Class 12th

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

Sol:

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

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

Sol:

$\begin{array}{l}Wehave,\left|\begin{array}{ccc}\hfill x\hfill & \hfill 3\hfill & \hfill 7\hfill \\ \hfill 2\hfill & \hfill x\hfill & \hfill 2\hfill \\ \hfill 7\hfill & \hfill 6\hfill & \hfill x\hfill \end{array}\right|=0\\ Expandingalong{R}_1\\ \Rightarrow x\left|\begin{array}{cc}\hfill x\hfill & \hfill 2\hfill \\ \hfill 6\hfill & \hfill x\hfill \end{array}\right|-3\left|\begin{array}{cc}\hfill 2\hfill & \hfill 2\hfill \\ \hfill 7\hfill & \hfill x\hfill \end{array}\right|+7\left|\begin{array}{cc}\hfill 2\hfill & \hfill x\hfill \\ \hfill 7\hfill & \hfill 6\hfill \end{array}\right|=0\\ \Rightarrow x\left(x^2-12\right)-3\left(2x-14\right)+7\left(12-7x\right)=0\\ \Rightarrow x^3-12x-6x+42+84-49x=0\\ \Rightarrow x^3-67x+126=0\dots \left(1\right)\\ Therootsoftheequationmaybethefactorsof126i.e.,2*7*9\\ 9isthegivenrootofthedeterminantputx=2ineq.\left(1\right)\\ {\left(2\right)}^3-67*2+126\Rightarrow 8-134+126=0\\ Hence,x=2istheotherroot.\\ Now,putx=7ineq.\left(1\right)\\ {\left(7\right)}^3-67*7+126\Rightarrow 343-469+126=0\\ Hence,x=7isalsotheotherrootofthedeterminant.\end{array}$

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

Sol:

$\begin{array}{l}Thesumoftheproductsofelementsofanyrowwiththeco-factorsofcorrespondingelements\\ isequaltothevalueofthedeterminantofthegivenmatrix.\\ Let\Delta =\left|\begin{array}{ccc}\hfill a_{11}\hfill & \hfill a_{12}\hfill & \hfill a_{13}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill & \hfill a_{23}\hfill \\ \hfill a_{31}\hfill & \hfill a_{32}\hfill & \hfill a_{33}\hfill \end{array}\right|\\ Expandingalong{R}_1\\ a_{11}\left|\begin{array}{cc}\hfill a_{22}\hfill & \hfill a_{23}\hfill \\ \hfill a_{32}\hfill & \hfill a_{33}\hfill \end{array}\right|-a_{12}\left|\begin{array}{cc}\hfill a_{21}\hfill & \hfill a_{23}\hfill \\ \hfill a_{31}\hfill & \hfill a_{33}\hfill \end{array}\right|+a_{13}\left|\begin{array}{cc}\hfill a_{21}\hfill & \hfill a_{22}\hfill \\ \hfill a_{31}\hfill & \hfill a_{32}\hfill \end{array}\right|\\ \Rightarrow a_{11}{M}_{11}+a_{12}{M}_{12}+a_{13}{M}_{13}\left(where{M}_{11},M_{12}and{M}_{13}aretheminorsofthecorrespondingelements\right)\end{array}$

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

Sol:

$\begin{array}{l}Theorderofamatrixis3*3\\ \therefore Totalnumberofelements=3*3=9\\ Hence,thenumberofminorsinthedeterminantis9.\end{array}$

This is a multiple choice answer as classified in NCERT Exemplar

(a, b, c) The length of the telescope tube L= f₀+f_e= 20+0.02= 20.02m

Also magnification is = 20/0.02= 1000

And image formed is inverted

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

Sol:

$\begin{array}{l}\underset{x\to 0^{-}}{lim}f\left(x\right)=\frac{e^{1/x}}{1+e^{1/x}}\\ =\underset{h\to 0}{lim}\frac{e^{\frac{1}{0-h}}}{1+e^{\frac{1}{0-h}}}=\underset{h\to 0}{lim}\frac{e^{-1/h}}{1+e^{-1/h}}\\ =\underset{h\to 0}{lim}\frac{1}{e^{1/h}\left(1-e^{-1/h}\right)}=\underset{h\to 0}{lim}\frac{1}{e^{1/h}-1}=\underset{h\to 0}{lim}\frac{1}{e^{1/0}-1}\\ =\underset{h\to 0}{lim}\frac{1}{e^{\infty }-1}=\underset{h\to 0}{lim}\frac{1}{0-1}=-1\left[?e^{\infty }=0\right]\\ \underset{x\to 0^{+}}{lim}f\left(x\right)=\frac{e^{1/x}}{1+e^{1/x}}=\underset{h\to 0}{lim}\frac{e^{\frac{1}{0+h}}}{1+e^{\frac{1}{0+h}}}=\underset{h\to 0}{lim}\frac{e^{1/h}}{1+e^{1/h}}\\ =\underset{h\to 0}{lim}\frac{1}{e^{-1/h}\left(1+e^{1/h}\right)}=\underset{h\to 0}{lim}\frac{1}{e^{-1/h}+1}=\underset{h\to 0}{lim}\frac{1}{e^{-1/0}+1}\\ =\underset{h\to 0}{lim}\frac{1}{e^{-\infty }+1}=\underset{h\to 0}{lim}\frac{1}{0+1}=1\left[?e^{-\infty }=0\right]\\ \underset{x\to 0}{lim}f\left(x\right)=0\\ As\underset{x\to 0^{-}}{lim}f\left(x\right)\ne \underset{x\to 0^{+}}{lim}f\left(x\right)\ne \underset{x\to 0}{lim}f\left(x\right)\\ Hence,f\left(x\right)isdiscontinuousatx=0.\end{array}$

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

Sol: $ForanysquarematrixA,{\left(A^2\right)}^{-1}={\left(A^{-1}\right)}^2.$

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

Sol:

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

Sol:

$\begin{array}{l}\underset{x\to a^{-}}{lim}f\left(x\right)=\left|x-a\right|sin\frac{1}{x-a}\\ =\underset{h\to 0}{lim}\left|a-h-a\right|sin\frac{1}{\left(a-h-a\right)}=\underset{h\to 0}{lim}h.sin\frac{1}{-h}\\ =\underset{h\to 0}{lim}-h.sin\frac{1}{h}\left[?sin\left(-\theta \right)=-sin\theta \right]\\ =0*\left[anumberoscillatebetween-1and1\right]=0\\ \underset{x\to a^{+}}{lim}f\left(x\right)=\left|x-a\right|sin\frac{1}{x-a}=\underset{h\to 0}{lim}\left|a+h-a\right|.sin\frac{1}{\left(a+h-a\right)}=\underset{h\to 0}{lim}h.sin\frac{1}{h}\\ =0*\left[anumberoscillatebetween-1and1\right]=0\\ \underset{x\to a}{lim}f\left(x\right)=0\\ As\underset{x\to a^{-}}{lim}f\left(x\right)=\underset{x\to a^{+}}{lim}f\left(x\right)=\underset{x\to a}{lim}f\left(x\right)=0\\ Hence,f\left(x\right)iscontinuousatx=a.\end{array}$

This is a multiple choice answer as classified in NCERT Exemplar

(a, b) A magnifying glass is used, as the object to be viewed can be brought closer to the eye than the normal near point. This results in a larger angle to be subtended by the object at the eye and hence, viewed in greater detail. Morever, the formation of a virtual erect and enlarged image, takes place.