\displaystyle \textbf{Question 1:}\ \ \text{Discuss the continuity of the function }f(x)\text{ at }x=\frac{1}{2},\text{ where}\qquad [\text{CBSE 2011}]
\displaystyle f(x)=\begin{cases} \frac{1}{2}-x\ ;\ 0\le x<\frac{1}{2}\\ 1\ ;\ x=\frac{1}{2}\\ \frac{3}{2}-x\ ;\ \frac{1}{2}<x\le 1 \end{cases}

\displaystyle \text{Answer:}
\displaystyle \text{We observe that:}

\displaystyle \text{(LHL at }x=\frac{1}{2}\text{)}=\lim_{x\to \frac{1}{2}^{-}}f(x)=\lim_{x\to \frac{1}{2}}\left(\frac{1}{2}-x\right)

\displaystyle \left[\because\ f(x)=\frac{1}{2}-x\text{ for }0\le x<\frac{1}{2}\right]

\displaystyle =\frac{1}{2}-\frac{1}{2}=0\qquad [\text{Using direct substitution method}]

\displaystyle \text{and,}\quad \text{(RHL at }x=\frac{1}{2}\text{)}=\lim_{x\to \frac{1}{2}^{+}}f(x)=\lim_{x\to \frac{1}{2}}\left(\frac{3}{2}-x\right)

\displaystyle \left[\because\ f(x)=\frac{3}{2}-x\text{ for }\frac{1}{2}<x\le 1\right]

\displaystyle =\frac{3}{2}-\frac{1}{2}=1\qquad [\text{Using direct substitution method}]

\displaystyle \text{Clearly,}\quad \lim_{x\to \frac{1}{2}^{-}}f(x)\neq \lim_{x\to \frac{1}{2}^{+}}f(x)

\displaystyle \text{Hence, }f(x)\text{ is not continuous at }x=\frac{1}{2}. \\ \text{Clearly, }f(x)\text{ has discontinuity of first kind at }x=\frac{1}{2}.

\displaystyle \textbf{Question 2:}\ \ \text{Show that the function }f(x)=2x-|x|\text{ is continuous at } \\ x=0.\qquad [\text{CBSE 2002}]

\displaystyle \text{Answer:}
\displaystyle \text{We have,}

\displaystyle f(x)=2x-|x|=\begin{cases} 2x-x,\ \text{if }x\ge 0\\ 2x-(-x),\ \text{if }x<0 \end{cases}

\displaystyle \Rightarrow\ f(x)=\begin{cases} x,\ \text{if }x\ge 0\\ 3x,\ \text{if }x<0 \end{cases}

\displaystyle \text{Now,}\quad \text{(LHL at }x=0\text{)}=\lim_{x\to 0^{-}}f(x)=\lim_{x\to 0^{-}}3x=3\times 0=0

\displaystyle \text{(RHL at }x=0\text{)}=\lim_{x\to 0^{+}}f(x)=\lim_{x\to 0^{+}}x=0

\displaystyle \text{and,}\quad f(0)=0

\displaystyle \therefore\ \lim_{x\to 0^{-}}f(x)=\lim_{x\to 0^{+}}f(x)=f(0)

\displaystyle \text{So, }f(x)\text{ is continuous at }x=0.

\displaystyle \textbf{Question 3:}\ \ \text{If the function }f(x)\text{ given by }f(x)=\begin{cases} 3ax+b,\ \text{if }x>1\\ 11,\ \text{if }x=1\\ 5ax-2b,\ \text{if }x<1 \end{cases}
\displaystyle \text{is continuous at }x=1,\text{ find the values of }a\text{ and }b. \ [\text{CBSE 2002, 2010, 2011, 2012}]

\displaystyle \text{Answer:}
\displaystyle \text{We observe that:}

\displaystyle \text{(LHL at }x=1\text{)}=\lim_{x\to 1^{-}}f(x)=\lim_{x\to 1^{-}}(5ax-2b)=5a-2b

\displaystyle \text{(RHL at }x=1\text{)}=\lim_{x\to 1^{+}}f(x)=\lim_{x\to 1^{+}}(3ax+b)=3a+b

