Class 12th

Option 'C' is correct as determinant is a number associated to a square matrix.

This is a Long Answer Type Questions as classified in NCERT Exemplar

in first case crowding spectrum is visible

If we want more wave to be modulated then more crowding will occur and more mixing up of signal.

But we want to accommodate this we use higher band width and frequency career wave

Then, KA = $k\left[\begin{array}{ccc}\hfill a_{11}\hfill & \hfill a_{12}\hfill & \hfill a_{13}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill & \hfill a_{23}\hfill \\ \hfill a_{31}\hfill & \hfill a_{32}\hfill & \hfill a_{33}\hfill \end{array}\right]$

$=\left[\begin{array}{ccc}\hfill k{a}_{11}\hfill & \hfill k{a}_{12}\hfill & \hfill k{a}_{13}\hfill \\ \hfill k{a}_4\hfill & \hfill k{a}_{22}\hfill & \hfill k{a}_{23}\hfill \\ \hfill k{a}_{31}\hfill & \hfill k{a}_{32}\hfill & \hfill k{a}_{33}\hfill \end{array}\right]$

$\therefore |KA|=\left|\begin{array}{ccc}\hfill {}_{a11}\hfill & \hfill k{a}_{12}\hfill & \hfill {}_{13}\hfill \\ \hfill {}_{a_{21}}\hfill & \hfill {}_{22}\hfill & \hfill k{a}_{23}\hfill \\ \hfill {}_{31}\hfill & \hfill {}_{a_{32}}\hfill & \hfill k_{a_{33}}\hfill \end{array}\right|$

$=k{}^3\left|\begin{array}{ccc}\hfill a_{11}\hfill & \hfill a_{12}\hfill & \hfill a_{13}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill & \hfill a_{22}\hfill \\ \hfill a_{31}\hfill & \hfill a_{32}\hfill & \hfill a_{33}\hfill \end{array}\right|$

$=k^3|A|$

So, option c is correct.

This is a Long Answer Type Questions as classified in NCERT Exemplar

Maximum voltage = 100/2 = 50V

And minimum voltage = 20/2 = 10V

Percentage modulation= max voltage -min voltage/ max voltage + min voltage $*100$

= $\frac{50-10}{50+10}*100$ = 66.67%

Peak career voltage= max voltage+ min voltage/2= 50+10/2=30V

Peak value of information voltage= 66.67/100 $*$  30= 20V

This is a Long Answer Type Questions as classified in NCERT Exemplar

height of satellite hs= 600km
As we know velocity = distance/time

 2x/4.0410^-3 = 3⁸

So x=606 km after solving 
According to Pythagoras theorem d²=x²-h²= 606²-600²= 7236

So d= 85.06km so total distance will be double from receiver to transmitter = 170.1km
And d = √2Rh
 h=7236/2*6400 = 565m

LHS = ${\left|\begin{array}{ccc}\hfill a^2+1\hfill & \hfill ab\hfill & \hfill ac\hfill \\ \hfill ab\hfill & \hfill b^2+1\hfill & \hfill bc\hfill \\ \hfill ca\hfill & \hfill cb\hfill & \hfill c^2+1\hfill \end{array}\right|}_{.}$

$=\frac{1}{abc}\left|\begin{array}{ccc}\hfill a\left(a^2+1\right)\hfill & \hfill a{b}^2\hfill & \hfill a{c}^2\hfill \\ \hfill a^{2}b\hfill & \hfill b\left(b^2+1\right)\hfill & \hfill b{c}^2\hfill \\ \hfill a^{2}c\hfill & \hfill b^{2}c\hfill & \hfill c\left(c^2+1\right)\hfill \end{array}\right|$

$=\frac{abc}{abc}\left[\begin{array}{ccc}\hfill a^2+1\hfill & \hfill b^2\hfill & \hfill c^2\hfill \\ \hfill a^2\hfill & \hfill b^2+1\hfill & \hfill c^2.\hfill \\ \hfill a^2\hfill & \hfill b^2\hfill & \hfill c^2+1\hfill \end{array}\right]$Taking a, b&c common from R₁, R₂&R₃

$\left[\begin{array}{ccc}\hfill 1+a^2+b^2+c^2\hfill & \hfill b^2\hfill & \hfill c^2\hfill \\ \hfill a^2+b^2+1+c^2\hfill & \hfill b^2+1\hfill & \hfill c^2\hfill \\ \hfill a^2+b^2+c^2+1\hfill & \hfill b^2\hfill & \hfill c^2+1\hfill \end{array}\right]$ C_1→ C₂ + C₃.

