NIMCET 2016 Question Paper with solutions

We have $\mathit{f}\left(\mathit{x}\right)={\mathit{x}}^2-\mathit{b}\mathit{x}+\mathit{c}$ and $\mathit{b}+\mathit{c}=35$
According to the condition $\mathit{b}+\mathit{c}=35$
Let assume that 2 and 11 be the prime root of above equation, then $\left(\mathit{x}-2\right)\left(\mathit{x}-11\right)=0$
$\Rightarrow {\mathit{x}}^2-2\mathit{x}-11\mathit{x}+22=0$
$\Rightarrow {\mathit{x}}^2-13\mathit{x}+22=0$
Which satisfies the condition $\mathit{b}+\mathit{c}=35$.
On comparing both equations, we get b = 13 and c =22
Global minimum f'(x)=0
$\Rightarrow 2\mathit{x}-13=0\Rightarrow \mathit{x}=\frac{13}{2}$

So, $\mathit{f}\left(\frac{13}{2}\right)={\left(\frac{13}{2}\right)}^2-13\times \frac{13}{2}+22$
$=\frac{169}{4}-\frac{169}{2}+22=-\frac{81}{4}$