Class 11th

Arithmetic mean is the measure of central tendency that is most affected by extreme items or outliers in the dataset. The reason behind this is since mean is calculated by summing up all the values and after that, it is divided by the number of values. Extreme values may disproprotionately impact the sum which in turn skews the mean and make it less represenative of dataset as whole.

Median is considered to be the most suitable average for qualitative measurement. It divides an entire frequency distribution into two haves. This is especially useful for ordinal data where values represent categories with meaningful order. However, it is not necessarily a linear scale. The median gives a cental value which is less influenced by extreme values or outliers. This is important while dealing with qualitative data which may not be either symmetrically scaled or evenly distributed.

The NO (Nitric Oxide) has 11 valence electrons while the NO? (Nitrosonium ion) has 10 valence electrons due to removal of one antibonding electron. Due to this, the bond order of NO increases from? 2.5 to 3 in the NO? (Nitrosonium ion). 

As per the NCERT, since the bond order of NO? (3) is higher than that of NO (2.5), the bond in Nitrosonium ion is stronger bond and the stronger the bond, the shorter bond length.

Hence bond length in NO? is shorter the NO.

Ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ passes through the points (7, 0) & (0,  $-2\sqrt[]{6}$ )

$\therefore a^2=49\&b^2=24$

$\therefore e=\sqrt[]{1-\frac{b^2}{a^2}}=\sqrt[]{1-\frac{24}{49}}=\frac{5}{7}$

A(2, 3, 9)

B(5, 2, 1)

C(1, $\lambda$ , 8)

D( $\lambda$ ,2, 3)

$\Delta =\left|\begin{array}{ccc}\hfill 3\hfill & \hfill -1\hfill & \hfill -8\hfill \\ \hfill -4\hfill & \hfill \lambda -2\hfill & \hfill 7\hfill \\ \hfill \lambda -2\hfill & \hfill -1\hfill & \hfill -6\hfill \end{array}\right|$           

    $\overrightarrow{AB}=\left(3,-1,-8\right)$

$\overrightarrow{AC}=\left(-4,\lambda -2,7\right)$            

$\overrightarrow{AD}=\left(\lambda -2,-1,-6\right)$

$\Delta =3\left[-6\lambda +12+7\right]-1\left(7\lambda -14-24\right)-8\left(4-{\left(\lambda -2\right)}^2\right)$

$=57-18\lambda \Rightarrow +38-32+8\left({\lambda }^2-4\lambda +4\right)$            

$=95-57\lambda +8{\lambda }^2$

$\Delta =0\Rightarrow {\lambda }_1{\lambda }_2=\frac{95}{8}$                       

x + 2y + z = 14

${\perp }^{r}linePQ:\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{1}=t$              

Q (1 + t, 2 + 2t, 3 + t)

x + 2y + z = 14 -> 1 + t + 4 + 4t + 3 + t = 14 Þ 6t = 6

t = 1

-> Q (2, 4, 4)

PQ = $\sqrt[]{1+4+1}=\sqrt[]{6}$  

$tan60°=\frac{PQ}{QR}\Rightarrow QR=\frac{PQ}{\sqrt[]{3}}=\sqrt[]{2}$

ar (PQR) = $\frac{1}{2}\sqrt[]{6}\cdot \sqrt[]{2}=\sqrt[]{3}$  

Circumcentre (D) $\equiv \left(5,\frac{\alpha }{4}\right)$  

${\left(5-\alpha \right)}^2+{\left(\frac{\alpha }{4}+2\right)}^2={\left(5-\alpha \right)}^2+{\left(\frac{\alpha }{4}-6\right)}^2...............\left(i\right)$

${\left(5-\frac{\alpha }{4}\right)}^2+{\left(\frac{\alpha }{4}+2\right)}^2....................\left(ii\right)$

 (i) -> $\frac{\alpha }{4}+2=\pm \left(\frac{\alpha }{4}-6\right)$  

(ii) -> 9 + 16 = 9 + 16

$\oplus \to x$

$\left(-\right)\to \frac{\alpha }{2}=4\Rightarrow \alpha =8$

ar   $\left(ABC\right)=24$

2S = 24

R = 5, r =   $\frac{\Delta }{s}=\frac{24}{12}=2$

$a_{n+2}{a}_{n+1}-a_{n+1}{a}_n=2$

Series will satisfy

$a_1{a}_2,a_2{a}_3,a_3{a}_4,......a_4{a}_5$

$\begin{array}{l}1.22.22.32.4\\ \frac{a_n+\frac{1}{a_{n+1}}}{a_{n+2}}=\frac{a_{n+2}-\frac{1}{a_{n+1}}}{a_{n+2}}\end{array}$

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

Now, proof = $\frac{30}{11}\frac{\left(2r+1\right)}{2\left(r+1\right)}$ $\frac{}{}\frac{}{}$

r = 1

$=\frac{\left(1.3.5......61\right)}{2^{30}\left(2.3......31\right)}$

$\frac{\overline{)61}}{2^{60}.\overline{)31}.\overline{)30}}=\alpha =-60$