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:
A
`x gt-1`
B
`x gt1`
C
`x lt-1`
D
`-1 le x le1`
Text Solution
Verified by Experts
The correct Answer is:
C
`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`
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