Maths

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.

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)

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

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

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

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

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

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

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

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

Slope of any point P (x, y) to y = f (x) is $\frac{dy}{dx}=-k\frac{y}{x}$

$\Rightarrow \frac{dy}{y}+k\frac{dx}{x}=0$

Solving the equation the curve is xky = c

It passes (1, 2) c = 2 xky = 2 again it passes (8, 1) 8k = 2 k = $\frac{1}{3}$

$\therefore$ the equation of curve is $x^{1/3}$ y = 2 …… (i)

$\therefore \left|y\left(\frac{1}{8}\right)\right|=\left|\frac{2}{{\left(\frac{1}{8}\right)}^{1/3}}\right|=4$

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 =   $\frac{\Delta }{s}=\frac{24}{12}=2$

$f\left(x\right)=e^x.{\displaystyle \underset{0}{\overset{x}{\int }}\frac{f\text{'}\left(t\right)}{e^t}dt}$

$f\text{'}\left(x\right)=e^x.{\displaystyle \underset{0}{\overset{x}{\int }}\frac{f\text{'}\left(t\right)}{e^t}dt+e^x.\frac{f\text{'}\left(x\right)}{e^x}-\left[\left(2x-1\right).e^x+\left(x^2-x+1\right).e^x\right]}$

$\begin{array}{l}f\left(x\right)=\left(2x+1\right).e^x-2e^x+c\\ f\left(x\right)=e^x\left(2x-1\right)c=0\end{array}$

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

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

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

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

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

Case I

= = 0 then

f (x) = yx

$g\left(x\right)=\frac{x}{y}$

$\frac{1}{x}{\displaystyle \sum _{i=1}^{n}f\left(a_i\right)\Rightarrow \frac{y}{x}\left(a_1+a_2+....+a_x\right)=0}$

$\Rightarrow f\left(g\left(0\right)\right)\Rightarrow f\left(0\right)$

=0