Question

$a_{ij}=\left\{\begin{array}{c}1,\\ -x,\\ 2x+1\end{array}\right.\begin{array}{c}i=j\\ |i-j|=1\\ otherwise\end{array}$
$A={\left[a_{ij}\right]}_{3\times 3}.f\left(x\right)=Det\left(A\right)$. Then find the sum of local maximum and minimum values of $fx$.

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

Answer 1

The correct option is D

$-\frac{88}{27}$

Explanation for the correct option:

Finding the sum of local maximum and minimum values of x:

It is given that $a_{ij}=\left\{\begin{array}{c}1,\\ -x,\\ 2x+1\end{array}\right.\begin{array}{c}i=j\\ |i-j|=1\\ otherwise\end{array}$

And $f\left(x\right)$ is the determinant of $a_{ij}$

$\begin{array}{rcl}f\left(x\right)& =& \left|\begin{array}{c}\begin{array}{ccc}\begin{array}{c}1\\ -x\\ 2x+1\end{array}& \begin{array}{c}-x\\ 1\\ -x\end{array}& \begin{array}{c}2x+1\\ -x\\ 1\end{array}\end{array}\end{array}\right|\\ & =& 4x^3–4x^2–4x\\ f’\left(x\right)& =& 4(3x^2–2x–1)\\ f’\left(x\right)& =& 4(x-1)\left(x+\frac{1}{3}\right)\\ f\left(1\right)& =& 4–4–4\\ & =& -4\\ f\left(\frac{-1}{3}\right)& =& \left(-\frac{4}{27}\right)–\left(\frac{4}{9}\right)+\left(\frac{4}{3}\right)\\ & =& \frac{(-4-12+36)}{27}\\ & =& \frac{20}{27}\\ \left(\frac{20}{27}\right)–4& =& \frac{(20-108)}{27}\\ & =& \frac{\mathbf{-}\mathbf{88}}{\mathbf{27}}\\ & & \end{array}$

Hence, the correct answer is option (D).