Question

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 values of $x$ for which the curve has a negative gradient are:

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

Answer 1

Answer: (C)

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

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