If a curve with equation of the form `y=ax^(4)+bx^(3)+cx+d` has zero gradient at the point (0, 1) and also touches the x-axis at the point `(-1,0)` then the value of x for which the curve has a negative gradient are

Options

A. ` x gt-1 `
B. ` x lt 1 `
C. ` xlt -1 `
D. ` -1 le x le 1 `

Answer

C

Explanation

We have , `(dy)/(dx)=0 ` at (0,1) and (-1,0)
` therefore c=0 " and " -4a+3b -c=0 rArr a=(3b)/(4) " and " c=0 " "...(i)`
Also, the curve passes through (0,1) and (-1,0)
` therefore d=1 and 0=a-b-c+d rArr a-b-c+1 =0 " "...(ii)`
From (i) and (ii), we get
`a=3, b=4,c=0 and d=1 `
` therefore y=3x^(4)+4x^(3)+1 rArr (dy)/(dx)=12x^(3)+12x^(2) `
Now,
`(dy)/(dx) lt 0 rArr 12x^(3)+12x^(2) lt 0 rArr x+1 lt 0 rArr x lt -1 `