Question

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

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

Answer 1

The correct option is B

$-3,-12,0$

Explanation for the correct option.

Step 1: Differentiate the curve

Given, an equation of the curve $y=2x^3+a{x}^2+bx+c....\left(1\right)$.
Also, it is passes through $(0,0)$.
$$\begin{gathered} \Rightarrow 0=2\left(0\right)+\mathrm{a}\left(0\right)+\mathrm{b}\left(0\right)+\mathrm{c} \\ \Rightarrow \mathrm{c}=0 \end{gathered}$$
On differentiating equation (1), we get
$\frac{dy}{dx}=6x^2+2ax+b$

Step 2: Form equations
As, the tangents at $\mathrm{x}=-1$ and $\mathrm{x}=2$ are parallel to $\mathrm{x}$-axis.

$\Rightarrow \frac{dy}{dx}=0$

$6x^2+2ax+b=0$

At, $x=-1$ we have:

$$\begin{gathered} 6(-1{)}^2+2a(-1)+b=0 \\ \Rightarrow 6-2a+b=0...(2) \end{gathered}$$

At, $x=2$ we have:

$$\begin{gathered} 6(2{)}^2+2\mathrm{a}(2)+\mathrm{b}=0 \\ \Rightarrow 24+4\mathrm{a}+\mathrm{b}=0...(3) \end{gathered}$$

Step 3: Find the value of $a\&b$

Solve equations (2) & (3) as Equation (3)$-$equation (2) we get:

$\begin{array}{rcl}6a+18& =& 0\\ a& =& -3\end{array}$

Put $a=-3$ in equation (2) we get:

$\begin{array}{rcl}6+6+b& =& 0\\ b& =& -12\end{array}$

$\Rightarrow a=-3,b=-12,c=0$

Hence, option B is correct.