\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.
\\

\displaystyle \textbf{Question 6. }\text{The function }f(x)=[x],\text{ where }[x]\text{ denotes the greatest integer less} \\ \text{than or equal to } x,\text{ is continuous at} \hspace{2.2cm} \text{[CBSE 2023]}
\displaystyle \text{(a) }x=1\qquad \text{(b) }x=1.5\qquad \text{(c) }x=-2\qquad \text{(d) }x=4
\displaystyle \text{Answer:}
\displaystyle \text{(b) We know that the greatest integer function }[x]\text{ is}
\displaystyle \text{continuous at all points except at integer points.}
\displaystyle \text{So, }f(x)\text{ is continuous at }x=1.5\text{ only.}
\\

\displaystyle \textbf{Question 7. }\text{The value of }k\text{ for which }f(x)=\begin{cases}3x+5,&x\geq2\\kx^2,&x<2\end{cases} \\ \text{ is a continuous function, is} \hspace{2.2cm} \text{[CBSE 2023]}
\displaystyle \text{(a) }-\frac{11}{4}\qquad \text{(b) }\frac{4}{11}\qquad \text{(c) }11\qquad \text{(d) }\frac{11}{4}
\displaystyle \text{Answer:}
\displaystyle \text{(d) Given, } f(x)=\begin{cases}3x+5, & x\geq2\\kx^2, & x<2\end{cases}
\displaystyle \text{Since, }f(x)\text{ is continuous.}
\displaystyle \therefore \text{ LHL}=\lim_{x\to2^-}f(x)=\lim_{h\to0}f(2-h)
\displaystyle =\lim_{h\to0}k(2-h)^2
\displaystyle =\lim_{h\to0}k(4+h^2-4h)
\displaystyle =k(4+0-4\times0)=4k
\displaystyle \text{RHL}=\lim_{x\to2^+}f(x)=\lim_{h\to0}f(2+h)
\displaystyle =\lim_{h\to0}(3(2+h)+5)
\displaystyle =\lim_{h\to0}(6+3h+5)
\displaystyle =6+3\times0+5=11
\displaystyle \text{Since, }f(x)\text{ is continuous at }x=2.
\displaystyle \therefore \text{ LHL}=\text{RHL}=f(2)
\displaystyle \Rightarrow 4k=11=3\times2+5
\displaystyle \Rightarrow k=\frac{11}{4}
\\

\displaystyle \textbf{Question 8. }\text{Find the value(s) of }\lambda,\text{ if the function} \hspace{2.2cm} \text{[CBSE 2023]}
\displaystyle f(x)=\begin{cases}\dfrac{\sin^2\lambda x}{x^2},&x\neq0\\1,&x=0\end{cases} \text{ is continuous at }x=0.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f(x)=\begin{cases}\frac{\sin^2\lambda x}{x^2}, & x\neq0\\ 1, & x=0\end{cases}
\displaystyle \lim_{x\to0}f(x)=\lim_{x\to0}\frac{\sin^2\lambda x}{x^2}=\lim_{x\to0}\lambda^2\left(\frac{\sin\lambda x}{\lambda x}\right)^2=\lambda^2
\displaystyle \text{and }f(0)=1
\displaystyle \text{Since, }f(x)\text{ is continuous at }x=0.
\displaystyle \therefore \lim_{x\to0}f(x)=f(0)
\displaystyle \Rightarrow \lambda^2=1\Rightarrow \lambda=-1,1
\\

\displaystyle \textbf{Question 9. }\text{Find the value of }k\text{ for which the function }f\text{ given by} \hspace{2.2cm} \text{[CBSE 2023]}
\displaystyle f(x)=\begin{cases}\dfrac{1-\cos x}{2x^2},&x\neq0\\k,&x=0\end{cases} \text{ is continuous at }x=0.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f(x)=\begin{cases}\frac{1-\cos x}{2x^2}, & \text{if }x\neq0\\ k, & \text{if }x=0\end{cases}\text{ is continuous at }x=0
\displaystyle \lim_{x\to0}f(x)=\lim_{h\to0}f(0+h)=\lim_{h\to0}f(h)
\displaystyle =\lim_{h\to0}\frac{1-\cos h}{2h^2}=\lim_{h\to0}\frac{2\sin^2(h/2)}{2h^2}
\displaystyle =\lim_{h\to0}\frac{\sin^2(h/2)}{h^2}=\lim_{h\to0}\frac14\left(\frac{\sin(h/2)}{h/2}\right)^2
\displaystyle =\frac14\times(1)^2=\frac14\text{ and }f(0)=k
\displaystyle \text{Since, }f(x)\text{ is continuous at }x=0.
\displaystyle \therefore \lim_{x\to0}f(x)=f(0)
\displaystyle \Rightarrow \frac14=k\Rightarrow k=\frac14
\\

