Continuity and Differentiability

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 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 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 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 Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}\underset{x\to 0^{-}}{lim}f\left(x\right)=\frac{1-cos2x}{x^2}=\underset{h\to 0}{lim}\frac{1-cos2\left(0-h\right)}{{\left(0-h\right)}^2}\\ =\underset{h\to 0}{lim}\frac{1-cos\left(-2h\right)}{h^2}=\underset{h\to 0}{lim}\frac{1-cos\left(2h\right)}{h^2}\\ =\underset{h\to 0}{lim}\frac{2si{n}^{2}h}{h^2}\left[?1-cos\theta =2si{n}^2\frac{\theta }{2}\right]\\ =\underset{h\to 0}{lim}\frac{2sinh}{h}.\frac{sinh}{h}=2.1.1=2\left[\underset{x\to 0}{lim}\frac{sinx}{x}=1\right]\\ \underset{x\to 0^{+}}{lim}f\left(x\right)=\frac{1-cos2x}{x^2}\\ =\underset{h\to 0}{lim}\frac{1-cos2\left(0+h\right)}{{\left(0+h\right)}^2}=\underset{h\to 0}{lim}\frac{1-cos\left(2h\right)}{h^2}\\ =\underset{h\to 0}{lim}\frac{2si{n}^{2}h}{h^2}=\underset{h\to 0}{lim}\frac{2sinh}{h}.\frac{sinh}{h}=2.1.1=2\\ \underset{x\to 0}{lim}f\left(x\right)=5\\ As\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}f\left(x\right)\ne \underset{x\to 0}{lim}f\left(x\right)\\ Hence,f\left(x\right)isdiscontinuousatx=0.\end{array}$

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

Sol:

$\begin{array}{l}\underset{x\to 2^{+}}{lim}f\left(x\right)=3x+5\\ =\underset{h\to 0}{lim}3\left(2+h\right)+5=11\\ \underset{x\to 2}{lim}f\left(x\right)=3x+5=3\left(2\right)+5=11\\ \underset{x\to 2^{-}}{lim}f\left(x\right)=x^2=\underset{h\to 0}{lim}{\left(2-h\right)}^2\\ =\underset{h\to 0}{lim}{\left(2\right)}^2+h^2-4h={\left(2\right)}^2=4\\ Since\underset{x\to 2^{+}}{lim}f\left(x\right)=\underset{x\to 2}{lim}f\left(x\right)\ne \underset{x\to 2^{-}}{lim}f\left(x\right)\\ Hence,f\left(x\right)isdiscontinuousatx=2.\end{array}$

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

Sol:

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