Ncert Solutions Maths class 12th

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

Sol:

$\begin{array}{l}\underset{x\to 4^{-}}{lim}f\left(x\right)=\frac{x-4}{\left|x-4\right|}+a=\underset{h\to 0}{lim}\frac{4-h-4}{\left|4-h-4\right|}+a\\ =\underset{h\to 0}{lim}\frac{-h}{h}+a=-1+a\\ \underset{x\to 4}{lim}f\left(x\right)=a+b\\ \underset{x\to 4^{+}}{lim}f\left(x\right)=\frac{x-4}{\left|x-4\right|}+b=\underset{h\to 0}{lim}\frac{4+h-4}{\left|4+h-4\right|}+b\\ =\underset{h\to 0}{lim}\frac{h}{h}+b=1+b\\ Asthefunctioniscontinuousatx=4.\\ \therefore \underset{x\to 4^{-}}{lim}f\left(x\right)=\underset{x\to 4}{lim}f\left(x\right)=\underset{x\to 4^{+}}{lim}f\left(x\right)\\ -1+a=a+b=1+b\\ \therefore -1+a=a+b\Rightarrow b=-1\\ 1+b=a+b\Rightarrow a=1\\ Hence,thevalueofa=1andb=-1.\end{array}$

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

Sol:

$\begin{array}{l}True.\\ Let\Delta =\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill \left(1+sin\theta \right)\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 1+cos\theta \hfill \end{array}\right|\\ C_1\to C_1-C_2,C_2\to C_2-C_3\\ =\left|\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill -sin\theta \hfill & \hfill sin\theta \hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill -cos\theta \hfill & \hfill 1+cos\theta \hfill \end{array}\right|\\ Expandingalong{C}_3\\ =1\left|\begin{array}{cc}\hfill -sin\theta \hfill & \hfill sin\theta \hfill \\ \hfill 0\hfill & \hfill -cos\theta \hfill \end{array}\right|=sin\theta cos\theta -0=sin\theta cos\theta \\ =\frac{1}{2}.2sin\theta cos\theta =\frac{1}{2}sin2\theta \\ =\frac{1}{2}\times 1=\frac{1}{2}\left[Maximumvalueofsin2\theta =1\right]\end{array}$

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

Sol:

$\begin{array}{l}True.\\ Giventhat\Delta =\left|\begin{array}{ccc}\hfill a\hfill & \hfill p\hfill & \hfill x\hfill \\ \hfill b\hfill & \hfill q\hfill & \hfill y\hfill \\ \hfill c\hfill & \hfill r\hfill & \hfill z\hfill \end{array}\right|=16\\ L.H.S.{\Delta }_1=\left|\begin{array}{ccc}\hfill p+x\hfill & \hfill a+x\hfill & \hfill a+p\hfill \\ \hfill q+y\hfill & \hfill b+y\hfill & \hfill b+q\hfill \\ \hfill r+z\hfill & \hfill c+z\hfill & \hfill c+r\hfill \end{array}\right|\\ C_1\to C_1+C_2+C_3\\ =\left|\begin{array}{ccc}\hfill 2p+2x+2a\hfill & \hfill a+x\hfill & \hfill a+p\hfill \\ \hfill 2q+2y+2b\hfill & \hfill b+y\hfill & \hfill b+q\hfill \\ \hfill 2r+2z+2c\hfill & \hfill c+z\hfill & \hfill c+r\hfill \end{array}\right|\\ =2\left|\begin{array}{ccc}\hfill p+x+a\hfill & \hfill a+x\hfill & \hfill a+p\hfill \\ \hfill q+y+b\hfill & \hfill b+y\hfill & \hfill b+q\hfill \\ \hfill r+z+c\hfill & \hfill c+z\hfill & \hfill c+r\hfill \end{array}\right|\left[Taking2commonfrom{C}_1\right]\\ C_1\to C_1-C_2\\ =2\left|\begin{array}{ccc}\hfill p\hfill & \hfill a+x\hfill & \hfill a+p\hfill \\ \hfill q\hfill & \hfill b+y\hfill & \hfill b+q\hfill \\ \hfill r\hfill & \hfill c+z\hfill & \hfill c+r\hfill \end{array}\right|\\ C_3\to C_3-C_1\\ =2\left|\begin{array}{ccc}\hfill p\hfill & \hfill a+x\hfill & \hfill a\hfill \\ \hfill q\hfill & \hfill b+y\hfill & \hfill b\hfill \\ \hfill r\hfill & \hfill c+z\hfill & \hfill c\hfill \end{array}\right|\\ Splittingup{C}_2\\ =2\left|\begin{array}{ccc}\hfill p\hfill & \hfill a\hfill & \hfill a\hfill \\ \hfill q\hfill & \hfill b\hfill & \hfill b\hfill \\ \hfill r\hfill & \hfill c\hfill & \hfill c\hfill \end{array}\right|+2\left|\begin{array}{ccc}\hfill p\hfill & \hfill x\hfill & \hfill a\hfill \\ \hfill q\hfill & \hfill y\hfill & \hfill b\hfill \\ \hfill r\hfill & \hfill z\hfill & \hfill c\hfill \end{array}\right|=2\left(0\right)+2\left|\begin{array}{ccc}\hfill p\hfill & \hfill x\hfill & \hfill a\hfill \\ \hfill q\hfill & \hfill y\hfill & \hfill b\hfill \\ \hfill r\hfill & \hfill z\hfill & \hfill c\hfill \end{array}\right|\\ =2\left|\begin{array}{ccc}\hfill p\hfill & \hfill x\hfill & \hfill a\hfill \\ \hfill q\hfill & \hfill y\hfill & \hfill b\hfill \\ \hfill r\hfill & \hfill z\hfill & \hfill c\hfill \end{array}\right|\Rightarrow 2\left|\begin{array}{ccc}\hfill a\hfill & \hfill p\hfill & \hfill x\hfill \\ \hfill b\hfill & \hfill q\hfill & \hfill y\hfill \\ \hfill c\hfill & \hfill r\hfill & \hfill z\hfill \end{array}\right|\left(C_1\leftrightarrow C_{3}and{C}_2\leftrightarrow C_3\right)\\ =2\times 16=32\end{array}$

