Differentiation

Edexcel IAL WMA12/01 /P2/June/Oct 2020/Q3 (Factor & Remainder Theorem, Differentiation)

f(x) = ax3 – x2 + bx + 4 

where a and b are constants.​​ 

When​​ f(x)​​ is divided by​​ (x+4), the remainder is –108​​ 

• Use the remainder theorem to show that​​ 

16a + b = 24 

(2)

Given also that​​ (2x–1)​​ is a factor of f(x),​​ 

• find the value of a and the value of b.​​ 

(3)

• Find f′(x).​​ 

(1)

• Hence find the exact coordinates of the stationary points of the curve with equation y = f(x).​​ 

(4)

SOLUTION

a- Using remainder theorem.

(Consider the flash card above, so on comparing​​ x+4​​ to​​ ax-b,​​ a=1​​ and​​ b=-4, and the remainder is​​ f-4. Using the fact that the remainder is​​ -108, substitute​​ x=-4​​ and solve the equation.)​​ 

x+4=0

x=-4

Where,​​ 

f(x) = ax3 – x2 + bx + 4

f-4=-108

a-43--42+b-4+4=-108

-64a-16-4b+4= -108

-64a-4b= -96

16a+b=24

b- Using the factor theorem.

Using the third point from the above flash card, since​​ (2x-1)​​ is the factor , so the remainder wpould be zero.​​ 

2x-1=0

x=12

Where,​​ 

fx=ax3+bx+4

f12=0

a123-122+b 12+4=0

18a-14+12 b+4=0 

a-2+4b+32=0

a+4b= -30

Solving above equation simultaneously with the equation derived in part a.​​ 

16a+b=24    (1)

a+4b= -30    (2)

Multiplying equation 1 by 4. And, subtracting them.​​ 

    64a+4b=96    a+4b= -30

63a=126

a=2    

Putting value of a in equation 2.​​ 

a+4b= -30

2+4b= -30

4b= -32

b= -8

Hence,​​ the​​ value​​ of​​ a​​ and​​ b is 2​​ &​​ -8.

c- To find​​ f'(x), simply diffrentiating the equation of​​ f(x). ​​ 

f(x) = ax3 – x2 + bx + 4

Now, putting the values of a and b.​​ 

 fx=ax3-x2+bx+4

fx=2x3-x2-8x+4

a=2              b=-8

f'x=6x2-2x-8 

d- To find the exact cordinates of stationary point, lets equate the​​ f'x​​ to zero.​​ 

(At stationary points, the gradient of the curve is equal to zero. Therefore, we will diffrentiate the equation of curve and equate the diffrentiated expression to zero. )

f'x=0

6x2-2x-8=0

2x2-x-4=0

x+13x-4= 0

x=-1                x=43

Now substituting the values of x one by one to find the value of respective y cordinate.​​ 

fx=2x3-x2-8x+4

when x= -1, 

         y=2-13--12-8-1+4= -2-1+8+4

y=9

So, one of the coordinate of the stationary point is​​ -1,    9.

When​​ x=43

 y=2433-432-843+4

y=-10027 

Whereas, the second coordinate of the stationary point is​​ 43, -10027.​​ 

Hence, the stationary point coordinates are​​ -1,    9​​ and​​ 43, -10027.​​ 

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