Question

A curve with equation of the form $y=a{x}^4+b{x}^3+cx+d$ has zero gradient at the point $(0,1)$ and also touches the $x$-axis at the point $(-1,0)$ then the values of $x$ for which the curve has a negative gradient are:

• A. $x>-1$
• B. $x<1$
• C. $x<-1$
• D. $-1\le x\le 1$

Answer 1

Answer: (C)

$\frac{dy}{dx}=4a{x}^3+3b{x}^2+c$
$at(0,1)$
$\frac{dy}{dx}=0\Rightarrow c=0\&d=1$

It touches $x$-axis at ${(-1,0)\Rightarrow \frac{dy}{dx}|}_{(-1,0)}=0$
$\Rightarrow -4a+3b=0$
so $\frac{dy}{dx}=4a\left(x^3+x^2\right)$
$\Rightarrow$ two points of extrema $\frac{d^{2}y}{d{x}^2}=4a\left(3x^2+2x\right)$
$\Rightarrow$ one point of inflection Hence negative gradient for $x<-1$