\displaystyle \textbf{Question 10. }\text{If the function }f\text{ defined as }f(x)=\begin{cases}\dfrac{x^2-9}{x-3},&x\neq3\\k,&x=3\end{cases}
\displaystyle \text{is continuous at }x=3,\text{ find the value of }k. \hspace{2.2cm} \text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f(x)=\begin{cases}\frac{x^2-9}{x-3}, & x\neq3\\ k, & x=3\end{cases}
\displaystyle \text{Since, }f(x)\text{ is continuous at }x=3.
\displaystyle \therefore \lim_{x\to3}f(x)=f(3)
\displaystyle \Rightarrow \lim_{x\to3}\frac{x^2-9}{x-3}=k
\displaystyle \Rightarrow \lim_{x\to3}\frac{(x-3)(x+3)}{(x-3)}=k
\displaystyle \Rightarrow \lim_{x\to3}(x+3)=k
\displaystyle \Rightarrow 3+3=k
\displaystyle \Rightarrow k=6
\\

\displaystyle \textbf{Question 11. }\text{Find the number of points of discontinuity of }f\text{ defined by } \\ f(x)=|x|-|x+1|. \hspace{2.2cm} \text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }f(x)=|x|-|x+1|
\displaystyle f(x)=\begin{cases}1, & x<-1\\ -2x-1, & -1\leq x<0\\ -1, & x\geq0\end{cases}
\displaystyle \text{Clearly, }f(x)\text{ is continuous for all values of }x.
\displaystyle \text{Hence, no discontinuous point exists.}
\\

\displaystyle \textbf{Question 12. }\text{Determine the value of }k\text{ for which the following function} \\ \text{is continuous at }x=3: \hspace{2.2cm} \text{[CBSE 2017]}
\displaystyle f(x)=\begin{cases}\dfrac{(x+3)^2-36}{x-3},&x\neq3\\k,&x=3\end{cases}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f(x)=\begin{cases}\frac{(x+3)^2-36}{x-3}, & x\neq3\\ k, & x=3\end{cases}
\displaystyle \text{Given, }f(x)\text{ is continuous at }x=3.
\displaystyle \text{Then, we have }\lim_{x\to3}f(x)=f(3)
\displaystyle \Rightarrow \lim_{x\to3}\frac{(x+3)^2-36}{x-3}=k
\displaystyle \Rightarrow \lim_{x\to3}\frac{(x+3)^2-6^2}{x-3}=k
\displaystyle \Rightarrow \lim_{x\to3}\frac{(x+3-6)(x+3+6)}{x-3}=k
\displaystyle \hspace{1cm}\left[\because a^2-b^2=(a-b)(a+b)\right]
\displaystyle \Rightarrow \lim_{x\to3}\frac{(x-3)(x+9)}{x-3}=k
\displaystyle \Rightarrow \lim_{x\to3}(x+9)=k
\displaystyle \Rightarrow 3+9=k\Rightarrow k=12
\\

\displaystyle \textbf{Question 13. }\text{Find the values of }p\text{ and }q\text{ for which}
\displaystyle f(x)=\begin{cases}\dfrac{1-\sin^3x}{3\cos^2x},&x<\dfrac{\pi}{2}\\p,&x=\dfrac{\pi}{2}\\\dfrac{q(1-\sin x)}{(\pi-2x)^2},&x>\dfrac{\pi}{2}\end{cases}
\displaystyle \text{is continuous at }x=\dfrac{\pi}{2}. \hspace{2.2cm} \text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f(x)=\begin{cases}\frac{1-\sin^3x}{3\cos^2x}, & \text{if }x<\frac{\pi}{2}\\ p, & \text{if }x=\frac{\pi}{2}\\ \frac{q(1-\sin x)}{(\pi-2x)^2}, & \text{if }x>\frac{\pi}{2}\end{cases}
\displaystyle \text{is continuous at }x=\frac{\pi}{2}
\displaystyle \text{Then, }(\text{LHL})_{x=\frac{\pi}{2}}=(\text{RHL})_{x=\frac{\pi}{2}}=f\left(\frac{\pi}{2}\right)\qquad \ldots(i)
\displaystyle \text{Now, LHL}=\lim_{x\to\frac{\pi}{2}^-}f(x)=\lim_{h\to0}f\left(\frac{\pi}{2}-h\right)
\displaystyle \left[\because x=\frac{\pi}{2}-h,\text{ when }x\to\frac{\pi}{2}^-,\text{ then }h\to0\right]
\displaystyle =\lim_{h\to0}\frac{1-\sin^3\left(\frac{\pi}{2}-h\right)}{3\cos^2\left(\frac{\pi}{2}-h\right)}=\lim_{h\to0}\frac{1-\cos^3h}{3\sin^2h}
\displaystyle \left[\because \cos\left(\frac{\pi}{2}-\theta\right)=\sin\theta,\ \sin\left(\frac{\pi}{2}-\theta\right)=\cos\theta\right]
\displaystyle =\lim_{h\to0}\frac{(1-\cos h)(1+\cos^2h+\cos h)}{3(1-\cos^2h)}
\displaystyle =\lim_{h\to0}\frac{(1-\cos h)(1+\cos^2h+\cos h)}{3(1-\cos h)(1+\cos h)}
\displaystyle =\lim_{h\to0}\frac{1+\cos^2h+\cos h}{3(1+\cos h)}
\displaystyle =\frac{1+\cos^20+\cos0}{3(1+\cos0)}
\displaystyle =\frac{1+1+1}{3(1+1)}=\frac12\qquad \ldots(ii)
\displaystyle \text{and RHL}=\lim_{x\to\frac{\pi}{2}^+}f(x)=\lim_{h\to0}f\left(\frac{\pi}{2}+h\right)
\displaystyle \left[\because x=\frac{\pi}{2}+h,\text{ when }x\to\frac{\pi}{2}^+,\text{ then }h\to0\right]
\displaystyle =\lim_{h\to0}\frac{q\left[1-\sin\left(\frac{\pi}{2}+h\right)\right]}{\left[\pi-2\left(\frac{\pi}{2}+h\right)\right]^2}
\displaystyle =\lim_{h\to0}\frac{q(1-\cos h)}{(-2h)^2}
\displaystyle =\lim_{h\to0}\frac{q(1-\cos h)}{4h^2}
\displaystyle =\lim_{h\to0}\frac{q\left(2\sin^2\frac{h}{2}\right)}{4h^2}\qquad \left[\because \cos x=1-2\sin^2\frac{x}{2}\right]
\displaystyle =\frac{q}{8}\lim_{h\to0}\left[\frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}\right]^2
\displaystyle =\frac{q}{8}\times1=\frac{q}{8}\qquad \left[\because \lim_{x\to0}\frac{\sin x}{x}=1\right]\qquad \ldots(iii)
\displaystyle \text{On substituting the values from Eqs. (ii) and (iii) into Eq. (i), we get}
\displaystyle \frac12=\frac{q}{8}=f\left(\frac{\pi}{2}\right)
\displaystyle \Rightarrow \frac12=\frac{q}{8}\text{ and }\frac12=p
\displaystyle \therefore q=4\text{ and }p=\frac12
\\

