overline{\mathrm{c}}=x \hat{\mathrm{i}}+\hat{\mathrm{k}} \end{aligned} $$

Volume of parallelopiped

$$ \begin{aligned} & =\left[\begin{array}{lll} \bar{a} & \bar{b} & \bar{c} \end{array}\right] \\ & =\left|\begin{array}{lll} 1 & x & 1 \\ 0 & 1 & x \\ x & 0 & 1 \end{array}\right| \\ & =1(1-0)-x\left(0-x^2\right)+1(0-x) \\ & =1+x^3-x \\ & \text { Let } \mathrm{f}(x)=1+x^3-x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \therefore \quad & \mathrm{f}^{\prime}(x)=3 x^2-1 \end{aligned} $$

Consider, $\mathrm{f}^{\prime}(x)=0$

$$ \begin{aligned} & \Rightarrow 3 x^2-1=0 \\ & \Rightarrow 3 x^2=1 \\ & \Rightarrow x^2=\frac{1}{3} \\ & \Rightarrow x= \pm \frac{1}{\sqrt{3}} \end{aligned} $$

Also, $\mathrm{f}^{\prime \prime}(x)=6 x$

When $x=\frac{1}{\sqrt{3}}, \mathrm{f}^{\prime \prime}(x)=\frac{6}{\sqrt{3}}>0$

$\therefore \mathrm{f}(x)$ is minimum at $x=\frac{1}{\sqrt{3}}$.

When $x=\frac{-1}{\sqrt{3}}, \mathrm{f}^{\prime \prime}(x)=\frac{-6}{\sqrt{3}}<0$

$\therefore \mathrm{f}(x)$ is maximum at $x=\frac{-1}{\sqrt{3}}$.

$\therefore \quad$ Maximum value $=1+\left(\frac{-1}{\sqrt{3}}\right)^3-\left(\frac{-1}{\sqrt{3}}\right)\ldots[\text { From (i) }]$

$$ \begin{aligned} & \Rightarrow \text { Max. value }=1+\frac{2}{3 \sqrt{3}} \\ & \therefore \quad \text { Minimum value }=1+\left(\frac{1}{\sqrt{3}}\right)^3-\left(\frac{1}{\sqrt{3}}\right) \\ & \Rightarrow \text { Min. value }=1-\frac{2}{3 \sqrt{3}} \ldots[\text { From (i)] }\end{aligned} $$

Vector Algebra

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