WMA11 October 2021

2. WMA11/01​​ Edexcel​​ IAL P1 Oct 2021,​​ Q2​​ (Differentiation)

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

Solutions relying on calculator technology are not acceptable.

A​​ curve has equation​​ 

y = 3x5 + 4x3 – x + 5

The points P and Q lie on the curve.​​ 

The gradient of the curve at both point P and point Q is 2​​ 

Find the x coordinates of P and Q.​​ 

(5)

SOLUTION

Differentiating the equation of a curve to find the expression of gradient of tangent of​​ a curve and also equating it to 2 as the gradient of the curve at both points is 2.​​ 

dydx=15x4+12x2-1

2=15x4+12x2-1

15x4+12x2-1=2

15x4+12x2-3=0 

Let b=x2, where b2=x4, Therefore, the above expression becomes​​ 

15b2+12b-3=0

5b2+4b-1=0

5b-1b+1=0

b=15        b=-1

Substituting back b=x2

x2=15x2= -1

x= ±15          x=-1

                               x=No Solution

Rationalising the denominator of​​ x= ±15

x=±15x55=±55

Hence, the x-cordonates of P &​​ Q are

x=55,  x=-55