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

Sol:

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

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

Sol:

$\begin{array}{l}True.\\ Let\Delta =\left|\begin{array}{ccc}\hfill x+a\hfill & \hfill p+u\hfill & \hfill l+f\hfill \\ \hfill y+b\hfill & \hfill q+v\hfill & \hfill m+g\hfill \\ \hfill z+c\hfill & \hfill r+w\hfill & \hfill n+h\hfill \end{array}\right|\\ Splittingup{C}_1\\ =\left|\begin{array}{ccc}\hfill x\hfill & \hfill p+u\hfill & \hfill l+f\hfill \\ \hfill y\hfill & \hfill q+v\hfill & \hfill m+g\hfill \\ \hfill z\hfill & \hfill r+w\hfill & \hfill n+h\hfill \end{array}\right|+\left|\begin{array}{ccc}\hfill a\hfill & \hfill p+u\hfill & \hfill l+f\hfill \\ \hfill b\hfill & \hfill q+v\hfill & \hfill m+g\hfill \\ \hfill c\hfill & \hfill r+w\hfill & \hfill n+h\hfill \end{array}\right|\\ Splittingup{C}_{2}inboth\text{\hspace{0.17em}determinants}\\ =\left|\begin{array}{ccc}\hfill x\hfill & \hfill p\hfill & \hfill l+f\hfill \\ \hfill y\hfill & \hfill q\hfill & \hfill m+g\hfill \\ \hfill z\hfill & \hfill r\hfill & \hfill n+h\hfill \end{array}\right|+\left|\begin{array}{ccc}\hfill x\hfill & \hfill u\hfill & \hfill l+f\hfill \\ \hfill y\hfill & \hfill v\hfill & \hfill m+g\hfill \\ \hfill z\hfill & \hfill w\hfill & \hfill n+h\hfill \end{array}\right|+\left|\begin{array}{ccc}\hfill a\hfill & \hfill p\hfill & \hfill l+f\hfill \\ \hfill b\hfill & \hfill q\hfill & \hfill m+g\hfill \\ \hfill c\hfill & \hfill r\hfill & \hfill n+h\hfill \end{array}\right|+\left|\begin{array}{ccc}\hfill a\hfill & \hfill u\hfill & \hfill l+f\hfill \\ \hfill b\hfill & \hfill v\hfill & \hfill m+g\hfill \\ \hfill c\hfill & \hfill w\hfill & \hfill n+h\hfill \end{array}\right|\\ Similarlybysplitting{C}_{3}ineach\text{\hspace{0.17em}determinants},wewillget\mathrm{8determinants}.\end{array}$

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

Sol:

