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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.

IIT JAM is the entrance exam for MSc admission at IIT Mandi. The last round closing ranks varied between 408 and 568 for students belonging to the General AI quota. Therefore, for admission in the last round, the maximum cutoff rank required was 568 for MSc admission. Refer to the table below to view the last round cutoff for 2025: 

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

Vidyavardhini's College of Engineering and Technology accepts JEE Main and MHT CET scores to shortlist candidates for admission to its BE programme. Candidates applying with JEE Main scores are required to clear the cut-off specified for their selected specialisations. The following table presents the specialisations-wise cut-off ranks for various BE specialisations offered at the college (for General category): 

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 Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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}Giventhat\left|\begin{array}{ccc}\hfill 1+x\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1+y\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 1+z\hfill \end{array}\right|=0\\ Takingx,yandzcommonfrom{R}_1,R_{2}and{R}_{3}respectively.\\ \Rightarrow xyz\left|\begin{array}{ccc}\hfill \frac{1}{x}+1\hfill & \hfill \frac{1}{x}\hfill & \hfill \frac{1}{x}\hfill \\ \hfill \frac{1}{y}\hfill & \hfill \frac{1}{y}+1\hfill & \hfill \frac{1}{y}\hfill \\ \hfill \frac{1}{z}\hfill & \hfill \frac{1}{z}\hfill & \hfill \frac{1}{z}+1\hfill \end{array}\right|=0\\ R_1\to R_1+R_2+R_3\\ \Rightarrow xyz\left|\begin{array}{ccc}\hfill \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+1\hfill & \hfill \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+1\hfill & \hfill \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+1\hfill \\ \hfill \frac{1}{y}\hfill & \hfill \frac{1}{y}+1\hfill & \hfill \frac{1}{y}\hfill \\ \hfill \frac{1}{z}\hfill & \hfill \frac{1}{z}\hfill & \hfill \frac{1}{z}+1\hfill \end{array}\right|=0\\ Taking\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+1commonfrom{R}_1\\ \Rightarrow xyz\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+1\right)\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill \frac{1}{y}\hfill & \hfill \frac{1}{y}+1\hfill & \hfill \frac{1}{y}\hfill \\ \hfill \frac{1}{z}\hfill & \hfill \frac{1}{z}\hfill & \hfill \frac{1}{z}+1\hfill \end{array}\right|=0\\ C_1\to C_1-C_2,C_2\to C_2-C_3\\ \Rightarrow xyz\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+1\right)\left|\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 1\hfill & \hfill \frac{1}{y}\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill \frac{1}{z}+1\hfill \end{array}\right|=0\\ Expandingalong{R}_1\\ \Rightarrow xyz\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+1\right)\left[1\left|\begin{array}{cc}\hfill -1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill -1\hfill \end{array}\right|\right]=0\\ \Rightarrow xyz\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+1\right)\left(1\right)=0\\ \Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+1=0andxyz\ne 0\left(x\ne y\ne z\ne 0\right)\\ \therefore x^{-1}+y^{-1}+z^{-1}=-1\\ Hence,thecorrectoptionis\left(d\right).\end{array}$

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}IfAandBaretwoinvertiblematricesthen\\ \left(a\right)adjA=\left|A\right|.A^{-1}iscorrect\\ \left(b\right)det{\left(A\right)}^{-1}={\left[det\left(A\right)\right]}^{-1}=\frac{1}{det\left(A\right)}iscorrect\\ \left(c\right)Also,{\left(AB\right)}^{-1}=B^{-1}{A}^{-1}iscorrect\\ \left(d\right){\left(A+B\right)}^{-1}=\frac{1}{\left|A+B\right|}.adj\left(A+B\right)\\ \therefore {\left(A+B\right)}^{-1}\ne B^{-1}+A^{-1}\\ Hence,thecorrectoptionis\left(d\right).\end{array}$

Yes, Amity School of Fashion Technology does offer an MBA course for the duration of two years at the postgraduate level. Additionally, the course is offered in full-time mode. The school offers an MBA course in one specialisations i.e., an MBA in Fashion Management. To get admission into this course, candidates are required to complete the basic eligibility criteria that state candidates must complete their graduation with a minimum aggregate of 50%. 

This is a multiple choice answer as classified in NCERT Exemplar

(a) The negative refractive index metamaterials are those in which incident ray from air (Medium 1) to them refract or bend differently to that of positive refractive index medium.

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Wehave,\\ A=\left[\begin{array}{ccc}\hfill 2\hfill & \hfill \lambda \hfill & \hfill -3\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 3\hfill \end{array}\right]\Rightarrow \left|A\right|=\left|\begin{array}{ccc}\hfill 2\hfill & \hfill \lambda \hfill & \hfill -3\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 3\hfill \end{array}\right|\\ Expandingalong{R}_1=2\left|\begin{array}{cc}\hfill 2\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill 3\hfill \end{array}\right|-\lambda \left|\begin{array}{cc}\hfill 0\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill 3\hfill \end{array}\right|-3\left|\begin{array}{cc}\hfill 0\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right|\\ =2\left(6-5\right)-\lambda \left(0-5\right)-3\left(0-2\right)\\ =2+5\lambda +6=8+5\lambda \\ If{A}^{-1}existsthen\left|A\right|\ne 0\\ \therefore 8+5\lambda \ne 0So\lambda \ne \frac{-8}{5}\\ Hence,thecorrectoptionis\left(d\right).\end{array}$