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If the curve $y=2 x^{3}+a x^{2}+b x+c$ 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,
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Verified Answer
The correct answer is:
$-3,-12$ and 0
$y=2 x^{3}+a x^{2}+b x+c...(i)$
Since, it passes through $(0,0)$,
$0=2(0)+a(0)+b(0)+c$
$\begin{aligned} c &=0 ...(ii)\\ \frac{d y}{d x} &=6 x^{2}+2 a x+b \end{aligned}$
Since, tangents at $x=-1$ and $x=2$ are parallel to $X$-axis.
$\therefore \quad \frac{d y}{d x}=0$
At
$\begin{aligned}
x &=-1 \\
6(-1)^{2}+2 a(-1)+b &=0 \\
6-2 a+b &=0...(iii)
\end{aligned}$
At $x=-2$
$\text { So, } \begin{aligned}
6(2)^{2}+2 a(2)+b &=0 \\
24+4 a+b &=0...(iv)
\end{aligned}$
By solving Eqs. (iii) and (iv), we get
$a=-3, b=-12$
Hence, $a=-3, b=-12, c=0$
Since, it passes through $(0,0)$,
$0=2(0)+a(0)+b(0)+c$
$\begin{aligned} c &=0 ...(ii)\\ \frac{d y}{d x} &=6 x^{2}+2 a x+b \end{aligned}$
Since, tangents at $x=-1$ and $x=2$ are parallel to $X$-axis.
$\therefore \quad \frac{d y}{d x}=0$
At
$\begin{aligned}
x &=-1 \\
6(-1)^{2}+2 a(-1)+b &=0 \\
6-2 a+b &=0...(iii)
\end{aligned}$
At $x=-2$
$\text { So, } \begin{aligned}
6(2)^{2}+2 a(2)+b &=0 \\
24+4 a+b &=0...(iv)
\end{aligned}$
By solving Eqs. (iii) and (iv), we get
$a=-3, b=-12$
Hence, $a=-3, b=-12, c=0$
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