\displaystyle \text{and,}\quad f(1)=11

\displaystyle \text{Since }f(x)\text{ is continuous at }x=1,

\displaystyle \therefore\ \lim_{x\to 1^{-}}f(x)=\lim_{x\to 1^{+}}f(x)=f(1)

\displaystyle \Rightarrow\ 5a-2b=3a+b=11

\displaystyle \Rightarrow\ 5a-2b=11\ \text{and}\ 3a+b=11

\displaystyle \Rightarrow\ a=3\ \text{and}\ b=2

\displaystyle \textbf{Question 4:}\ \ \text{Let }f(x)=\begin{cases} \frac{1-\cos 4x}{x^{2}},\ \text{if }x<0\\ a,\ \text{if }x=0\\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4},\ \text{if }x>0 \end{cases}
\displaystyle \text{Determine the value of }a\text{ so that }f(x)\text{ is continuous at }x=0. \ [\text{CBSE 2010, 2012, 2013}]

\displaystyle \text{Answer:}
\displaystyle \text{For }f(x)\text{ to be continuous at }x=0,\text{ we must have}

\displaystyle \lim_{x\to 0^{-}}f(x)=\lim_{x\to 0^{+}}f(x)=f(0)

\displaystyle \Rightarrow\ \lim_{x\to 0^{-}}f(x)=\lim_{x\to 0^{+}}f(x)=a\qquad (i)

\displaystyle \text{Now,}\quad \lim_{x\to 0^{-}}f(x)=\lim_{x\to 0}\frac{1-\cos 4x}{x^{2}}\qquad \left[\because\ f(x)=\frac{1-\cos 4x}{x^{2}}\text{ for }x<0\right]

\displaystyle =\lim_{x\to 0}\frac{2\sin^{2}2x}{x^{2}}

\displaystyle =2\lim_{x\to 0}\left(\frac{\sin 2x}{x}\right)^{2}=2\times 4\lim_{x\to 0}\left(\frac{\sin 2x}{2x}\right)^{2}=8(1)^{2}=8\qquad (ii)

\displaystyle \text{and,}\quad \lim_{x\to 0^{+}}f(x)=\lim_{x\to 0}\frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}\qquad \left[\because\ f(x)=\frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}\text{ for }x>0\right]

\displaystyle =\lim_{x\to 0}\frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}\cdot\frac{\sqrt{16+\sqrt{x}}+4}{\sqrt{16+\sqrt{x}}+4}

\displaystyle =\lim_{x\to 0}\frac{\sqrt{x}\left(\sqrt{16+\sqrt{x}}+4\right)}{16+\sqrt{x}-16}

\displaystyle =\lim_{x\to 0}\left(\sqrt{16+\sqrt{x}}+4\right)=4+4=8\qquad (iii)

\displaystyle \text{From }(i),(ii)\text{ and }(iii),\text{ we get }a=8.

\displaystyle \textbf{Question 5:}\ \ \text{Discuss the continuity of the function }f(x)\text{ given by } \\ f(x)=\begin{cases} 2x-1,\ \text{if }x<0\\ 2x+1,\ \text{if }x\ge 0 \end{cases}\qquad [\text{CBSE 2002}]

\displaystyle \text{Answer:}
\displaystyle \text{When }x<0,\text{ we have }f(x)=2x-1.

\displaystyle \text{Clearly, }f(x)\text{ is a polynomial function for }x<0.\text{ So, }f(x)\text{ is continuous for all }x<0.

\displaystyle \text{When }x>0,\text{ we have }f(x)=2x+1.

\displaystyle \text{Clearly, }f(x)\text{ is a polynomial function for }x>0.\text{ So, it is continuous for all }x>0.

\displaystyle \text{Let us now consider the point }x=0.\text{ At }x=0,\text{ we have}

\displaystyle \lim_{x\to 0^{-}}f(x)=\lim_{x\to 0}(2x-1)=-1\ \text{and}\ \lim_{x\to 0^{+}}f(x)=\lim_{x\to 0}(2x+1)=1.

\displaystyle \therefore\ \lim_{x\to 0^{-}}f(x)\neq \lim_{x\to 0^{+}}f(x).

\displaystyle \text{So, }f(x)\text{ is not continuous at }x=0.\text{ Hence, }f(x)\text{ is everywhere continuous except at }x=0.

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