Mathematics - Application of Derivative Question with Solution | TestHub

We have,

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

Then,

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

$\Rightarrow \left|A\right|=\left(1-x^2\right)+x\left(-x+2x^2+x\right)+\left(2x+1\right)\left(x^2-2x-1\right)$

$\Rightarrow \left|A\right|=1-x^2+2x^3+\left(2x^3-4x^2-2x\right)+\left(x^2-2x-1\right)$

$\Rightarrow \left|A\right|=f\left(x\right)=4x^3-4x^2-4x$

$\Rightarrow f^{\text{'}}\left(x\right)=4\left(3{\mathrm{x}}^2-2\mathrm{x}-1\right)$

For maxima or minima, $f^{\text{'}}\left(x\right)=0$

$\Rightarrow 3x^2-2x-1=0$

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

Now,

$f^{\text{'}\text{'}}\left(x\right)=4\left(6x-2\right)$

$f^{\text{'}\text{'}}\left(1\right)=4\left(6·1-2\right)=16>0$( point of minima)

$f^{\text{'}\text{'}}\left(-\frac{1}{3}\right)=4\left(6·\left(-\frac{1}{3}\right)-2\right)=-16<0$ ( point of maxima)

Hence, required sum is

$f\left(1\right)+f\left(-\frac{1}{3}\right)$

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