Class 11th

  $f\left(x\right)=\{\begin{array}{l}-2x^2+3,-2<x<-1\\ x^2,-1\le x<0\\ \frac{\sqrt[]{3}}{2}-1,0\le x<1\\ x^2-3+\frac{1}{\sqrt[]{2}},1\le x<2\end{array}$

$f\left(-1^{-}\right)=f\left(-1^{+}\right)=f\left(-1\right)=1$                

$f\left(1^{+}\right)=f\left(1\right)=\frac{1}{\sqrt[]{2}}-2$                

Points of discontinuity

x = 0, 1

According to ideal gas law we know that

PV = nRT

According to question, we can write

Assuming Tension in metal wire suspended from roof varies linearly and 0 and T0 (developed due its own weight) are tensions are ends of wire of length $\mathcal{l}.$ $$ So

$T_1\alpha \left({\mathcal{l}}_1-\mathcal{l}\right),and{T}_2\alpha \left({\mathcal{l}}_2-\mathcal{l}\right)$

$\frac{T_1}{T_2}=\frac{{\mathcal{l}}_1-\mathcal{l}}{{\mathcal{l}}_2-\mathcal{l}}\Rightarrow \mathcal{l}=\frac{T_2{\mathcal{l}}_1-T_1{\mathcal{l}}_2}{T_2-T_1}$

ar (ABC) = 4ar (DEF)

$=4*\frac{1}{2}*\left|2\left(2-5\right)+1\left(5-3\right)+7\left(3-2\right)\right|=2\left|-6+2+7\right|=6$

$K_T=\frac{m{v}^2}{2}and{K}_R=\frac{l{\omega }^2}{2}=\frac{l{v}^2}{2R^2}$

According to question, we can write

$K_R=\frac{1}{2}{K}_T\Rightarrow \frac{l{v}^2}{2R^2}=\frac{1}{2}*\frac{m{v}^2}{2}\Rightarrow l=\frac{m{R}^2}{2}\Rightarrow Solidcylinder$

$\gamma =1+\frac{2}{f}\Rightarrow \frac{2}{f}=\gamma -1\Rightarrow f=\frac{2}{\gamma -1}$

Let the distance travelled is x, so

$$ $\left|{\overrightarrow{v}}_{Eg}\right|=\frac{x}{t_2}\Rightarrow$ speed of escalator for ground

${\stackrel{}{}}_{}$ $\left|{\overrightarrow{v}}_{BE}\right|=\frac{x}{t_1}\Rightarrow$ speed of boy concerning the escalator

${\stackrel{}{}}_{}$ $\left|{\overrightarrow{v}}_{Bg}\right|=\frac{x}{t_1}+\frac{x}{t_2}\Rightarrow$ speed of boy concerning ground

The time taken by him to walk up the moving escalator = t =  $\frac{x}{\left|{\overrightarrow{v}}_{Bg}\right|}$ $\frac{}{{\stackrel{}{}}_{}}$

$\Rightarrow t=\frac{x}{\frac{x}{t_1}+\frac{x}{t_2}}=\frac{t_1{t}_2}{t_1+t_2}$