Question

If the curve $y=2x^3+a{x}^2+bx+c$ passes through the origin and the tangents drawn to it at $x=-1$ and $x=2$ are parallel to the $x$-axis, then the values of $a,b$ and $c$ are respectively

• A. $12,-3$ and $0$
• B. $-3,-12$ and $0$
• C. $-3,12$ and $0$
• D. $3,-12$ and $0$

Answer 1

Answer: (B)

$-3,-12$ and $0$

Given, equation of curve is
$y=2x^3+a{x}^2+bx+c$ ....(i)
Since, it is passes through $(0,0)$
$\Rightarrow 0=2\left(0\right)+a\left(0\right)+b\left(0\right)+c$
$\Rightarrow c=0$ ....(ii)
On differentiating equation (i), we get
$\frac{dy}{dx}=6x^2+2ax+b$
Since, the tangents at $x=-1$ and $x=2$ are parallel to $x$-axis.
Thus $\frac{dy}{dx}=0$
At $x=-1$,
$6{(-1)}^2+2a(-1)+b=0$
$\Rightarrow 6-2a+b=0$ ....(iii)
At $x=2$,
$6{\left(2\right)}^2+2a\left(2\right)+b=0$
$\Rightarrow 24+4a+b=0$ ....(iv)
On solving equations (iii) and (iv), we get
$a=-3,b=-12$ and $c=0$