Class 12th

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}GiventhatA=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right]\\ A^2=A.A=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 0+1\hfill & \hfill 0+0\hfill \\ \hfill 0+0\hfill & \hfill 1+0\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\\ Hence,thecorrectoptionis(d).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Asweknowthattheadditionandsubtractionoftwomatricesisonlypossiblewhentheyhave\\ sameorder.Itisalsogiventhatm=n.\\ \therefore Orderof\left(5A-2B\right)is3*n.\\ Hence,thecorrectoptionis(d).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Giventhat:A=\frac{1}{\pi }\left[\begin{array}{cc}\hfill si{n}^{-1}\left(x\pi \right)\hfill & \hfill ta{n}^{-1}\left(\frac{x}{\pi }\right)\hfill \\ \hfill si{n}^{-1}\left(\frac{x}{\pi }\right)\hfill & \hfill co{t}^{-1}\left(\pi x\right)\hfill \end{array}\right]\\ andB=\frac{1}{\pi }\left[\begin{array}{cc}\hfill -co{s}^{-1}\left(x\pi \right)\hfill & \hfill ta{n}^{-1}\left(\frac{x}{\pi }\right)\hfill \\ \hfill si{n}^{-1}\left(\frac{x}{\pi }\right)\hfill & \hfill -ta{n}^{-1}\left(\pi x\right)\hfill \end{array}\right]\\ A-B=\frac{1}{\pi }\left[\begin{array}{cc}\hfill si{n}^{-1}\left(x\pi \right)\hfill & \hfill ta{n}^{-1}\left(\frac{x}{\pi }\right)\hfill \\ \hfill si{n}^{-1}\left(\frac{x}{\pi }\right)\hfill & \hfill co{t}^{-1}\left(\pi x\right)\hfill \end{array}\right]-\frac{1}{\pi }\left[\begin{array}{cc}\hfill -co{s}^{-1}\left(x\pi \right)\hfill & \hfill ta{n}^{-1}\left(\frac{x}{\pi }\right)\hfill \\ \hfill si{n}^{-1}\left(\frac{x}{\pi }\right)\hfill & \hfill -ta{n}^{-1}\left(\pi x\right)\hfill \end{array}\right]\\ =\frac{1}{\pi }\left[\begin{array}{cc}\hfill si{n}^{-1}\left(x\pi \right)+co{s}^{-1}\left(x\pi \right)\hfill & \hfill ta{n}^{-1}\left(\frac{x}{\pi }\right)-ta{n}^{-1}\left(\frac{x}{\pi }\right)\hfill \\ \hfill si{n}^{-1}\left(\frac{x}{\pi }\right)-si{n}^{-1}\left(\frac{x}{\pi }\right)\hfill & \hfill co{t}^{-1}\left(\pi x\right)+ta{n}^{-1}\left(\pi x\right)\hfill \end{array}\right]\\ =\frac{1}{\pi }\left[\begin{array}{cc}\hfill \frac{\pi }{2}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \frac{\pi }{2}\hfill \end{array}\right]\left[\begin{array}{l}?si{n}^{-1}x+co{s}^{-1}x=\frac{\pi }{2}\\ ta{n}^{-1}x+co{t}^{-1}x=\frac{\pi }{2}\end{array}\right]\\ =\frac{1}{\pi }*\frac{\pi }{2}\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=\frac{1}{2}I\\ Hence,thecorrectoptionis(d).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Giventhat:\left[\begin{array}{cc}\hfill 2x+y\hfill & \hfill 4x\hfill \\ \hfill 5x-7\hfill & \hfill 4x\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 7\hfill & \hfill 7y-13\hfill \\ \hfill y\hfill & \hfill x+6\hfill \end{array}\right]\\ Equatingthecorrespondingelements,weget,\\ 2x+y=7\dots \left(i\right)\\ and4x=7y-13\dots \left(ii\right)\\ fromeqn.\left(ii\right)4x-x=6\\ 3x=6\\ \therefore x=2\\ fromeqn.\left(i\right)2*2+y=7\\ 4+y=7\therefore y=7-4=3\\ Hence,thecorrectoptionis(b).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Totalnumberofpossiblematricesoforder3*3witheachentry0or2=2^{3*3}=2^9=512.\\ Hence,thecorrectoptionis(d).\end{array}$

Sol. (a) In extraction of iron lime stone is added on a flux

$\begin{array}{rr}& {\mathrm{C}\mathrm{a}\mathrm{C}\mathrm{O}}_3?\mathrm{C}\mathrm{a}\mathrm{O}+{\mathrm{C}\mathrm{O}}_2\\ & {\mathrm{C}\mathrm{O}}_2+{\mathrm{S}\mathrm{i}\mathrm{O}}_2?{\mathrm{C}\mathrm{a}\mathrm{S}\mathrm{i}\mathrm{O}}_3\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}GiventhatA=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 4\hfill \\ \hfill 0\hfill & \hfill 4\hfill & \hfill 0\hfill \\ \hfill 4\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]\\ Here,numberofcolumnsandthenumberofrowsareequali.e.,3.So,Aisasquarematrix.\\ Hence,thecorrectoptionis(a).\end{array}$

Modulation Index,  $\mu =\frac{A_m}{A_C}$

Variation = $2A_m=8\Rightarrow A_m=4v$

$A_m+A_c=9$

$A_C=9-A_m=5v$

$\therefore \mu =\frac{4}{5}=0.8$

 $\Delta E_1=-\frac{E_0}{4}+\frac{E_0}{1}=\frac{3}{4}{E}_0$

$\Delta E_2=0-\left(-E_0\right)=E_0$

$\frac{\Delta E_1}{\Delta E_2}=\frac{3}{4}$