\displaystyle \textbf{Question 14. }\text{If }f(x)=\begin{cases}\dfrac{\sin(a+1)x+2\sin x}{x},&x<0\\2,&x=0\\\dfrac{\sqrt{1+bx}-1}{x},&x>0\end{cases}
\displaystyle \text{is continuous at }x=0,\text{ then find the values of }a\text{ and }b. \hspace{2.2cm} \text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Given,}
\displaystyle f(x)=\begin{cases}\frac{\sin(a+1)x+2\sin x}{x}, & x<0\\ 2, & x=0\\ \frac{\sqrt{1+bx}-1}{x}, & x>0\end{cases}
\displaystyle \text{is continuous at }x=0
\displaystyle \therefore (\text{LHL})_{x=0}=(\text{RHL})_{x=0}=f(0)\qquad \ldots(i)
\displaystyle \text{Now, LHL}=\lim_{x\to0^-}f(x)=\lim_{h\to0}f(0-h)
\displaystyle \left[\because x=0-h,\text{ when }x\to0^-,\text{ then }h\to0\right]
\displaystyle =\lim_{h\to0}\frac{\sin[(a+1)(0-h)]+2\sin(0-h)}{0-h}
\displaystyle =\lim_{h\to0}\frac{-\sin(a+1)h-2\sin h}{-h}\qquad \left[\because \sin(-\theta)=-\sin\theta\right]
\displaystyle =\lim_{h\to0}\frac{\sin(a+1)h+2\sin h}{h}
\displaystyle =\lim_{h\to0}\frac{\sin(a+1)h}{h}+\lim_{h\to0}\frac{2\sin h}{h}
\displaystyle =\lim_{h\to0}\frac{\sin(a+1)h}{(a+1)h}\times(a+1)+2\lim_{h\to0}\frac{\sin h}{h}
\displaystyle =1\times(a+1)+2\times1\qquad \left[\because \lim_{x\to0}\frac{\sin x}{x}=1\right]
\displaystyle =a+1+2=a+3\qquad \ldots(ii)
\displaystyle \text{and RHL}=\lim_{x\to0^+}f(x)=\lim_{h\to0}f(0+h)
\displaystyle \left[\because x=0+h,\text{ when }x\to0^+,\text{ then }h\to0\right]
\displaystyle =\lim_{h\to0}\frac{\sqrt{1+b(0+h)}-1}{0+h}=\lim_{h\to0}\frac{\sqrt{1+bh}-1}{h}
\displaystyle =\lim_{h\to0}\frac{\sqrt{1+bh}-1}{h}\times\frac{\sqrt{1+bh}+1}{\sqrt{1+bh}+1}
\displaystyle \text{[multiplying numerator and denominator by }\sqrt{1+bh}+1\text{]}
\displaystyle =\lim_{h\to0}\frac{(1+bh)-1}{h(\sqrt{1+bh}+1)}
\displaystyle =\lim_{h\to0}\frac{bh}{h(\sqrt{1+bh}+1)}
\displaystyle =\lim_{h\to0}\frac{b}{\sqrt{1+bh}+1}=\frac{b}{\sqrt{1+0}+1}=\frac{b}{2}\qquad \ldots(iii)
\displaystyle \text{From Eqs. (i), (ii) and (iii), we get}
\displaystyle a+3=\frac{b}{2}=2\qquad \left[\because f(0)=2\right]
\displaystyle \Rightarrow a+3=2\text{ and }\frac{b}{2}=2
\displaystyle \therefore a=-1\text{ and }b=4
\\

