NIMCET 2013 Question Paper with solutions

Let's say that the function is $\mathit{y}=\mathit{f}\left(\mathit{x}\right)=2{\mathit{x}}^3-21{\mathit{x}}^2+36\mathit{x}-20$
$\therefore {\mathit{f}}^{\prime }\left(\mathit{x}\right)=\frac{\mathit{d}}{\mathit{d}\mathit{x}}\mathit{f}\left(\mathit{x}\right)=\frac{\mathit{d}}{\mathit{d}\mathit{x}}\left(2{\mathit{x}}^3-21{\mathit{x}}^2+36\mathit{x}-20\right)=6{\mathit{x}}^2-42\mathit{x}+36$
And, ${\mathit{f}}^{\prime }\left(\mathit{x}\right)=\frac{{\mathit{d}}^2}{\mathit{d}{\mathit{x}}^2}\mathit{f}\left(\mathit{x}\right)=\frac{\mathit{d}}{\mathit{d}\mathit{x}}\left[\frac{\mathit{d}}{\mathit{d}\mathit{x}}\mathit{f}\left(\mathit{x}\right)\right]=\frac{\mathit{d}}{\mathit{d}\mathit{x}}\left(6{\mathit{x}}^2-42\mathit{x}+36\right)=12\mathit{x}-42$
For maxima\/minima points, ${\mathit{f}}^{\prime }\left(\mathit{x}\right)=0$.
$\Rightarrow 6{\mathit{x}}^2-42\mathit{x}+36=0$
$\Rightarrow {\mathit{x}}^2-7\mathit{x}+6=0$
$\Rightarrow {\mathit{x}}^2-6\mathit{x}-\mathit{x}+6=0$
$\Rightarrow \mathit{x}\left(\mathit{x}-6\right)-\left(\mathit{x}-6\right)=0$
$\Rightarrow \left(\mathit{x}-6\right)\left(\mathit{x}-1\right)=0$
$\Rightarrow \mathit{x}-6=0\mathit{O}\mathit{R}\times -1=0$
$\Rightarrow \mathit{x}=6$ OR $\mathit{x}=1$
Now, let's check these points for maxima\/minima by inspecting the values of ${\mathit{f}}^{\prime \prime }\left(\mathit{x}\right)$ at these points.
${\mathit{f}}^{\prime }\left(6\right)=12\left(6\right)-42=72-42=30.$
${\mathit{f}}^{\prime }\left(1\right)=12\left(1\right)-42=12-42=-30.$
Since, ${\mathit{f}}^{\prime }\left(6\right)=30>0$, it is the point of minimum value.
And the minimum value is $\mathit{f}\left(6\right)$ :
$=2{\left(6\right)}^3-21{\left(6\right)}^2+36\left(6\right)-20$
$=432-756+216-20$
$=-128$