# Solve the Following Questions

Question:

Let $\mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right]$ be a $3 \times 3$ matrix, where

$\mathrm{a}_{\mathrm{ij}}=\left\{\begin{array}{ccc}1 & , & \text { if } \mathrm{i}=\mathrm{j} \\ -\mathrm{x} & , & \text { if }|\mathrm{i}-\mathrm{j}|=1 \\ 2 \mathrm{x}+1 & , & \text { otherwise. }\end{array}\right.$

Let a function $\mathrm{f}: \mathbf{R} \rightarrow \mathbf{R}$ be defined as $\mathrm{f}(\mathrm{x})=\operatorname{det}(\mathrm{A})$. Then the sum of maximum and minimum values of f on $\mathbf{R}$ is equal to:

1. $-\frac{20}{27}$

2. $\frac{88}{27}$

3. $\frac{20}{27}$

4. $-\frac{88}{27}$

Correct Option: , 4

Solution:

$A=\left[\begin{array}{ccc}1 & -x & 2 x+1 \\ -x & 1 & -x \\ 2 x+1 & -x & 1\end{array}\right]$

$|A|=4 x^{3}-4 x^{2}-4 x=f(x)$

$f^{\prime}(x)=4\left(3 x^{2}-2 x-1\right)=0$

$\Rightarrow x=1 ; x=\frac{-1}{3}$

$\therefore \underbrace{f(1)=-4}_{\min } ; f(\underbrace{\left.-\frac{1}{3}\right)=\frac{20}{27}}_{\max }$

Sum $=-4+\frac{20}{27}=-\frac{88}{27}$