\displaystyle \textbf{Question 1. }\text{If }f(x)=\frac{4-x^{2}}{4x-x^{3}},\text{ then the function is } \ \
\displaystyle \text{(a) discontinuous at only one point} \ \
\displaystyle \text{(b) discontinuous at exactly two points} \ \
\displaystyle \text{(c) discontinuous at exactly three points} \ \
\displaystyle \text{(d) discontinuous at exactly four points} \ \
\displaystyle \text{Answer:}
\displaystyle \text{(c) } \text{Here, } f(x)=\frac{4-x^{2}}{4x-x^{3}}
\displaystyle \text{This is a rational function, so it will be discontinuous at points, where }
\displaystyle \text{denominator becomes zero. }
\displaystyle 4x-x^{3}=0
\displaystyle \Rightarrow x(4-x^{2})=0
\displaystyle \Rightarrow x(2+x)(2-x)=0
\displaystyle \therefore x=0, x=-2 \text{ and } x=2
\displaystyle \text{Hence, the function } f(x)=\frac{4-x^{2}}{4x-x^{3}} \text{ will be discontinuous at }
\displaystyle \text{exactly three points } 0, -2 \text{ and } 2.
\\

\displaystyle \textbf{Question 2. }\text{The set of points, where the function }f(x)=x|x|\text{ is} \ \
\displaystyle \text{differentiable in  } \ \
\displaystyle \text{(a) }(-\infty,\infty) \ \
\displaystyle \text{(b) }(-\infty,0)\cup(0,\infty) \ \
\displaystyle \text{(c) }(0,\infty) \ \
\displaystyle \text{(d) }[0,\infty) \ \
\displaystyle \text{Answer:}
\displaystyle \text{(a) } \text{Given, function can be written as }
\displaystyle f(x)=\begin{cases} x^{2}, & \text{if } x>0 \\ -x^{2}, & \text{if } x<0 \\ 0, & \text{if } x=0 \end{cases}
\displaystyle \text{Now, it is clear that the function } f(x) \text{ is differentiable at all non-zero }
\displaystyle \text{values of } x.
\displaystyle \text{Also, } f'(0)=f(0)=0
\displaystyle \Rightarrow f(x) \text{ is differentiable at } x=0
\displaystyle \therefore f(x)=x|x| \text{ is differentiable in } (-\infty,\infty).
\\

\displaystyle \textbf{Question 3. }\text{Show below is a function which is differentiable at} \ \
\displaystyle x=3 \ \
\displaystyle f(x)=\begin{cases}-\log(4-x), & x<3 \\ p-qx, & x\geq 3\end{cases} \ \
\displaystyle \text{Find the values of }p\text{ and }q,\text{ where they are real} \ \
\displaystyle \text{numbers. Show your steps and give a valid reason.  } \ \
\displaystyle \text{Answer:}
\displaystyle \text{We have, } f(x)=\begin{cases} -\ln(4-x), & x<3 \\ p-qx, & x \ge 3 \end{cases}
\displaystyle \therefore f(x) \text{ is continuous at } x=3
\displaystyle \therefore \text{LHL = RHL}
\displaystyle \text{LHL}=\lim_{x \to 3^{-}}[-\ln(4-x)]
\displaystyle =\lim_{h \to 0}[-\ln(4-(3-h))]
\displaystyle =\lim_{h \to 0}[-\ln(1+h)]=0
\displaystyle \text{RHL}=\lim_{x \to 3^{+}}(p-qx)
\displaystyle =\lim_{h \to 0}[p-q(3+h)]
\displaystyle =\lim_{h \to 0}(p-3q-qh)
\displaystyle =p-3q
\displaystyle \therefore p-3q=0
\displaystyle \therefore f(x) \text{ is differentiable at } x=3
\displaystyle \therefore \text{LHD = RHD}
\displaystyle \text{LHD}=\lim_{h \to 0}\frac{f(3-h)-f(3)}{-h}
\displaystyle =\lim_{h \to 0}\frac{-\ln(4-(3-h))-(p-3q)}{-h}
\displaystyle =\lim_{h \to 0}\frac{-\ln(1+h)}{-h}
\displaystyle =\lim_{h \to 0}\frac{\ln(1+h)}{h}=1
\displaystyle \text{RHD}=\lim_{h \to 0}\frac{f(3+h)-f(3)}{h}
\displaystyle =\lim_{h \to 0}\frac{p-q(3+h)-(p-3q)}{h}
\displaystyle =\lim_{h \to 0}\frac{p-3q-qh-p+3q}{h}
\displaystyle =\lim_{h \to 0}\frac{-qh}{h}
\displaystyle =-q
\displaystyle \therefore -q=1 \Rightarrow q=-1
\displaystyle \therefore p-3(-1)=0 \Rightarrow p+3=0 \Rightarrow p=-3
\\

