Draw a rough sketch of the curve $\mathrm{y}=|\mathrm{x}-5|$. Find the area under the curve and line x = 3 and x = 6.

OR

Find the value(s) of x for which y = [x(x – 2)]2 is an increasing function.

Given, curve y =|x – 5| is an absolute function 
$\therefore \mathrm{y}=\left\{\begin{array}{l}\mathrm{x}-5,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}\ge 5\\ 5-\mathrm{x},\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}<5\end{array}\right.$

$\therefore \text{ Area of bounded region }={\int }_3^6 |\mathrm{x}-5|\mathrm{dx}$
$\begin{array}{l}={\int }_3^5 (5-\mathrm{x})\mathrm{dx}+{\int }_5^6 (\mathrm{x}-5)\mathrm{dx}\\ ={\left[5\mathrm{x}-\frac{{\mathrm{x}}^2}{2}\right]}_3^5+{\left[\frac{{\mathrm{x}}^2}{2}-5\mathrm{x}\right]}_5^6\\ =\left(5\times 5-\frac{5^2}{2}\right)-\left(5\times 3-\frac{3^2}{2}\right)+\left[\frac{6^2}{2}-5\times 6-\left(\frac{5^2}{2}-5\times 5\right)\right]\\ =\left(25-\frac{25}{2}\right)-\left(15-\frac{9}{2}\right)+\left[18-30-\left(\frac{25}{2}-25\right)\right]\\ =\frac{25}{2}-\frac{21}{2}+\left[-12-\left(-\frac{25}{2}\right)\right]\\ =\frac{4}{2}+\left[-12+\frac{25}{2}\right]=2+\frac{1}{2}=\frac{5}{2}\text{ sq units }\end{array}$

 OR

Given function is $\mathrm{y}=\left[\mathrm{x}\right(\mathrm{x}-2){]}^2$
On differentiating both sides w.r.t x, we get
$\begin{array}{l}\frac{\mathrm{dy}}{\mathrm{dx}}=2\left({\mathrm{x}}^2-2\mathrm{x}\right)\frac{d}{dx}\left({\mathrm{x}}^2-2\mathrm{x}\right)\\ =2\left({\mathrm{x}}^2-2\mathrm{x}\right)(2\mathrm{x}-2)=4\mathrm{x}(\mathrm{x}-2)(\mathrm{x}-1)\end{array}$
$\mathrm{On} \mathrm{putting} \frac{\mathrm{dy}}{\mathrm{dx}}=0, \mathrm{we} \mathrm{get}$
$4\mathrm{x}(\mathrm{x}-2)(\mathrm{x}-1)=0\Rightarrow \mathrm{x}=0,1\text{ and }2$

Now, we find interval in which function  is strictly increasing or strictly decreasing.
 

$$\begin{gathered} \mathrm{Hence}, \mathrm{y} \mathrm{is} \mathrm{strictly} \mathrm{increasing} \mathrm{in} (0,1) \mathrm{and} (2,\infty ). \mathrm{Also}, \mathrm{y} \mathrm{is} \mathrm{apolynomial} \mathrm{function},\mathrm{so} \mathrm{it} \mathrm{continuous} \mathrm{at} \mathrm{x}=0,1 \mathrm{and} 2. \\ \mathrm{Hence}, \mathrm{y} \mathrm{is} \mathrm{increasing} \mathrm{in} [0,1]\cup [2,\infty ). \end{gathered}$$