This is a Long Answer Type Questions as classified in NCERT Exemplar

as we know that I= I_{0 $e^{-\propto x}$}

And I= 25%of I₀=

I=I₀/4

I₀/4= I_{0 $e^{-\propto x}$}

I₀ cancel from both sides

¼= $e^{-\propto x}$

Taking log on both sides log1 -log4= - $\propto x$ loge

X= log4/ $\propto$

LHS = ${\left|\begin{array}{ccc}\hfill 1+a^2-b^2\hfill & \hfill 2ab\hfill & \hfill -2b\hfill \\ \hfill 2ab\hfill & \hfill 1-a^2+b^2\hfill & \hfill 2a\hfill \\ \hfill 2b\hfill & \hfill -2a\hfill & \hfill 1-a^2-b^2\hfill \end{array}\right|}_{.}$

$=\left|\begin{array}{ccc}\hfill 1+a^2-b^2-b(-2b)\hfill & \hfill 2ab+a(-2b)\hfill & \hfill -2b\hfill \\ \hfill 2ab-b(2a)\hfill & \hfill 1-a^2+b^2+a(2a)\hfill & \hfill 2a\hfill \\ \hfill 2b-b\left(1-a^2-b^2\right)\hfill & \hfill -2a+a\left(1-a^2-b^2\right)\hfill & \hfill 1-a^2-b^2\hfill \end{array}\right|$

$=\left|\begin{array}{ccc}\hfill 1+a^2-b^2+2b^2\hfill & \hfill 2ab-2ab\hfill & \hfill -2b\hfill \\ \hfill 2ab-2ab\hfill & \hfill 1-a^2+b^2+2a^2\hfill & \hfill 2a\hfill \\ \hfill 2b-b+a^{2}b+b^3\hfill & \hfill -2a+a-a^3-a{b}^2\hfill & \hfill 1-a^2-b^2\hfill \end{array}\right|$

$=\left|\begin{array}{ccc}\hfill 1+a^2+b^2\hfill & \hfill 0\hfill & \hfill -2b\hfill \\ \hfill 0\hfill & \hfill 1+a^2+b^2\hfill & \hfill 2a\hfill \\ \hfill b\left(1+a^2+b^2\right)\hfill & \hfill -a\left(1+a^2+b^2\right)\hfill & \hfill 1-a^2-b^2\hfill \end{array}\right|$

= (1 + a² + b²)² [(1 -a² + b²) - 2a (-a)]

= (1 + a² + b²)² (1 -a² + b² + 2a²)

= (1 + a² + b²)² (1 + a² + b²)

= (1 + a² + b²)³ = R.H.S.

LHS = $\left|\begin{array}{ccc}\hfill 1\hfill & \hfill x\hfill & \hfill x^2\hfill \\ \hfill x^2\hfill & \hfill 1\hfill & \hfill x\hfill \\ \hfill x\hfill & \hfill x^2\hfill & \hfill 1\hfill \end{array}\right|$

$=\left|\begin{array}{ccc}\hfill 1+x^2+x\hfill & \hfill x+1+x^2\hfill & \hfill x^2+x+1\hfill \\ \hfill x^2\hfill & \hfill 1\hfill & \hfill x\hfill \\ \hfill x\hfill & \hfill x^2\hfill & \hfill 1\hfill \end{array}\right|{}_{\mathrm{}}$R_1→ R₁ + R₂ + R₃

= (1 + x² + x) (1 -x)² [ (1 + x)* 1 - (-x) x].

= (1 + x² + x) (1 -x)² (1 + x + x²).

= { (1 + x² + x) (1 -x)}²

= {1 -x + x²-x³ + x-x²}²

= (1 -x³)² = R.H.S.

