Ncert Solutions Maths class 11th

A  (A -> B) -> B is a tautology

According to question,

$\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{1}{4}$

$\frac{1}{a}+\frac{1}{b}=\frac{1}{2}........(i)$           

Equation of required line is $\frac{x}{a}+\frac{y}{b}=1$  

Obviously B (2, 2) satisfying condition (i)

$h=\frac{a\left(1+t^2\right)}{2}.......(i)$

k = at        ……… (ii)

From (i) & (ii) $\frac{2h}{a}=1+\frac{k^2}{a^2}$  

$\therefore$ required locus of Q is y² = 2a (x – a/2)

Equation of directrix x – a/2 = -a/2-> x = 0

I (6)    F (8)

Case I       2            4

Case II      3            6

Case III     4            8

Total =   

$x^2+y^2-2x-6y+6=0$  centre (1, 3)

$\begin{array}{l}r=\sqrt[]{1+9-6}=2\\ CM=\sqrt[]{1+4}=\sqrt[]{5}\end{array}$

$r=\sqrt[]{5+4}=3$

$\frac{4}{sinx}+\frac{1}{1-sinx}=a\forall x\in \left(0,\frac{\pi }{2}\right)$  

 Let sin x = t, t $\in$ (0, 1)

$g\left(t\right)=\frac{4}{t}+\frac{1}{1-t}$

$g\text{'}\left(t\right)=0\Rightarrow t=\frac{2}{3}$

$g\text{'}\text{'}\left(\frac{2}{3}\right)>0$

$\therefore g{\left(t\right)}_{minimum}=\frac{4}{2/3}+\frac{1}{1-2/3}=9$  

 Minimum value of a for which solution exist = 9

  $z+\alpha \left|z-1\right|+2i=0$

Let z = x + iy

$\Rightarrow x+i\left(y+2\right)=-\alpha \sqrt[]{{\left(x-1\right)}^2+y^2}$            

 for $\alpha \in Ry+2=0\Rightarrow y=-2$  

$x^2={\alpha }^2\left[{\left(x-1\right)}^2+4\right]$      = 0

$\frac{1}{{\alpha }^2}\ge \frac{4}{5}\Rightarrow {\alpha }^2\le \frac{5}{4}\Rightarrow \frac{-\sqrt[]{5}}{2}\le \alpha \le \frac{\sqrt[]{5}}{2}$

$4\left(p^2+q^2\right)=4\left(\frac{5}{4}+\frac{5}{4}\right)=10$

  ${\displaystyle \sum _{r=1}^{n}ta{n}^{-1}\frac{1}{1+r+r^2}={\displaystyle \sum _{r=1}^{n}ta{n}^{-1}\frac{\left(r+1\right)}{1+r\left(r+1\right)}}}$

$=ta{n}^{-1}2-ta{n}^{-1}+......+ta{n}^{-1}\left(n+1\right)-ta{n}^{-1}n$            

For n  $\infty$ value = $\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$  

$\therefore tan\left(\frac{\pi }{4}\right)=1$           

$M=\left[\begin{array}{ccc}\hfill a_1\hfill & \hfill a_2\hfill & \hfill a_3\hfill \\ \hfill b_1\hfill & \hfill b_2\hfill & \hfill c_3\hfill \\ \hfill c_1\hfill & \hfill c_2\hfill & \hfill c_3\hfill \end{array}\right]$    

$M^{T}M=\left[\begin{array}{ccc}\hfill a_1\hfill & \hfill b_1\hfill & \hfill c_1\hfill \\ \hfill a_2\hfill & \hfill b_2\hfill & \hfill c_2\hfill \\ \hfill a_3\hfill & \hfill b_3\hfill & \hfill c_3\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill a_1\hfill & \hfill a_2\hfill & \hfill a_3\hfill \\ \hfill b_1\hfill & \hfill b_2\hfill & \hfill b_3\hfill \\ \hfill c_1\hfill & \hfill c_2\hfill & \hfill c_3\hfill \end{array}\right]$

$Tr\left(M^{T}M\right)=a_1^2+b_1^2+c_1^2+a_2^2+b_2^2+c_2^2+a_3^2+b_3^2+c_3^2=7$           

all   $a_i,b_i,c_i\in \left\{0,1,2\right\}for$ i = 1, 2, 3

Case 1 7 one's and two zeroes which can occur in ways

Case 2 One 2 three 1's five zeroes =

total such matrices = 504 + 36 = 540

  ${\displaystyle \sum _{r=1}^{n}ta{n}^{-1}\frac{1}{1+r+r^2}={\displaystyle \sum _{r=1}^{n}ta{n}^{-1}\frac{\left(r+1\right)}{1+r\left(r+1\right)}}}$

$=ta{n}^{-1}2-ta{n}^{-1}+......+ta{n}^{-1}\left(n+1\right)-ta{n}^{-1}n$            

For n  $\infty$ value = $\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$  

$\therefore tan\left(\frac{\pi }{4}\right)=1$