Class 11th

As slope of line joining (1, 2) and (3, 6) is 2 given diameter is parallel to side

  $?a=\sqrt[]{{\left(3?1\right)}^2+{\left(6?2\right)}^2}=\sqrt[]{20}\text{?}\text{?}and$

$\frac{b}{2}=\frac{4}{\sqrt[]{5}}?b=\frac{8}{\sqrt[]{5}}$

Area

$ab=2\sqrt[]{5}.\frac{8}{\sqrt[]{5}}=16$

  $T_r=\frac{r}{{\left(2r^2\right)}^2+1}$

$=\frac{r}{{\left(2r^2+1\right)}^2-{\left(2r\right)}^2}$

$=\frac{1}{4}\frac{4r}{\left(2r^2+2r+1\right)\left(2r^2-2r+1\right)}$

$S_{10}=\frac{1}{4}{\displaystyle \sum _{r=1}^{10}\left(\frac{1}{\left(2r^2-2r+1\right)}-\frac{1}{\left(2r^2+2r+1\right)}\right)}$

$\Rightarrow S_{10}=\frac{1}{4}.\frac{220}{221}=\frac{55}{221}=\frac{m}{n}$

$\therefore m+m=276$

  $T_{r+1}{=}^{60}{C}_r{\left(x^{\frac{1}{2}}\right)}^{60-r}{\left(x^{-\frac{1}{3}}\right)}^r{\left(5^{\frac{-1}{4}}\right)}^{60-r}{\left(5^{\frac{1}{2}}\right)}^r$

for

$x^{10}\frac{60-r}{2}-\frac{r}{3}=10$    

$\Rightarrow 180-3r-2r=60$             

->r = 24

k = 3 + exponent of 5 in 

= $3+\left(\left[\frac{60}{5}\right]+\left[\frac{60}{5^2}\right]-\left[\frac{24}{5}\right]-\left[\frac{24}{5^2}\right]-\left[\frac{36}{5}\right]-\left[\frac{36}{5^2}\right]\right)$  

= 3 + (12 + 2 – 4 – 0 – 7 – 1)

= 3 + 2 = 5

For an acid- base titration, Methly orange exist at end point as quinonoid form.

Let e^x = t then equation reduces to

$t^2-11t-\frac{45}{t}+\frac{81}{2}=0$                

$\Rightarrow 2t^3-22t^2+81t-45=0$                     ……. (i)

If roots of

$e^{2x}-11e^x-45e^{-x}+\frac{81}{2}=0$            

$\Rightarrow {\alpha }_1+{\alpha }_2+{\alpha }_3=\mathcal{l}n45\Rightarrow p=45$                

The two main types of potential energy are:

1. Gravitational Potential Energy - Energy stored due to an object's position in a gravitational field (PE = mgh).
2. Elastic Potential Energy - Energy stored in deformed elastic objects like springs or stretched materials (PE = ½kx²).

The best way to find potential energy is to use the relationship 

${}_{}$ $U_f-U_i=-{\int }_i^f\overrightarrow{F}\cdot d\overrightarrow{r}$

and integrating the conservative force over the path.

For common cases, apply the standard potential energy formulas:

• PE = mgh for gravity
• PE = ½kx² for springs. 

Potential energy is best defined as a stored energy in a system. This energy is stored due to the configuration or position of the objects within a conservative force field. Potential energy in physics tells us that it's the capacity of the work done when the system is allowed to move from that position to the reference point.  

  $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\Rightarrow \frac{{\left(-4\sqrt[]{\frac{2}{5}}\right)}^2}{a^2}+\frac{32}{b^2}=1$                

$\Rightarrow \frac{32}{5a^2}+\frac{9}{b^2}=1$ ……. (i)

From (i)

$\frac{6}{b^2}+\frac{9}{b^2}=1\Rightarrow b^2=15\&a^2=16$

$a^2+b^2=15+16=31$