The curve y x = a x 3 + b x 2 + c x + 5 touches the x -axis at the point P - 2 , 0 and cuts the y -axis at the point Q , where y ' is equal to 3 . Then the local maximum value of y x is

Step-by-step solution

Given that y x = a x 3 + b x 2 + c x + 5 pass through - 2 , 0 so 8 a - 4 b + 2 c = 5       . . . i Since the curve touches x -axis at - 2 , 0 , so its slope would be 0 i.e y ' - 2 = 0 ⇒ 3 a x 2 + 2 b x + c x = - 2 = 0 12 a - 4 b + c = 0         . . . i i Also given, for x = 0 ,   y ' ( x ) = 3 c = 3           . . . i i i Solving eq. i ,   i i   &   i i i , we get, a = - 1 2 ,   b = - 3 4 For local maxima y ' x