(i) LHS = $\left|\begin{array}{ccc}\hfill a-b-c\hfill & \hfill 2a\hfill & \hfill 2a\hfill \\ \hfill 2b\hfill & \hfill b-c-a\hfill & \hfill 2b\hfill \\ \hfill 2c\hfill & \hfill 2c\hfill & \hfill c-a-b\hfill \end{array}\right|$

$=\left|\begin{array}{ccc}\hfill a-b-c+2b+2c\hfill & \hfill 2a+b-c-a+2c\hfill & \hfill 2a+2b+c-ab\hfill \\ \hfill 2b\hfill & \hfill b-c-a\hfill & \hfill 2b\hfill \\ \hfill 2c\hfill & \hfill 2c\hfill & \hfill c-a-b\hfill \end{array}\right|$R_1→ R₁ + R₂ + R₃

$=\left|\begin{array}{ccc}\hfill a+b+c\hfill & \hfill a+b+c\hfill & \hfill a+b+c\hfill \\ \hfill 2b\hfill & \hfill b-c-a\hfill & \hfill 2b\hfill \\ \hfill 2c\hfill & \hfill 2c\hfill & \hfill c-a-b\hfill \end{array}\right|$

= (a + b + c) $\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 2b\hfill & \hfill b-c-a\hfill & \hfill 2b\hfill \\ \hfill 2c\hfill & \hfill 2c\hfill & \hfill c-a-b\hfill \end{array}\right|$ Taking (a + b + c) common from R₁

= (a + b+ c) $\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1-1\hfill & \hfill 1-1\hfill \\ \hfill 2b\hfill & \hfill b-c-a-2b\hfill & \hfill 2b-2b\hfill \\ \hfill 2c\hfill & \hfill 2c-2c\hfill & \hfill c-a-b-2c\hfill \end{array}\right|\begin{array}{cc}\hfill {}_{}\to {}_2-{}_1\hfill & \hfill {}_{}\to {}_3-{}_1\hfill \end{array}$

= (a + b + c) $\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 2b\hfill & \hfill -b-c-a\hfill & \hfill 0\hfill \\ \hfill 2c\hfill & \hfill 0\hfill & \hfill -c-a-b\hfill \end{array}\right|.$

= (a + b + c) * 1. $\left|\begin{array}{cc}\hfill -(a+b+c)\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -(a+b+c)\hfill \end{array}\right|$ Expand along R₁

= (a + b + c){(a + b + c)²- 0}

= (a + b +c)³ = R.H.S

(ii) LHS = ${\left|\begin{array}{ccc}\hfill x+y+2z\hfill & \hfill x\hfill & \hfill y\hfill \\ \hfill z\hfill & \hfill y+z+2x\hfill & \hfill y\hfill \\ \hfill z\hfill & \hfill x\hfill & \hfill z+x+2y\hfill \end{array}\right|}_{\cdot }$

$=|\begin{array}{ccc}\hfill x+y+2z+x+y\hfill & \hfill x\hfill & \hfill y\hfill \\ \hfill z+y+z+2x+y\hfill & \hfill y+z+2x\hfill & \hfill y\hfill \\ \hfill z+x+z+x+2y\hfill & \hfill x\hfill & \hfill z+x+2y\hfill \end{array}$
C_1→ C₁ + C₂ + C₃.

$={\left|\begin{array}{ccc}\hfill 2(x+y+z)\hfill & \hfill x\hfill & \hfill y\hfill \\ \hfill 2(x+y+z)\hfill & \hfill y+z+2x\hfill & \hfill y\hfill \\ \hfill 2(x+y+z)\hfill & \hfill x\hfill & \hfill z+x+2y\hfill \end{array}\right|}_{.}$

= 2 (k + y + z) $\left|\begin{array}{ccc}\hfill 1\hfill & \hfill x\hfill & \hfill y\hfill \\ \hfill 1\hfill & \hfill y+z+2x\hfill & \hfill y\hfill \\ \hfill 1\hfill & \hfill x\hfill & \hfill z+x+2y\hfill \end{array}\right|$ Taking 2