Ncert Solutions Maths class 11th

$\overline{x}=\frac{{\displaystyle \sum _{i=1}^{10}{x}_i}}{10}=15;\frac{{\displaystyle \sum _{i=1}^{10}{x_i}^2}}{10}-{\left(\overline{x}\right)}^2=15$

$\Rightarrow \Sigma x_i=150;\Sigma {x_i}^2=2400$

Actual mean $\overline{x}=\frac{\Sigma x_i+15-25}{10}=\frac{140}{10}=14$

Actual variance =  $\frac{\Sigma {x_i}^2+15^2-25^2}{10}-{\left(14\right)}^2$

$=\frac{2400-400}{10}-196$

$\begin{array}{l}{\sigma }^2=4\\ \sigma =2\end{array}$

Circle passes through (6, 1)

12 g – 19 c = 43               …. (i)

Centre lies on x – 2xy = 8

->g + 6c = 8                     …. (ii)

From (i) & (ii), c = 1, 9 = 2

Length of x – intercept -  $2\sqrt[]{g^2-C}$

$\frac{{\Delta }_1}{{\Delta }_2}=\frac{\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill x\hfill & \hfill -4x-x\hfill & \hfill 1\hfill \\ \hfill -4\hfill & \hfill 3\hfill & \hfill 1\hfill \end{array}\right|}{\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill -4\hfill & \hfill 3\hfill & \hfill 1\hfill \\ \hfill -2\hfill & \hfill -5\hfill & \hfill 1\hfill \end{array}\right|}=\frac{4}{7}$

->14 x – 35 y = -95        …. (ii)

Solve (i) & (ii), x =    $\frac{-20}{7},y=\frac{-11}{7}$

$ar\Delta AQR$

$=\frac{1}{2}*1*1=\frac{1}{2}$                

Solve tan 2a = $\frac{h}{b}$  

$tan\alpha =\frac{2h}{\sqrt[]{7}h+b}$

tangent at (2t², 4t) is ty = x + 2t²,

It passes through (5, 7)

$2t^2-7t+5=0\Rightarrow t=1,\frac{5}{2}$                

$P\left(2t^2,4t\right)$ will be (2, 4), $\left(\frac{25}{2},0\right)$  

$\frac{S_5}{S_9}=\frac{5}{17}\Rightarrow d=4a$  

110 < a₁₅ < 120

110 < a + 14d < 120

110 < 57a < 120

->a = 2, d = 8

Use binomial theorem

2023 = 7 * 289

$2021^{2022}+2022^{2021}={\left(2022-2\right)}^{2022}+{\left(2023-1\right)}^{2021}$

$=7P_1+2^{2022}+7P_2-1$

$=7\left(P_1+P_2\right)+1+7P_3-1$                          

Let the equation of circle be

$x\left(x-\frac{1}{2}\right)+y^2+\lambda y=0$               

$\Rightarrow x^2+y^2-\frac{1}{2}x+\lambda y=0$

Radius = $\sqrt[]{\frac{1}{16}+\frac{{\lambda }^2}{4}}=2$  

$\Rightarrow {\lambda }^2=\frac{63}{4}\Rightarrow {\left(x-\frac{1}{4}\right)}^2+{\left(y+\frac{\lambda }{2}\right)}^2=4$   

$?$ This circle and parabola

$y-\alpha ={\left(x-\frac{1}{4}\right)}^2$ touch each other, so

$\alpha =-\frac{\lambda }{2}+2\Rightarrow \alpha -2=-\frac{\lambda }{2}\Rightarrow {\left(\alpha -2\right)}^2=\frac{{\lambda }^2}{4}=\frac{63}{16}$  

${\left(4\alpha -8\right)}^2=63$  

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

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

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

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

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

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

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

$\therefore m+m=276$