\displaystyle \textbf{Question 15. }\text{Find the value of }k,\text{ so that the function}
\displaystyle f(x)=\begin{cases}\dfrac{1-\cos4x}{8x^2},&x\neq0\\k,&x=0\end{cases}
\displaystyle \text{is continuous at }x=0. \hspace{2.2cm} \text{[CBSE 2014C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f(x)=\begin{cases}\frac{1-\cos4x}{8x^2}, & \text{if }x\neq0\\ k, & \text{if }x=0\end{cases}
\displaystyle \text{is continuous at }x=0.
\displaystyle \text{Then, }(\text{LHL})_{x=0}=(\text{RHL})_{x=0}=f(0)\qquad \ldots(i)
\displaystyle \text{Now, LHL}=\lim_{x\to0^-}f(x)
\displaystyle =\lim_{x\to0^-}\frac{1-\cos4x}{8x^2}
\displaystyle =\lim_{h\to0}\frac{1-\cos4(0-h)}{8(0-h)^2}
\displaystyle \left[\because x=0-h,\text{ when }x\to0^-,\text{ then }h\to0\right]
\displaystyle =\lim_{h\to0}\frac{1-\cos4h}{8h^2}\qquad \left[\because \cos(-\theta)=\cos\theta\right]
\displaystyle =\lim_{h\to0}\frac{2\sin^22h}{8h^2}\qquad \left[\because 1-\cos2\theta=2\sin^2\theta\right]
\displaystyle =\lim_{h\to0}\frac{\sin^22h}{4h^2}=\lim_{h\to0}\left(\frac{\sin2h}{2h}\right)^2=1
\displaystyle \qquad \left[\because \lim_{x\to0}\frac{\sin x}{x}=1\right]\qquad \ldots(ii)
\displaystyle \text{On substituting this value in Eq. (i), we get}
\displaystyle 1=f(0)\Rightarrow1=k\qquad \left[\because f(0)=k,\text{ given}\right]
\displaystyle \text{Hence, for }k=1,\text{ the given function }f(x)\text{ is continuous at }x=0.
\\

\displaystyle \textbf{Question 16. }\text{If }f(x)=\begin{cases}\dfrac{1-\cos4x}{x^2},&\text{when }x<0\\a,&\text{when }x=0\\\dfrac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4},&\text{when }x>0\end{cases}
\displaystyle \text{and }f\text{ is continuous at }x=0,\text{ then find the value of }a. \hspace{2.2cm} \text{[CBSE 2013C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f(x)=\begin{cases}\frac{1-\cos4x}{x^2}, & \text{when }x<0\\ a, & \text{when }x=0\\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & \text{when }x>0\end{cases}
\displaystyle \text{Since, }f(x)\text{ is continuous at }x=0.
\displaystyle \therefore (\text{LHL})_{x=0}=(\text{RHL})_{x=0}=f(0)\qquad \ldots(i)
\displaystyle \text{Now, (LHL)}_{x=0}=\lim_{x\to0^-}f(x)=\lim_{x\to0^-}\frac{1-\cos4x}{x^2}
\displaystyle =\lim_{h\to0}\frac{1-\cos4(0-h)}{(0-h)^2}
\displaystyle \left[\because x=0-h,\text{ when }x\to0^-,\text{ then }h\to0\right]
\displaystyle =\lim_{h\to0}\frac{1-\cos4h}{h^2}\qquad \left[\because \cos(-\theta)=\cos\theta\right]
\displaystyle =\lim_{h\to0}\frac{2\sin^22h}{h^2}\qquad \left[\because 1-\cos2\theta=2\sin^2\theta\right]
\displaystyle =2\lim_{h\to0}\left(\frac{\sin2h}{2h}\right)^2\times4
\displaystyle =2\times(1)^2\times4\qquad \left[\because \lim_{x\to0}\frac{\sin x}{x}=1\right]
\displaystyle =8
\displaystyle \text{Now, from Eq. (i), we have}
\displaystyle (\text{LHL})_{x=0}=f(0)
\displaystyle \Rightarrow 8=a\qquad \left[\because f(0)=a\text{ (given)}\right]
\displaystyle \therefore a=8
\\

