Physics NCERT Exemplar Solutions Class 12th Chapter Five

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Sol:

$\begin{array}{l}Giventhatf\left(x\right)=\left|sinx+cosx\right|atx=\pi \\ Putg\left(x\right)=sinx+cosxandh\left(x\right)=\left|x\right|\\ \therefore h\left[g\left(x\right)\right]=h\left(sinx+cosx\right)=\left|sinx+cosx\right|\\ Now,g\left(x\right)=sinx+cosxisacontinuousfunctionsincesinxandcosxaretwocontinuous\\ functionsatx=\pi .\\ Weknowthateverymodulusfunctionisacontinuousfunctioneverywhere.\\ Hence,f\left(x\right)=\left|sinx+cosx\right|isacontinuousfunctionatx=\pi .\end{array}$

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

Sol:

$\begin{array}{l}Wehavef\left(x\right)=\frac{1}{t^2+t-2}\\ \Rightarrow f\left(t\right)=\frac{1}{{\left(\frac{1}{x-1}\right)}^2+\frac{1}{x-1}-2}\left[Puttingt=\frac{1}{x-1}\right]\\ =\frac{1}{\frac{1+x-1-2{\left(x-1\right)}^2}{{\left(x-1\right)}^2}}=\frac{{\left(x-1\right)}^2}{x-2x^2-2+4x}\\ =\frac{{\left(x-1\right)}^2}{-2x^2+5x-2}=\frac{{\left(x-1\right)}^2}{-\left(2x^2-5x+2\right)}\\ =\frac{{\left(x-1\right)}^2}{-\left[2x^2-4x-x+2\right]}=\frac{{\left(x-1\right)}^2}{-\left[2x\left(x-2\right)-1\left(x-2\right)\right]}\\ =\frac{{\left(x-1\right)}^2}{-\left(x-2\right)\left(2x-1\right)}=\frac{{\left(x-1\right)}^2}{\left(2-x\right)\left(2x-1\right)}\\ So,iff\left(t\right)isdiscontinuous,then2-x=0\therefore x=2\\ and2x-1=0\therefore x=\frac{1}{2}\\ Hence,therequiredofdiscontinuityare2and\frac{1}{2}.\end{array}$

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

Sol:

$\begin{array}{l}f\left(x\right)=\frac{1}{x+2}\\ f\left[f\left(x\right)\right]=\frac{1}{f\left(x\right)+2}=\frac{1}{\frac{1}{x+2}+2}=\frac{1}{\frac{1+2x+4}{x+2}}=\frac{x+2}{2x+5}\\ \therefore f\left[f\left(x\right)\right]=\frac{x+2}{2x+5}\\ Thisfunctionwillnotbedefinedandcontinuouswhere2x+5=0\\ \Rightarrow x=\frac{-5}{2}.\\ Hence,x=\frac{-5}{2}isthepointofdiscontinuity.\end{array}$

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

Sol:

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

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

Sol:

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

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

Sol:

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

Sol:

$\begin{array}{l}f\left(x\right)=\frac{2^{x+2}-16}{4^x-16}=\frac{2^2.2^x-16}{{\left(2^x\right)}^2-{\left(4\right)}^2}=\frac{4\left(2^x-4\right)}{\left(2^x-4\right)\left(2^x+4\right)}\\ =\frac{4}{\left(2^x+4\right)}\\ \underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}\frac{4}{\left(2^{2-h}+4\right)}=\frac{4}{2^2+4}=\frac{4}{4+4}=\frac{4}{8}=\frac{1}{2}\\ \underset{x\to 2}{lim}f\left(x\right)=k\\ Asthefunctioniscontinuousatx=2\\ \therefore \underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{x\to 2}{lim}f\left(x\right)\\ \therefore k=\frac{1}{2}\\ Hence,thevalueofkis\frac{1}{2}.\end{array}$