WMA11 June 2021

6.​​ WMA11/01​​ Edexcel​​ IAL P1 June 2021 Q6​​ (Integration​​ & Differentiation:​​ Equations of Tangents)

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

Given that​​ 

• C passes through the point P(8, 2)​​ 

• fʹx=323x2+3-2 x3 

(a) find the equation of the tangent to C at P. Write your answer in the form y = mx + c, where m and c are constants to be found.​​ 

(3)

(b) Find, in simplest form, f(x).​​ 

(5)

SOLUTION​​ 

a-​​ 

To find the equation of the tangent, we will first find slope/gradient of C at point P by substituting the x-coordinate in the f’(x) expression and then use it alongwith point P coordinates to get the equation of tangent throght point-slope formula.

So let’s first find the gradient of tangent of curve.​​ 

mT=f'8=32382+3-238

mT=323 x 64+3-2 x 2 

mT=16+3-4

mT=16-1

mT= -56

Now, use point slope formula, where point P is (8, 2) and mT=-5/6.

y=mx+c

y-y1=mx-x1

y-2=-56 x-8

y-2= -56x+203

y=-56x+263

b-

Remember,​​ 

fx=∫f'x.dx

So, we need to integrate the f’(x) expression to get f(x).

f'x=323x-2+3-2x13

∫f'x=∫323x-2+3-2x13.dx 

fx=323×-1x-2+1+3x-243x13+1+C

fx=32x-1-3+3x-2x4343+C

fx= -323x +3x-32 x43+c

Since point P(8, 2) lies on the curve, it must satisfy the​​ equation.​​ 

2=-3238+38-32  843+c

2=-43+24-32 16+c

2=-43+24-24+c

C=2+43

C=63+43

C=103

Putting the value of C in the expression of f(x).

fx= -323x+3x  -32x43+103