Class 12th

Given,

Area of triangle = 35 sq. Units

$\Rightarrow \frac{1}{2}\left|\begin{array}{ccc}\hfill 2\hfill & \hfill -6\hfill & \hfill 1\hfill \\ \hfill 5\hfill & \hfill 4\hfill & \hfill 1\hfill \\ \hfill k\hfill & \hfill 4\hfill & \hfill 1\hfill \end{array}\right|=35.$

∴ Option D is correct.

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

Maximum amplitude = Am+Ac =12 while minimum amplitude is Ac-Am=3

Using elimination method = 2Ac= 18

Ac=9 and Am= 6V

So modulating index m = 6/9= 2/3

(i) Let P (x, y) be any point on line joining A (1, 2) & B (3, 6)

Then, area of triangle (ABP) = 0 {the point are collinear

$\frac{1}{2}\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 6\hfill & \hfill 1\hfill \\ \hfill x\hfill & \hfill y\hfill & \hfill 1\hfill \end{array}\right|=0$

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

As frequency of B is more than A so it has more refractive index also and if a wave have higher refractive index then it has less angle of refraction. So wave B travel more in ionosphere.

(i) Area of the triangle = 4 sq. units (given)

$\Rightarrow \frac{1}{2}\left|\begin{array}{ccc}\hfill k\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 4\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 1\hfill \end{array}\right|=4$

(ii) Area of the triangle = 4 sq units

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

When we send a signal to be transmitted through sky waves must have a frequency range of 1710 kHz to 40 MHz.

But, the frequency of TV signals are 60 MHz which is beyond the required range.

So, sky waves will not be suitable for transmission of TV signals of 60 MHz frequency.

The area of triangle from by the given points is area ( ΔABC) = $\frac{1}{2}$ $\left|\begin{array}{ccc}\hfill a\hfill & \hfill b+c\hfill & \hfill 1\hfill \\ \hfill b\hfill & \hfill c+a\hfill & \hfill 1\hfill \\ \hfill c\hfill & \hfill a+b\hfill & \hfill 1\hfill \end{array}\right|$

$=\frac{1}{2}\left|\begin{array}{ccc}\hfill a+b+c+1\hfill & \hfill b+c\hfill & \hfill 1\hfill \\ \hfill b+c+a+1\hfill & \hfill c+a\hfill & \hfill 1\hfill \\ \hfill c+a+b+1\hfill & \hfill a+b\hfill & \hfill 1\hfill \end{array}\right|$C_1→ C₁ + C₂ + C₃

$=\frac{1}{2}(a+b+c+1)\left|\begin{array}{ccc}\hfill 1\hfill & \hfill b+c\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill c+a\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill a+b\hfill & \hfill 1\hfill \end{array}\right|$ Taking (a + b + c + 1) common from C₁

$=\frac{(a+b+c+1)}{2}*0\left\{?c=c_3\right\}$

= 0

Hence the points A, B C are collinear.

(i) Area of triangle is given by,

Δ = $\frac{1}{2}$ $\left|\begin{array}{ccc}\hfill x_1\hfill & \hfill y_1\hfill & \hfill 1\hfill \\ \hfill x_2\hfill & \hfill y_2\hfill & \hfill 1\hfill \\ \hfill x_3\hfill & \hfill y_3\hfill & \hfill 1\hfill \end{array}\right|=\frac{1}{2}\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 6\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 4\hfill & \hfill 3\hfill & \hfill 1\hfill \end{array}\right|$

$=\frac{1}{2}|[-3+18]|=\frac{15}{2}$ =7.5sq. units.

(ii) Area of the triangle is given by,

Δ = $\frac{1}{2}$ $\left|\begin{array}{ccc}\hfill 2\hfill & \hfill 7\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 10\hfill & \hfill 8\hfill & \hfill 1\hfill \end{array}\right|$

$=\frac{1}{2}|[2(1-8)-7(1-10)+1(8-10)]|$

$=\frac{1}{2}|(2*(-7)-7*(-9)+21*(-2)]|$

$=\frac{1}{2}|[-14+63-2]|=\frac{1}{2}\left|47\right|$

= $\frac{47}{2}$ =23.5 sq. units

(iii) Area of triangle is given by,

Δ = $\frac{1}{2}$ ${\left|\begin{array}{ccc}\hfill -2\hfill & \hfill -3\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill -8\hfill & \hfill 1\hfill \end{array}\right|}_{.}$

$=\frac{1}{2}|[-2*10+3(4)+(-24+2)]|$

$=\frac{1}{2}|[-20+12-22]|$

$=\frac{1}{2}|-30|=\frac{30}{2}$  = 15 sq. units.

Frequency of career wave is 20MHz

Bandwidth require for modulation is 3kHz/2= 1.5kHz

For demodulating we need to reciprocal it

1/f= 1/20MHz= 0.5 $*$ 10^-7s

For modulation = 1/1.5KHz= 0.7 $*$ 10³s

According to first option =RC= 1 $*$ 0.01= 10^-5s

So it will demodulated

According to 2^nd option- RC= 10^-4s

So it can also be demodulated

According to third option – RC= 10^-8s

So it cannot be modulated

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