Question

Let $A=\left[a_{ij}\right]$ be a $3\times 3$ matrix, where
$a_{ij}=\{\begin{array}{ccc}1& ,& \text{if}i=j\\ -x& ,& \text{if}|i-j|=1\\ 2x+1& ,& \text{otherwise}\end{array}$
Let a function $f:R\to R$ be defined as $f\left(x\right)=det\left(A\right).$ Then the sum of maximum and minimum values of $f$ on $R$ is equal to

• A. $-\frac{20}{27}$
• B. $-\frac{88}{27}$
• C. $\frac{88}{27}$
• D. $\frac{20}{27}$

Answer 1

The correct option is B $-\frac{88}{27}$

BONUS QUESTION

$f\left(x\right)=\left|\begin{array}{ccc}1& -x& 2x+1\\ -x& 1& -x\\ 2x+1& -x& 1\end{array}\right|\Rightarrow f\left(x\right)=4x^3-4x^2-4x$
Since, $f\left(x\right)$ is a cubic function and its range is $R$(all real numbers)
Hence, maximum and minimum values are not defined.

Possible solution for modified question:

$f\left(x\right)=\left|\begin{array}{ccc}1& -x& 2x+1\\ -x& 1& -x\\ 2x+1& -x& 1\end{array}\right|=4x^3-4x^2-4x$

$f^{\prime }\left(x\right)=12x^2-8x-4=0\Rightarrow x=\frac{-1}{3},1$

$\therefore$ Sum of local maximum and local minimum values of $f\left(x\right)=$ $f\left(\frac{-1}{3}\right)+f\left(1\right)=\frac{20}{27}-4=-\frac{88}{27}$