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: `xgeq-1` b. `x<1` c. `x<-1` d. `-1lt=xlt=1`

Options

A. xgt-I
B. xlt1
C. xlarr1
D. `-1le xle 1`

Answer

C

Explanation

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