WMA11 Jan 2023

• WMA11/01 (Edexcel) IAL P1 January 2023, Q11, Integration, 2nd Derivatives, Equation of Normal

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

Given that

• f''x=4x+1x

• the point P has x coordinate 4 and lies on C

• the tangent to C at P has equation y = 3x + 4

• find an equation of the normal to C at P

(2)

• find f(x), writing your answer in simplest form.

(6)

SOLUTION

a-​​ 

On comparing the equation of tangent at point C (y = 3x + 4) to the general equation of a straight line, the gradient of the tangent is 3.​​ 

Therefore,​​ 

mT×mN=-1 

3×mN=-1

 mN =-13

At point P, the y-coordinate may be found by substituting the value of x as 4 in the equation of a tangent.​​ 

y=3x+4

y=3(4)+4

y=16

Therefore, the point P coordinates are (4, 16).​​ 

Now, using the point slope formula to find the equation of normal at point P.​​ 

y-y1=mx-x1 

y-16= -13x-4

y-16= -13x+43

y=-13x +43+16 

y= -13 x+523 

b-​​ 

We know integration is anti-derivative. So, if we want to find the f(x), we need to integrate the double derivative twice.​​ 

f''x=4x+x-12

f'x=∫4x+x-12dx

f'x=4x22+x12 12+c

f'x=2x2+2x12 +c

At point C, the gradient of tangent is 3 when x = 4. This will help us in getting the value of c.​​ 

f'4=3

242+24 +c=3

32+4+c=3

36+c=3

c= -33

f'x=2x2 +2x12-33

Integrating​​ f’(x)​​ to get​​ f(x).

fx=∫2x2+2x12-33 dx 

fx=2x33+2x32.32-33x+d

fx=23x3+43x32-33x+d

Where, at x=4, the value of y-cordinate was given to us as 16:​​ f4=16.

23  43+43 43-33 4+d=16

-2363 +d=16

d=16+2363 

d= 2843 

fx=23 x3+43x32-33x+2843