WMA11 January 2020

11. WMA11/01​​ Edexcel IAL P1​​ 2020, Q11 (Integration)

A curve has equation y = f(x), where​​ 

fʹʹx=6x3+x    x>0

The point P(4, –50) lies on the curve.​​ 

Given that fʹ(x) = –4 at P,​​ 

(a) find the​​ equation of the normal at P, writing your answer in the form y = mx + c, where m and c are constants,​​ 

(3)

(b) find f(x).​​ 

(8)

SOLUTION​​ 

a-​​ 

To find the equation of normal use the point-slope formula, where point the normal through is (4, -50) & the gradient can be found in the following way

mT=-4

mT×mN=-1

-4×mN=-1

mN=14

Now, using the gradient of normal and point​​ ​​ (4, -50)​​ ​​ to find out the equation of normal.

y-y1=mNx-x1

y--50=14x-4

y+50=14x-1

y=14x-51

b-​​ 

integrating twice the expression of f’’(x) to find out the expression of f(x). Always remember, that integration is​​ anti-derivative or reverse of differentiation.​​ 

f'x=∫6x32+xdx

f'x= 6x-12-12+x22+c

f'x=-12x-12+12x2+C

f'x=-12x +12x2+c

To find the value of c, using the given information that fʹ(x) = –4 at P​​ (4, –50). So substituting the value pf x as in above equation of​​ fʹ(x).

-124+1242+c= -4

-5+8+c= -4

2+c= -4

c=-6

So the equation of​​ fʹ(x)​​ comes out to be

f'x= -12x-12+12x2-6 

fx=∫f'xdx

fx=∫-12x-12+12x2-6dx

fx=-12x1212+12x33-6x+d

fx= -24x12+16x3-6x+d

To find the value of d, subtituting the point P coordinates​​ (4, –50) as the point lies on the curve and therefore, it satisfies theequation oif curve found above.​​ 

-50= -244+1643-64+d

-50= -48+323=24+d

-50= -72+323+d

22-323=d

 d=343

fx=-24x12+16x3-6x+343