Math, asked by zainabkamranabbasi, 6 hours ago

How to solve this question?

Attachments:

Answers

Answered by maneezareyan
0

Answer:

which there is not question

Answered by ArunSivaPrakash
1

Given:  The curve y=ax^{3} -4x +1 is parallel to line y= 50x-1 at x=3

To find: (I) Gradient, m at  x=4,

             (II) Tangent at x=-3 is parallel to line y=50x-1

Solution:   x=3

on comparing the line y= 50x-1 with y=mx+c, slope m = 50

m= \frac{dy}{dx}=\frac{d}{dx}(ax^{3}-4x+1 ) = 3ax^{2} -4  -----(1) , since x= 3 given

                                         27a-4=50a=2, (since m=50)

from equation (1) m=3X2Xx^{2}-4 =6x^{2}-4

(I) at x=4, m=6X(4)^{2}-4  =   96-4=92

(II)at x=-3, m= 6x^{2} -4=6(-3)^{2} -4=6X9-4=54-4=50

so slope at x=-3 is equal to the slope of the line y= 50x-1,

hence tanget at x = -3 is parallel to the line y = 50x - 1,

hence proved.

Similar questions