies on C​​ 

(a)​​ 

• find the value of the gradient at P

• Hence find the equation of the tangent to C at P, giving your answer in the form ax + by + c = 0 where a, b and c are integers to be found.​​ 

(4)

(b) Find​​ f(x), simplifying your answer.​​ 

(7)

SOLUTION

i-​​ Subtituting the value of x=4 in the expression of f’(x), to find the gradient of the curve at point P. ​​ 

f'4=4+3244=498

f'4=498

ii-​​ To find the equation of tangent, we will use point-slope formula, where point is P​​ (4, 20)​​ ​​ and slope or​​ gradient of equation is 49/8.

y-y1=mx-x1

y-20=498x-4

8y-160=49x-4

8y-160=49x-196

0=49x-8y-36

49x-8y-36=0

b-​​ To find f(x), we need to integrate f’(x), as integration is anti-derivative.  ​​​​ 

y=∫f'xdx

∫x+32xxdx=∫x2+6x+9x32dx

y=∫x2x32+6xx32+9x32dx

y=∫x12+6x-12+9x-32dx

y=x3232+6x1212 +9x-12-12+c 

y=23x32 +12x12-18x-12+c

To find the value of c, substituting the value of point P as it lies on the curve so it must satisfy the equation.​​ 

y=23(4)32 +12(4)12-18(4)-12+c

20=2343+124-184+c 

20=238+24-9+c

20=163+15+c

20=163+15+c

c= 5-163

c=153-163

 c=-13

So, the​​ expression of f(x) will be

fx=23x32+12x12-18x-12-13