Class 12th

This is a  Fill in the blanks as classified in NCERT Exemplar

Sol:

$\begin{array}{l}WeknowthatforaninvertiblematrixAofanyorder,\\ \left|A^{-1}\right|=\frac{1}{\left|A\right|}\end{array}$

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

Sol:

$\begin{array}{l}\underset{x\to 0^{-}}{lim}f\left(x\right)=\left|x\right|cos\frac{1}{x}\\ =\underset{h\to 0}{lim}\left|0-h\right|cos\frac{1}{\left(0-h\right)}=\underset{h\to 0}{lim}h.cos\frac{1}{h}=0\left[?cos\frac{1}{x}oscillatebetween-1and1\right]\\ \underset{x\to 0^{+}}{lim}f\left(x\right)=\left|x\right|cos\frac{1}{x}=\underset{h\to 0}{lim}\left|0+h\right|cos\frac{1}{\left(0+h\right)}=\underset{h\to 0}{lim}h.cos\frac{1}{h}=0\\ \underset{x\to 0}{lim}f\left(x\right)=0\\ \therefore \underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0}{lim}f\left(x\right)=0\\ Hence,f\left(x\right)iscontinuousatx=0.\end{array}$

This is a multiple choice answer as classified in NCERT Exemplar

(a, d) Alexandar's dark band lies between the primary and secondary rainbows, forms due to light Scattered into this region interfere destructively.

Since, primary rainbows subtends an angle nearly 41° to 42° at observer's eye, whereas, secondary rainbows subtends an angle nearly 51° to 54° at observer's eye w.r.t. incident light ray. So, the scattered rays with respect to the incident light of the sun lies between approximately 42°and50°.

This is a  Fill in the blanks as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Weknowthatforamatrixoforder3*3,\\ \left|KA\right|=K^3\left|A\right|\\ \therefore \left|3A\right|=3^3\left|A\right|=27\left|A\right|\end{array}$

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

Sol:

$\begin{array}{l}\underset{x\to 4^{-}}{lim}f\left(x\right)=\frac{\left|x-4\right|}{2\left(x-4\right)}\left[\begin{array}{l}forx<4,\left|x-4\right|=-\left(x-4\right)\\ forx>4,\left|x-4\right|=\left(x-4\right)\end{array}\right]\\ =\underset{h\to 0}{lim}\frac{-\left[4-h-4\right]}{2\left[4-h-4\right]}=\underset{h\to 0}{lim}\frac{h}{-2h}=-\frac{1}{2}\\ \underset{x\to 4^{+}}{lim}f\left(x\right)=\frac{\left|x-4\right|}{2\left(x-4\right)}=\underset{h\to 0}{lim}\frac{\left[4+h-4\right]}{2\left[4+h-4\right]}=\underset{h\to 0}{lim}\frac{h}{2h}=\frac{1}{2}\\ \underset{x\to 4}{lim}f\left(x\right)=0\\ \therefore \underset{x\to 4^{-}}{lim}f\left(x\right)\ne \underset{x\to 4^{+}}{lim}f\left(x\right)\ne \underset{x\to 4}{lim}f\left(x\right)\\ Hence,f\left(x\right)isdiscontinuousatx=4.\end{array}$

This is a multiple choice answer as classified in NCERT Exemplar

(d) For µ = 16., the critical angle, µ = 1/ sin C, we have C = 38.7°, when viewed from AD, as long as angle of incidence on AD of the ray emanating from pin is greater than the critical angle, the light suffers from total internal reflection and cannot be seen through AD.

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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

Sol:

$\begin{array}{l}f\left(x\right)=\frac{2x^2-3x-2}{x-2}\\ =\frac{2x^2-4x+x-2}{x-2}=\frac{2x\left(x-2\right)+1\left(x-2\right)}{x-2}\\ =\frac{\left(2x+1\right)\left(x-2\right)}{x-2}=2x+1\\ \underset{x\to 2^{-}}{lim}f\left(x\right)=2x+1=\underset{h\to 0}{lim}2\left(2-h\right)+1=4+1=5\\ \underset{x\to 2^{+}}{lim}f\left(x\right)=2x+1=\underset{h\to 0}{lim}2\left(2+h\right)+1=4+1=5\\ \underset{x\to 2}{lim}f\left(x\right)=5\\ As\underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{x\to 2^{+}}{lim}f\left(x\right)=\underset{x\to 2}{lim}f\left(x\right)=5\\ Hence,f\left(x\right)iscontinuousatx=2.\end{array}$

This is a multiple choice answer as classified in NCERT Exemplar

(a, b, c) When immersed object is seen from close to the edge of the trough the object looks distorted because the apparent depth of the points close to the edge are nearer the surface of the water compared to the points away from the edge.

The angle subtended by the image of the object at the eye is smaller than the actual angle subtended by the object in air and some of the points of the object far away from the edge may not be visible because of total internal reflection.

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Let\Delta =\left|\begin{array}{ccc}\hfill x\hfill & \hfill x+y\hfill & \hfill x+2y\hfill \\ \hfill x+2y\hfill & \hfill x\hfill & \hfill x+y\hfill \\ \hfill x+y\hfill & \hfill x+2y\hfill & \hfill x\hfill \end{array}\right|\\ C_1\to C_1+C_2+C_3\\ =\left|\begin{array}{ccc}\hfill 3x+3y\hfill & \hfill x+y\hfill & \hfill x+2y\hfill \\ \hfill 3x+3y\hfill & \hfill x\hfill & \hfill x+y\hfill \\ \hfill 3x+3y\hfill & \hfill x+2y\hfill & \hfill x\hfill \end{array}\right|\\ =\left(3x+3y\right)\left|\begin{array}{ccc}\hfill 1\hfill & \hfill x+y\hfill & \hfill x+2y\hfill \\ \hfill 1\hfill & \hfill x\hfill & \hfill x+y\hfill \\ \hfill 1\hfill & \hfill x+2y\hfill & \hfill x\hfill \end{array}\right|\left[Taking\left(3x+3y\right)commonfrom{C}_1\right]\\ R_1\to R_1-R_2,R_2\to R_2-R_3\\ \Rightarrow =\left(3x+3y\right)\left|\begin{array}{ccc}\hfill 0\hfill & \hfill y\hfill & \hfill y\hfill \\ \hfill 0\hfill & \hfill -2y\hfill & \hfill y\hfill \\ \hfill 1\hfill & \hfill x+2y\hfill & \hfill x\hfill \end{array}\right|\\ Expandingalong{C}_1\\ \Rightarrow =3\left(x+y\right)\left[1\left|\begin{array}{cc}\hfill y\hfill & \hfill y\hfill \\ \hfill -2y\hfill & \hfill y\hfill \end{array}\right|\right]\\ \Rightarrow =3\left(x+y\right)\left(y^2+2y^2\right)\Rightarrow 3\left(x+y\right)\left(3y^2\right)\Rightarrow 9y^2\left(x+y\right)\\ Hence,thecorrectoptionis\left(b\right).\end{array}$