Maths NCERT Exemplar Solutions Class 11th Chapter Five

I (6)    F (8)

Case I       2            4

Case II      3            6

Case III     4            8

Total =   

  ${\displaystyle \underset{-a}{\overset{a}{\int }}\left(\left|x\right|+\left|x-2\right|\right)dx=22,a>2}$

${\displaystyle \underset{-a}{\overset{0}{\int }}\left(-2x+2\right)dx+{\displaystyle \underset{0}{\overset{2}{\int }}\left(x-x+2\right)dx+{\displaystyle \underset{2}{\overset{a}{\int }}\left(2x-2\right)dx=22}}}$

$\Rightarrow 2a^2+2=20\Rightarrow a^2=9\Rightarrow a=3$

$\therefore {\displaystyle \underset{3}{\overset{-3}{\int }}\left(x+\left[x\right]\right)dx=-{\displaystyle \underset{-3}{\overset{3}{\int }}\left(2x-\left\{x\right\}\right)dx=}}-{\displaystyle \underset{-3}{\overset{3}{\int }}2xdx+6}{{\displaystyle \underset{0}{\overset{1}{\int }}xdx=6.\frac{x^2}{2}}|}_0^1=3$           

$P=\left[\begin{array}{ccc}\hfill 3\hfill & \hfill -1\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill \alpha \hfill \\ \hfill 3\hfill & \hfill -5\hfill & \hfill 0\hfill \end{array}\right]andQ=\left[q_{ij}\right]\Rightarrow PQ=k{l}_3$           

$q_{23}=-\frac{k}{8}and\left|Q\right|=\frac{k^2}{2}$            

  $PQ=k{l}_3\Rightarrow P^{-1}=\frac{Q}{k}={\left(\begin{array}{ccc}\hfill 3\hfill & \hfill -1\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill \alpha \hfill \\ \hfill 3\hfill & \hfill -5\hfill & \hfill 0\hfill \end{array}\right)}^{-1}$

$\left|p\right|\left|Q\right|=\left(k{l}_3\right)\Rightarrow 8.\frac{k^2}{2}=k^3$

$k\ne 0\Rightarrow k=4$

$\therefore {\alpha }^2+k^2=1+16=17.$         

$\overrightarrow{c}=\alpha \overrightarrow{a}+\beta \overrightarrow{b}.....\left(i\right)$

$\overrightarrow{a}.\overrightarrow{c}=7\overrightarrow{b}.\overrightarrow{c}=0$           

$\overrightarrow{a}=-\widehat{i}+\widehat{j}+\widehat{k}\Rightarrow \left|\overrightarrow{a}\right|=\sqrt[]{3}$          

$\overrightarrow{b}=2\widehat{i}+\widehat{k}\Rightarrow \left|\overrightarrow{b}\right|=\sqrt[]{5}\overrightarrow{a}.\overrightarrow{b}=-2+1=-1$           

From   $\left(i\right)\overrightarrow{a}.\overrightarrow{c}=\alpha {\left|\overrightarrow{a}\right|}^2-\beta$

$3\alpha -\beta =7..........\left(ii\right)$        

$\overline{b}.\overline{c}=\alpha \overline{b}.\overline{a}+\beta {\left|\overrightarrow{b}\right|}^2\Rightarrow -\alpha +5\beta =0.......\left(iii\right)$     

Solving $\alpha =\frac{5}{2}and\beta =\frac{1}{2}$  

$\Rightarrow 2{\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|}^2=75$       

Let P (B₁) = a      P (B₂) = b            P (B₃) = c

Given a (1 – b) (1 – c) = a . (i)

b (1 – a) (1 – c) = b              . (ii)

c (1 – b) (1 – a) = $\gamma$             . (iii)

(1 – a) (1 – b) (1 – c) = p     . (iv)

$\left(\alpha -2\beta \right)p=\alpha \beta$  

->a – ab – 2b + 2ab = ab Þ a = 2b . (v)

Again $\left(\beta -3\gamma \right)p=2\beta \gamma$  

->b – bc – 3c + 3bc = 2bc Þ b = 3c   . (vi)

$\Rightarrow \frac{P\left(B_1\right)}{P\left(B_3\right)}=\frac{a}{c}=\frac{2b}{b/3}=6$      

$x^2+y^2-2x-6y+6=0$  centre (1, 3)

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

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

Sum of elements $A\cap \left(B\cup C\right)=274*400$  

In set B numbers of the form 9k + 2 are {101, 109, .992}

$\therefore sum=\frac{100}{2}\left(101+992\right)=\frac{100*1093}{2}......\left(i\right)$          

Another possible number is 9k + 5 forms are {104, .995}

$\therefore sum=\frac{100}{2}\left(104+995\right)=\frac{100}{2}*1099.........\left(ii\right)$  

$\therefore Total=\frac{100}{2}*\left[1093+1099\right]=100*1096=274*4*100=274*400$           

$\therefore$  possible value of $\mathcal{l}$ = 5

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Letz=1+2i\\ \therefore \left|z\right|=\sqrt[]{{\left(1\right)}^2+{\left(2\right)}^2}=\sqrt[]{5}\\ Nowf\left(z\right)=\frac{7-z}{1-z^2}\\ =\frac{7-\left(1+2i\right)}{1-{\left(1+2i\right)}^2}=\frac{7-1-2i}{1-1-4i^2-4i}=\frac{6-2i}{4-4i}\\ =\frac{3-i}{2-2i}=\frac{3-i}{2-2i}*\frac{2+2i}{2+2i}=\frac{6+6i-2i-2i^2}{4-4i^2}\\ =\frac{6+4i+2}{4+4}=\frac{8+4i}{8}=1+\frac{1}{2}i\\ So,\left|f\left(z\right)\right|=\sqrt[]{{\left(1\right)}^2+{\left(\frac{1}{2}\right)}^2}=\sqrt[]{1+\frac{1}{4}}=\frac{\sqrt[]{5}}{2}=\frac{\left|z\right|}{2}\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Letz=-x+0iandx<0\\ \therefore \left|z\right|=\sqrt[]{{\left(-1\right)}^2+{\left(0\right)}^2}=1,x<0\\ Since,thepoint\left(-x,0\right)liesonthenegativesideoftherealaxis\left(?x<0\right)\\ \therefore Principleargument\left(z\right)=\pi \\ Hence,thecorrectoptionis\left(c\right).\end{array}$