\displaystyle \textbf{Question 17. }\text{Find the value of }k,\text{ for which}
\displaystyle f(x)=\begin{cases}\dfrac{\sqrt{1+kx}-\sqrt{1-kx}}{x},&\text{if }-1\leq x<0\\\dfrac{2x+1}{x-1},&\text{if }0\leq x<1\end{cases}
\displaystyle \text{is continuous at }x=0. \hspace{2.2cm} \text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f(x)=\begin{cases}\frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}, & \text{if }-1\leq x<0\\ \frac{2x+1}{x-1}, & \text{if }0\leq x<1\end{cases}
\displaystyle \text{is continuous at }x=0.
\displaystyle \text{Now, }f(0)=\frac{2\cdot0+1}{0-1}=\frac{1}{-1}=-1\qquad \ldots(i)
\displaystyle \text{and LHL}=\lim_{h\to0}f(0-h)
\displaystyle =\lim_{h\to0}\frac{\sqrt{1-kh}-\sqrt{1+kh}}{-h}
\displaystyle =\lim_{h\to0}\frac{\sqrt{1-kh}-\sqrt{1+kh}}{-h}\times\frac{\sqrt{1-kh}+\sqrt{1+kh}}{\sqrt{1-kh}+\sqrt{1+kh}}
\displaystyle =\lim_{h\to0}\frac{(1-kh)-(1+kh)}{-h(\sqrt{1-kh}+\sqrt{1+kh})}
\displaystyle \hspace{1cm}\left[\because (a+b)(a-b)=a^2-b^2\right]
\displaystyle =\lim_{h\to0}\frac{-2kh}{-h(\sqrt{1-kh}+\sqrt{1+kh})}
\displaystyle =\lim_{h\to0}\frac{2k}{\sqrt{1-kh}+\sqrt{1+kh}}=\frac{2k}{1+1}=\frac{2k}{2}=k
\displaystyle \text{Since, }f(x)\text{ is continuous at }x=0.
\displaystyle \therefore f(0)=\text{LHL}\Rightarrow-1=k
\displaystyle \therefore k=-1
\\

\displaystyle \textbf{Question 18. }\text{Find the value of }k,\text{ so that the following function} \\ \text{is continuous at }x=2. \hspace{2.2cm} \text{[CBSE 2012C]}
\displaystyle f(x)=\begin{cases}\dfrac{x^3+x^2-16x+20}{(x-2)^2},&x\neq2\\k,&x=2\end{cases}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f(x)=\begin{cases}\frac{x^3+x^2-16x+20}{(x-2)^2}, & x\neq2\\ k, & x=2\end{cases}
\displaystyle \text{is continuous at }x=2.
\displaystyle \text{Now, we have }f(2)=k
\displaystyle \text{and }\lim_{x\to2}f(x)=\lim_{x\to2}\frac{x^3+x^2-16x+20}{(x-2)^2}
\displaystyle =\lim_{x\to2}\frac{(x-2)(x^2+3x-10)}{(x-2)^2}
\displaystyle =\lim_{x\to2}\frac{(x-2)(x+5)(x-2)}{(x-2)^2}
\displaystyle =\lim_{x\to2}(x+5)=2+5=7
\displaystyle \text{Since, }f(x)\text{ is continuous at }x=2.
\displaystyle \therefore \lim_{x\to2}f(x)=f(2)\Rightarrow k=7
\\

\displaystyle \textbf{Question 19. }\text{Find the value of }k,\text{ so that the function }f\text{ defined by}
\displaystyle f(x)=\begin{cases}\dfrac{k\cos x}{\pi-2x},&x\neq\dfrac{\pi}{2}\\3,&x=\dfrac{\pi}{2}\end{cases}
\displaystyle \text{is continuous at }x=\dfrac{\pi}{2}. \hspace{2.2cm} \text{[CBSE 2012C; CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f(x)=\begin{cases}\frac{k\cos x}{\pi-2x}, & \text{if }x\neq\frac{\pi}{2}\\ 3, & \text{if }x=\frac{\pi}{2}\end{cases}
\displaystyle \text{is continuous at }x=\frac{\pi}{2}
\displaystyle \text{Then, at }x=\frac{\pi}{2},\ \text{LHL}=\text{RHL}=f\left(\frac{\pi}{2}\right)\qquad \ldots(i)
\displaystyle \text{Now, LHL}=\lim_{x\to\frac{\pi}{2}^-}f(x)=\lim_{x\to\frac{\pi}{2}^-}\frac{k\cos x}{\pi-2x}
\displaystyle =\lim_{h\to0}\frac{k\cos\left(\frac{\pi}{2}-h\right)}{\pi-2\left(\frac{\pi}{2}-h\right)}
\displaystyle \left[\because x=\frac{\pi}{2}-h,\text{ when }x\to\frac{\pi}{2}^-,\text{ then }h\to0\right]
\displaystyle =\lim_{h\to0}\frac{k\sin h}{\pi-\pi+2h}\qquad \left[\because \cos\left(\frac{\pi}{2}-\theta\right)=\sin\theta\right]
\displaystyle =\lim_{h\to0}\frac{k\sin h}{2h}
\displaystyle =\frac{k}{2}\lim_{h\to0}\frac{\sin h}{h}
\displaystyle =\frac{k}{2}\times1\qquad \left[\because \lim_{h\to0}\frac{\sin h}{h}=1\right]
\displaystyle \Rightarrow \text{LHL}=\frac{k}{2}
\displaystyle \text{Also, from the given function, we get}
\displaystyle f\left(\frac{\pi}{2}\right)=3
\displaystyle \text{Now, from Eq. (i), we have}
\displaystyle \text{LHL}=f\left(\frac{\pi}{2}\right)\Rightarrow\frac{k}{2}=3
\displaystyle \Rightarrow k=6
\\