$\begin{array}{l}True.\\ Let\Delta =\left|\begin{array}{ccc}\hfill sinA\hfill & \hfill cosA\hfill & \hfill sinA+cosB\hfill \\ \hfill sinB\hfill & \hfill cosA\hfill & \hfill sinB+cosB\hfill \\ \hfill sinC\hfill & \hfill cosA\hfill & \hfill sinC+cosB\hfill \end{array}\right|\\ Splittingup{C}_3\\ =\left|\begin{array}{ccc}\hfill sinA\hfill & \hfill cosA\hfill & \hfill sinA\hfill \\ \hfill sinB\hfill & \hfill cosA\hfill & \hfill sinB\hfill \\ \hfill sinC\hfill & \hfill cosA\hfill & \hfill sinC\hfill \end{array}\right|+\left|\begin{array}{ccc}\hfill sinA\hfill & \hfill cosA\hfill & \hfill cosB\hfill \\ \hfill sinB\hfill & \hfill cosA\hfill & \hfill cosB\hfill \\ \hfill sinC\hfill & \hfill cosA\hfill & \hfill cosB\hfill \end{array}\right|\\ =0+\left|\begin{array}{ccc}\hfill sinA\hfill & \hfill cosA\hfill & \hfill cosB\hfill \\ \hfill sinB\hfill & \hfill cosA\hfill & \hfill cosB\hfill \\ \hfill sinC\hfill & \hfill cosA\hfill & \hfill cosB\hfill \end{array}\right|\left[?C_{1}and{C}_{3}areidentical\right]\\ =cosAcosB\left|\begin{array}{ccc}\hfill sinA\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill sinB\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill sinC\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right|\left[TakingcosAandcosBcommonfrom{C}_{2}and{C}_{3}respectively\right]\\ =cosAcosB\left(0\right)\left[?C_{2}and{C}_{3}areidentical\right]\\ =0\end{array}$

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

Sol:

$\begin{array}{l}\underset{x\to 0^{-}}{lim}f\left(x\right)=\frac{1-coskx}{xsinx}=\underset{h\to 0}{lim}\frac{1-cosk\left(0-h\right)}{\left(0-h\right)sin\left(0-h\right)}\\ =\underset{h\to 0}{lim}\frac{1-cos\left(-kh\right)}{\left(-h\right)sin\left(-h\right)}=\underset{h\to 0}{lim}\frac{1-coskh}{hsinh}\left[\begin{array}{l}?sin\left(-\theta \right)=-sin\theta \\ cos\left(-\theta \right)=cos\theta \end{array}\right]\\ =\underset{h\to 0}{lim}\frac{2si{n}^2\frac{kh}{2}}{hsinh}=\underset{\begin{array}{l}h\to 0\\ kh\to 0\end{array}}{lim}\frac{2sin\frac{kh}{2}}{\frac{kh}{2}}\times \frac{kh}{2}\times \frac{sin\frac{kh}{2}}{\frac{kh}{2}}\times \frac{kh}{2}.\frac{1}{h.\frac{sinh}{h}.h}\\ =2.1.\frac{kh}{2}.1.\frac{kh}{2}.\frac{1}{h^2}.1\left[\begin{array}{l}\underset{h\to 0}{lim}\frac{sinh}{h}=1and\\ \underset{kh\to 0}{lim}\frac{sinkh}{kh}=1\end{array}\right]\\ =\frac{k^2}{2}\\ \underset{x\to 0}{lim}f\left(x\right)=\frac{1}{2}\\ \therefore \underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0}{lim}f\left(x\right)\\ \therefore \frac{k^2}{2}=\frac{1}{2}\Rightarrow k^2=1\Rightarrow k=\pm 1\\ Hence,thevalueofkis\pm 1.\end{array}$

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

Sol:

$\begin{array}{l}False.\\ Since?adjA?=?A{?}^{n-1}wherenistheorderofthesquarematrix.\end{array}$

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

Sol:

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

Sol:

$\begin{array}{l}True.\\ Let\Delta =\left|\begin{array}{ccc}\hfill x+1\hfill & \hfill x+2\hfill & \hfill x+a\hfill \\ \hfill x+2\hfill & \hfill x+3\hfill & \hfill x+b\hfill \\ \hfill x+3\hfill & \hfill x+4\hfill & \hfill x+c\hfill \end{array}\right|\\ R_2\to 2R_2-\left(R_1+R_3\right)\\ =\left|\begin{array}{ccc}\hfill x+1\hfill & \hfill x+2\hfill & \hfill x+a\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 2b-\left(a+c\right)\hfill \\ \hfill x+3\hfill & \hfill x+4\hfill & \hfill x+c\hfill \end{array}\right|\\ a,b,careinA.P.\\ \therefore b-a=c-b\Rightarrow 2b=a+c\\ =\left|\begin{array}{ccc}\hfill x+1\hfill & \hfill x+2\hfill & \hfill x+a\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill x+3\hfill & \hfill x+4\hfill & \hfill x+c\hfill \end{array}\right|=0\end{array}$