A curve with equation of the form `y=a x^4+b x^3+c x+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 gt1`
C. `x lt-1`
D. `-1 le x le1`

Answer

C

Explanation

`y=ax^(4)+bx^(3)+cx+d`
`rArr" "dy//dx=4ax^(3)+3bx^(2)+c`
Point (0, 1) satisfies the curve `rArr d=1`
Also `((dy)/(dx))_("(0,1)")=0 rArr c=0`
`(-1,0)` satisfies the curve `rArr a-b-c+1=0`
Also `((dy)/(dx))_("(-1,0)")=0 rArr c=0`
`rArr" "-4a+3b=0 and a-b=-1`
`rArr" "a=3, b=4`
`rArr" The curve is "y=3x^(4)+4x^(3)+1`
`rArr" "dy//dx=12x^(3)+12x^(2)`
For `dy//dx lt` we have `x^(3)+x^(2)lt 0 rArr x^(2)(x+1)lt0 rArr xlt-1`