\displaystyle \textbf{Question 20. }\text{Find the value of }a\text{ for which the function }f\text{ is defined as}
\displaystyle f(x)=\begin{cases}a\sin\dfrac{\pi}{2}(x+1),&x\leq0\\\dfrac{\tan x-\sin x}{x^3},&x>0\end{cases}
\displaystyle \text{is continuous at }x=0. \hspace{2.2cm} \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }f(x)=\begin{cases}a\sin\frac{\pi}{2}(x+1), & x\leq0\\ \frac{\tan x-\sin x}{x^3}, & x>0\end{cases}
\displaystyle \text{is continuous at }x=0.
\displaystyle \text{Then, LHL}=\text{RHL}=f(0)\qquad \ldots(i)
\displaystyle \text{Now, LHL}=\lim_{x\to0^-}a\sin\frac{\pi}{2}(x+1)
\displaystyle \Rightarrow \text{LHL}=\lim_{h\to0}a\sin\frac{\pi}{2}(-h+1)
\displaystyle \left[\because x=0-h,\text{ when }x\to0^-,\text{ then }h\to0\right]
\displaystyle =a\sin\frac{\pi}{2}=a\qquad \ldots(ii)
\displaystyle f(0)=a\sin\frac{\pi}{2}=a\qquad \ldots(iii)
\displaystyle \text{Now, we need to evaluate RHL at }x=a.
\displaystyle \left[\because \text{LHL}=f(0)=a\text{ and from this, we cannot find the value of }a\right]
\displaystyle \text{Here, RHL}=\lim_{x\to0^+}\frac{\tan x-\sin x}{x^3}
\displaystyle \Rightarrow \text{RHL}=\lim_{h\to0}\frac{\tan h-\sin h}{h^3}
\displaystyle \left[\because x=0+h=h,\text{ when }x\to0^+,\text{ then }h\to0\right]
\displaystyle =\lim_{h\to0}\frac{\frac{\sin h}{\cos h}-\sin h}{h^3}
\displaystyle =\lim_{h\to0}\frac{\sin h-\sin h\cos h}{h^3\cos h}
\displaystyle =\lim_{h\to0}\frac{\sin h(1-\cos h)}{h^3\cos h}
\displaystyle =\lim_{h\to0}\frac{\sin h}{h}\cdot\lim_{h\to0}\frac{1-\cos h}{h^2}\cdot\lim_{h\to0}\frac{1}{\cos h}
\displaystyle =1\times\lim_{h\to0}\frac{1-\cos h}{h^2}\times1
\displaystyle \left[\because \lim_{h\to0}\frac{\sin h}{h}=1\text{ and }\lim_{h\to0}\frac{1}{\cos h}=\frac{1}{\cos0}=1\right]
\displaystyle =\lim_{h\to0}\frac{1-\cos h}{h^2}
\displaystyle =\lim_{h\to0}\frac{2\sin^2\frac{h}{2}}{h^2}\qquad \left[\because 1-\cos x=2\sin^2\frac{x}{2}\right]
\displaystyle =\lim_{h\to0}\frac{2\sin^2\frac{h}{2}}{\frac{h^2}{4}\times4}
\displaystyle =\lim_{h\to0}\frac12\times\frac{\sin^2\frac{h}{2}}{\frac{h^2}{4}}
\displaystyle =\frac12\lim_{h\to0}\left(\frac{\sin\frac{h}{2}}{\frac{h}{2}}\right)^2=\frac12\times1
\displaystyle =\frac12\qquad \left[\because \lim_{x\to0}\frac{\sin x}{x}=1\right]
\displaystyle \therefore \text{RHL}=\frac12\qquad \ldots(iv)
\displaystyle \text{On substituting the values from Eqs. (ii), (iii) and (iv) to Eq. (i), we get}
\displaystyle \Rightarrow a=\frac12
\\

\displaystyle \textbf{Question 21. }\text{If the function }f(x)\text{ given by}
\displaystyle f(x)=\begin{cases}3ax+b,&x>1\\11,&x=1\\5ax-2b,&x<1\end{cases}
\displaystyle \text{is continuous at }x=1,\text{ then find the values of }a\text{ and }b. \hspace{0.2cm} \text{[CBSE 2011; CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }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.
\displaystyle \therefore \text{LHL}=\text{RHL}=f(1)\qquad \ldots(i)
\displaystyle \text{Now, LHL}=\lim_{x\to1^-}f(x)=\lim_{x\to1^-}(5ax-2b)
\displaystyle =\lim_{h\to0}\left[5a(1-h)-2b\right]
\displaystyle \left[\because x=1-h,\text{ when }x\to1^-,\text{ then }h\to0\right]
\displaystyle =\lim_{h\to0}(5a-5ah-2b)=5a-2b\qquad \ldots(ii)
\displaystyle \text{and RHL}=\lim_{x\to1^+}f(x)=\lim_{x\to1^+}(3ax+b)
\displaystyle =\lim_{h\to0}\left[3a(1+h)+b\right]
\displaystyle \left[\because x=1+h,\text{ when }x\to1^+,\text{ then }h\to0\right]
\displaystyle =\lim_{h\to0}(3a+3ah+b)=3a+b\qquad \ldots(iii)
\displaystyle \text{Also, given that }f(1)=11
\displaystyle \text{On substituting these values in Eq. (i), we get}
\displaystyle 5a-2b=3a+b=11
\displaystyle \Rightarrow 3a+b=11\qquad \ldots(ii)
\displaystyle \text{and }5a-2b=11\qquad \ldots(iii)
\displaystyle \text{On subtracting }3\times\text{Eq. (iii) from }5\times\text{Eq. (ii), we get}
\displaystyle 15a+5b-15a+6b=55-33
\displaystyle \Rightarrow 11b=22\Rightarrow b=2
\displaystyle \text{On putting the value of }b\text{ in Eq. (ii), we get}
\displaystyle 3a+2=11\Rightarrow3a=9\Rightarrow a=3
\displaystyle \text{Hence, }a=3\text{ and }b=2.
\\

