Continuity and Differentiability

$f\left(x\right)=\left[x-1\right]cos\left(\frac{2x-}{2}\right)\pi$           

$Ifx=k,k\in I$

then f(x) = 0 as $cos\left(\frac{2k-1}{2}\right)\pi =0,\forall k\in I$  

LHL = $\underset{h\to 0}{Lt}\left[k-h-1\right]cos\left(\frac{2k-2h-1}{2}\right)\pi =\underset{h\to 0}{Lt}\left(k-2\right)cos\left(\frac{2k-1}{2}\right)\pi \to 0$  

$RHL=\underset{h\to 0}{Lt}\left[k+h-1\right]cos\left(\frac{2k+2h-1}{2}\right)\pi =\underset{h\to 0}{Lt}\left(k-1\right)cos\left(\frac{2k-1}{2}\right)\pi \to 0$        

$\therefore f\left(x\right)$ is continuous $\forall x\in R$  

 $f\left(x\right)=\left\{\begin{array}{cc}\hfill min\left\{\left|x\right|,2-x^2\right\},\hfill & \hfill -2\le x\le 2\hfill \\ \hfill \left[\left|x\right|\right],\hfill & \hfill 2\le \left|x\right|\le 3\hfill \end{array}\right\}$

Number of points where f is not differentiable = 5

$l={\displaystyle \underset{0}{\overset{2}{\int }}f\left(x\right)dx={\left[xf\left(x\right)\right]}_0^2-{\displaystyle \underset{0}{\overset{2}{\int }}xf\text{'}\left(x\right)dx=2e^2-{\displaystyle \underset{0}{\overset{2}{\int }}xf\text{'}\left(x\right)dx}}}$        .(A)

Put   $l_1={\displaystyle \underset{0}{\overset{2}{\int }}xf\text{'}\left(x\right)dx}$            .(i)

Using properties ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx={\displaystyle \underset{a}{\overset{b}{\int }}f\left(a+b-x\right)dx}}$  

$l_1={\displaystyle \underset{0}{\overset{2}{\int }}\left(2-x\right)f\text{'}\left(2-x\right)dx={\displaystyle \underset{0}{\overset{2}{\int }}\left(2-x\right)f\text{'}\left(x\right)dx}}$    .(ii)

Adding (i) and (ii) we get

  $2l_1=2{\displaystyle \underset{0}{\overset{2}{\int }}f\text{'}\left(x\right)dx\Rightarrow l_1={\left[f\left(x\right)\right]}_0^2}$         

f(2) – f(0) = e² – 1

From (A) l = 2e² – e² + 1 = e² + 1

Critical point of function are x = $\frac{-1}{2},$ 1 and -2 but x = -2 is making zero .

$\therefore nondifferentiableatx=\frac{-1}{2},1$      

$f:\mathbf{R}\to \mathbf{R}$

$$$(\mathrm{s}\mathrm{i}\mathrm{n}?\mathrm{x}\mathrm{c}\mathrm{o}\mathrm{s}?\mathrm{y})\left(f\right(2\mathrm{x}+2\mathrm{y})-f(2\mathrm{x}-2\mathrm{y}\left)\right)=(\mathrm{c}\mathrm{o}\mathrm{s}?\mathrm{x}$ Put $\mathrm{s}\mathrm{i}\mathrm{n}?\mathrm{y}\left)\right(f(2\mathrm{x}+2\mathrm{y})+f(2\mathrm{x}-2\mathrm{y}))$

$$$\mathrm{x},\mathrm{y}\in \mathbf{R}$ Put $f^{\text{'}}\left(0\right)=\frac{1}{2}$

$24f^{\text{'}\text{'}}\left(\frac{5\pi }{3}\right)$

$\mathrm{}(\mathrm{s}\mathrm{i}\mathrm{n}?\mathrm{x}\mathrm{c}\mathrm{o}\mathrm{s}?\mathrm{y})\left(\mathrm{f}\right(2\mathrm{x}+2\mathrm{y})-\mathrm{f}(2\mathrm{x}-2\mathrm{y}\left)\right)=(\mathrm{c}\mathrm{o}\mathrm{s}?\mathrm{x}\mathrm{s}\mathrm{i}\mathrm{n}?\mathrm{y})$

$\left(f\right(2x+2y)+f(2x-2y\left)\right)$

$\mathrm{f}(2\mathrm{x}+2\mathrm{y})(\mathrm{s}\mathrm{i}\mathrm{n}?(\mathrm{x}-\mathrm{y}\left)\right)=\mathrm{f}(2\mathrm{x}-2\mathrm{y})\mathrm{s}\mathrm{i}\mathrm{n}?(\mathrm{x}+\mathrm{y})$

: $?2$ adj $(3\mathrm{A}$  adj(2A))|
${=2}^3.?3\mathrm{A}$ adj(2A)| ${\left.\right|}^2$

${=2}^3\cdot {\left(3^3\right)}^2\cdot |\mathrm{A}{|}^2\cdot |\mathrm{a}\mathrm{d}\mathrm{j}\left(2\mathrm{A}\right){|}^2$  intersect the line $=2^3\cdot 3^6\cdot |\mathrm{A}{|}^2\cdot {\left(|2\mathrm{A}{|}^2\right)}^2$ at the point

$=2^3\cdot 3^6\cdot |\mathrm{A}{}^{}$

We can convert 50! In terms of prime factor: $\mathrm{}{2}^{\alpha }\cdot 3^{\beta }\cdot 5^{\gamma }$ & using the greatest integer function.

n $=\left. [\frac{50}{3}\right]+\left. [\frac{50}{3^2}\right]+\left. [\frac{50}{3^3}\right]+\left. [\frac{50}{3^4}\right]$
$=16+5+1+0$ The maximum value of n is 22.

$f\left(x\right)=[\begin{array}{l}3\left(1-\frac{\left|x\right|}{2}\right);\left|x\right|\le 2\\ 0;\left|x\right|>2\end{array}$

$\therefore g\left(x\right)=f\left(x+2\right)-f\left(x-2\right)[\begin{array}{l}0x<-4\\ 3\left(1-\frac{\left|x-2\right|}{2}\right)-4\le x\le 0\\ -3\left(1-\frac{\left|x-2\right|}{2}\right)0<x\le 4\\ 0x>4.\end{array}$

g(x) is continuous every where but not differentiable

at x = -4, -2, 2 and 4

$\therefore n=0\&m=4\Rightarrow n+m=4$

f (t) = t³ – 6t² + 9t + 3

$\Rightarrow f\text{'}\left(t\right)=3t^2-12t+9=3\left(t-1\right)\left(t-3\right)$

$\therefore f\text{'}\left(t\right)=0\Rightarrow t=1,3$

$\Rightarrow f\left(1\right)=1,f\left(3\right)=-3$

$g\left(x\right)=\{\begin{array}{l}f\left(x\right),0\le x<1\\ 1,1\le x\le 3g\left(x\right)iscontinuous\\ 4-x,3<x\le 4\end{array}$

Hence g (x) is not differentiable at only x = 3.

Kindly consider the following figure