\displaystyle \textbf{Question 22. }\text{Find the values of }a\text{ and }b\text{ such that the following function } \\ f(x)\text{ is a continuous function.}
\displaystyle f(x)=\begin{cases}5,&x\leq2\\ax+b,&2<x<10\\21,&x\geq10\end{cases} \hspace{2.2cm} \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f(x)=\begin{cases}5, & x\leq2\\ ax+b, & 2<x<10\\ 21, & x\geq10\end{cases}
\displaystyle \text{is a continuous function. So, it is continuous at }x=2\text{ and at }x=10.
\displaystyle \text{By definition,}
\displaystyle (\text{LHL})_{x=2}=(\text{RHL})_{x=2}=f(2)\qquad \ldots(i)
\displaystyle \text{and }(\text{LHL})_{x=10}=(\text{RHL})_{x=10}=f(10)\qquad \ldots(ii)
\displaystyle \text{Now, let us calculate LHL and RHL at }x=2.
\displaystyle \text{LHL}=\lim_{x\to2^-}f(x)=\lim_{x\to2^-}5=5
\displaystyle \text{and RHL}=\lim_{x\to2^+}f(x)=\lim_{x\to2^+}(ax+b)
\displaystyle =\lim_{h\to0}[a(2+h)+b]
\displaystyle \left[\because x=2+h,\text{ when }x\to2^+,\text{ then }h\to0\right]
\displaystyle =\lim_{h\to0}(2a+ah+b)
\displaystyle =2a+b
\displaystyle \text{From Eq. (i), we have}
\displaystyle \text{LHL}=\text{RHL}
\displaystyle \Rightarrow 2a+b=5\qquad \ldots(iii)
\displaystyle \text{Now, we have to find LHL and RHL at }x=10.
\displaystyle \text{LHL}=\lim_{x\to10^-}f(x)=\lim_{x\to10^-}(ax+b)
\displaystyle =\lim_{h\to0}[a(10-h)+b]
\displaystyle \left[\because x=10-h,\text{ when }x\to10^-,\text{ then }h\to0\right]
\displaystyle =\lim_{h\to0}(10a-ah+b)
\displaystyle \Rightarrow \text{LHL}=10a+b
\displaystyle \text{and RHL}=\lim_{x\to10^+}f(x)=\lim_{x\to10^+}21=21
\displaystyle \text{Now, from Eq. (ii), we have}
\displaystyle \text{LHL}=\text{RHL}
\displaystyle \Rightarrow 10a+b=21\qquad \ldots(iv)
\displaystyle \text{On subtracting Eq. (iii) from Eq. (iv), we get}
\displaystyle 8a=16\Rightarrow a=2
\displaystyle \text{On putting }a=2\text{ in Eq. (iv), we get}
\displaystyle 20+b=21\Rightarrow b=1
\displaystyle \text{Hence, }a=2\text{ and }b=1.
\\

\displaystyle \textbf{Question 23. }\text{Find the relationship between }a\text{ and }b,\text{ so that the function }f\text{ defined by}
\displaystyle f(x)=\begin{cases}ax+1,&x\leq3\\bx+3,&x>3\end{cases}
\displaystyle \text{is continuous at }x=3. \hspace{2.2cm} \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f(x)=\begin{cases}ax+1, & \text{if }x\leq3\\ bx+3, & \text{if }x>3\end{cases}
\displaystyle \text{is continuous at point }x=3.
\displaystyle \text{Then, }\text{LHL}=\text{RHL}=f(3)\qquad \ldots(i)
\displaystyle \text{Now, LHL}=\lim_{x\to3^-}f(x)=\lim_{x\to3^-}(ax+1)
\displaystyle =\lim_{h\to0}[a(3-h)+1]
\displaystyle \left[\because x=3-h,\text{ when }x\to3^-,\text{ then }h\to0\right]
\displaystyle =\lim_{h\to0}(3a-ah+1)
\displaystyle \Rightarrow \text{LHL}=3a+1
\displaystyle \text{and RHL}=\lim_{x\to3^+}f(x)=\lim_{x\to3^+}(bx+3)
\displaystyle =\lim_{h\to0}[b(3+h)+3]
\displaystyle \left[\because x=3+h,\text{ when }x\to3^+,\text{ then }h\to0\right]
\displaystyle =\lim_{h\to0}(3b+bh+3)
\displaystyle \Rightarrow \text{RHL}=3b+3
\displaystyle \text{From Eq. (i), we have}
\displaystyle \text{LHL}=\text{RHL}\Rightarrow3a+1=3b+3
\displaystyle \text{Then, }3a-3b=2,\text{ which is the required relation between }a\text{ and }b.
\\

\displaystyle \textbf{Question 24. }\text{Find the value of }k,\text{ so that the function }f\text{ defined by } \\ f(x)=\begin{cases}kx+1,&x\leq\pi\\\cos x,&x>\pi\end{cases}
\displaystyle \text{is continuous at }x=\pi. \hspace{2.2cm} \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }f(x)=\begin{cases}kx+1, & \text{if }x\leq\pi\\ \cos x, & \text{if }x>\pi\end{cases}
\displaystyle \text{is continuous at }x=\pi.
\displaystyle \text{Then, }\text{LHL}=\text{RHL}=f(\pi)\qquad \ldots(i)
\displaystyle \text{Now, LHL}=\lim_{x\to\pi^-}f(x)=\lim_{x\to\pi^-}(kx+1)
\displaystyle =\lim_{h\to0}[k(\pi-h)+1]
\displaystyle \left[\because x=\pi-h,\text{ when }x\to\pi^-,\text{ then }h\to0\right]
\displaystyle =\lim_{h\to0}(k\pi-kh+1)
\displaystyle \Rightarrow \text{LHL}=k\pi+1
\displaystyle \text{and RHL}=\lim_{x\to\pi^+}f(x)=\lim_{x\to\pi^+}\cos x
\displaystyle =\lim_{h\to0}\cos(\pi+h)
\displaystyle \left[\because x=\pi+h,\text{ when }x\to\pi^+,\text{ then }h\to0\right]
\displaystyle =\cos\pi
\displaystyle =-1\qquad \left[\because \cos\pi=-1\right]
\displaystyle \text{Now, from Eq. (i), we have}
\displaystyle \text{LHL}=\text{RHL}
\displaystyle \Rightarrow k\pi+1=-1\Rightarrow k\pi=-2
\displaystyle \therefore k=-\frac{2}{\pi}
\\

\displaystyle \textbf{Question 25. }\text{For what values of }\lambda,\text{ is the function}
\displaystyle f(x)=\begin{cases}\lambda(x^2-2x),&x\leq0\\4x+1,&x>0\end{cases}
\displaystyle \text{continuous at }x=0? \hspace{2.2cm} \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }f(x)=\begin{cases}\lambda(x^2-2x), & \text{if }x\leq0\\ 4x+1, & \text{if }x>0\end{cases}
\displaystyle \text{is continuous at }x=0.
\displaystyle \text{Then, }(\text{LHL})_{x=0}=(\text{RHL})_{x=0}=f(0)\qquad \ldots(i)
\displaystyle \text{Now, }f(0)=\lambda[0-0]^2=0,
\displaystyle \text{LHL}=\lim_{x\to0^-}f(x)=\lim_{h\to0}f(0-h)
\displaystyle =\lambda\lim_{h\to0}[(0-h)-2(0-h)]
\displaystyle =\lambda\times0\times0=0
\displaystyle \text{and RHL}=\lim_{x\to0^+}f(x)=\lim_{h\to0}f(0+h)
\displaystyle =\lim_{h\to0}[4(0+h)+1]=1
\displaystyle \therefore \text{LHL}\neq\text{RHL, which is a contradiction to Eq. (i).}
\displaystyle \therefore \text{There is no value of }\lambda\text{ for which }f(x)\text{ is continuous}
\displaystyle \text{at }x=0.
\\

\displaystyle \textbf{Question 26. }\text{Find the value of }a,\text{ if the function }f(x)\text{ defined by}
\displaystyle f(x)=\begin{cases}2x-1,&x<2\\a,&x=2\\x+1,&x>2\end{cases}
\displaystyle \text{is continuous at }x=2.\text{ Also, discuss the continuity of }f(x)\text{ at }x=3. \hspace{0.2cm} \text{[CBSE 2011C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f(x)=\begin{cases}2x-1, & x<2\\ a, & x=2\\ x+1, & x>2\end{cases}\text{ is continuous at }x=2.
\displaystyle \therefore (\text{LHL})_{x=2}=(\text{RHL})_{x=2}=f(2)\qquad \ldots(i)
\displaystyle \text{Now, }f(2)=a
\displaystyle \text{and LHL}=\lim_{x\to2^-}f(x)=\lim_{x\to2^-}(2x-1)
\displaystyle =\lim_{h\to0}[2(2-h)-1]=3
\displaystyle \left[\because x=2-h,\text{ when }x\to2^-,\text{ then }h\to0\right]
\displaystyle \text{From Eq. (i), we have}
\displaystyle \text{LHL}=f(2)\Rightarrow a=3
\displaystyle \text{Now, let us check the continuity at }x=3.
\displaystyle \text{Consider, }\lim_{x\to3}f(x)=\lim_{x\to3}(x+1)
\displaystyle \hspace{3cm}\left[\because f(x)=x+1\text{ for }x>2\right]
\displaystyle =4=f(3)\qquad \left[\because f(3)=3+1=4\right]
\displaystyle \therefore f(x)\text{ is continuous at }x=3.
\\

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.