\displaystyle \textbf{Question 4. }\text{Prove that the function }f(x)=|x-1|,\ x\in R,\text{ is} \ \
\displaystyle \text{continuous at }x=1\text{ but not differentiable. } \ \
\displaystyle \text{Answer:}
\displaystyle \text{We have, } f(x)=|x-1|=\begin{cases} x-1, & x \ge 1 \\ 1-x, & x<1 \end{cases}
\displaystyle \text{Test for continuity at } x=1
\displaystyle \text{LHL}=\lim_{x \to 1^{-}} f(x)=\lim_{h \to 0} f(1-h)
\displaystyle =\lim_{h \to 0}(1-(1-h))=\lim_{h \to 0}h=0
\displaystyle \text{RHL}=\lim_{x \to 1^{+}} f(x)=\lim_{h \to 0} f(1+h)
\displaystyle =\lim_{h \to 0}((1+h)-1)=\lim_{h \to 0}h=0
\displaystyle \text{and } f(1)=0
\displaystyle \therefore \text{LHL = RHL = } f(1)=0
\displaystyle \text{Hence, } f(x) \text{ is continuous at } x=1
\displaystyle \text{Test for differentiability at } x=1
\displaystyle \text{LHD}=\lim_{h \to 0}\frac{f(1-h)-f(1)}{-h}
\displaystyle =\lim_{h \to 0}\frac{1-(1-h)-0}{-h}=\lim_{h \to 0}\frac{h}{-h}=-1
\displaystyle \text{RHD}=\lim_{h \to 0}\frac{f(1+h)-f(1)}{h}
\displaystyle =\lim_{h \to 0}\frac{(1+h)-1-0}{h}=\lim_{h \to 0}\frac{h}{h}=1
\displaystyle \therefore LHD \ne RHD
\displaystyle \text{Hence, } f \text{ is not differentiable at } x=1
\\

\displaystyle \textbf{Question 5. }\text{Find the values of }p\text{ and }q\text{ for which} \ \
\displaystyle f(x)=\begin{cases}\frac{(1-\sin^{3}x)}{3\cos^{2}x}, & x<\frac{\pi}{2} \\ p, & x=\frac{\pi}{2} \\ \frac{q(1-\sin x^{2})}{(\pi-2x)^{2}}, & x>\frac{\pi}{2}\end{cases} \ \
\displaystyle \text{is continuous at }x=\frac{\pi}{2}.\ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } f(x)=\begin{cases} \frac{1-\sin^{3}x}{3\cos^{2}x}, & x<\frac{\pi}{2} \\ p, & x=\frac{\pi}{2} \\ \frac{q(1-\sin x)}{(\pi-2x)^{2}}, & x>\frac{\pi}{2} \end{cases}
\displaystyle \text{is continuous at } x=\frac{\pi}{2}
\displaystyle \therefore \lim_{x \to \frac{\pi}{2}^{-}} f(x)=\lim_{x \to \frac{\pi}{2}^{+}} f(x)=f\left(\frac{\pi}{2}\right)
\displaystyle \text{Now, LHL}=\lim_{x \to \frac{\pi}{2}^{-}} f(x)=\lim_{h \to 0} f\left(\frac{\pi}{2}-h\right)
\displaystyle =\lim_{h \to 0}\frac{1-\sin^{3}\left(\frac{\pi}{2}-h\right)}{3\cos^{2}\left(\frac{\pi}{2}-h\right)}
\displaystyle =\lim_{h \to 0}\frac{1-\cos^{3}h}{3\sin^{2}h}
\displaystyle =\lim_{h \to 0}\frac{(1-\cos h)(1+\cos^{2}h+\cos h)}{3(1-\cos^{2}h)}
\displaystyle =\lim_{h \to 0}\frac{(1+\cos^{2}h+\cos h)}{3(1+\cos h)}=\frac{1+1+1}{3(1+1)}=\frac{3}{6}=\frac{1}{2}
\displaystyle =\lim_{h \to 0}\frac{1+\cos^{2}h+\cos h}{3(1+\cos h)}
\displaystyle =\frac{1+\cos^{2}0+\cos 0}{3(1+\cos 0)}
\displaystyle =\frac{1+1+1}{3(1+1)}=\frac{1}{2}
\displaystyle \text{and RHL}=\lim_{x \to \frac{\pi}{2}^{+}} f(x)=\lim_{h \to 0} f\left(\frac{\pi}{2}+h\right)
\displaystyle \text{Put } x=\frac{\pi}{2}+h, \text{ when } x \to \frac{\pi}{2}^{+}, \text{ then } h \to 0
\displaystyle =\lim_{h \to 0}\frac{q\left[1-\sin\left(\frac{\pi}{2}+h\right)\right]} {[\pi-2\left(\frac{\pi}{2}+h\right)]^{2}}
\displaystyle =\lim_{h \to 0}\frac{q(1-\cos h)}{(\pi-\pi-2h)^{2}} =\lim_{h \to 0}\frac{q(1-\cos h)}{4h^{2}}
\displaystyle =\lim_{h \to 0}\frac{q\left(2\sin^{2}\frac{h}{2}\right)}{4h^{2}}
\displaystyle =\frac{q}{8}\lim_{h \to 0}\frac{\sin^{2}\frac{h}{2}}{\left(\frac{h}{2}\right)^{2}}
\displaystyle =\frac{q}{8}\times (1)^{2}=\frac{q}{8}
\displaystyle \text{From Eq. (i), } \frac{1}{2}=\frac{q}{8}=p
\displaystyle \Rightarrow \frac{1}{2}=\frac{q}{8} \text{ and } \frac{1}{2}=p
\displaystyle \therefore q=4 \text{ and } p=\frac{1}{2}
\\

\displaystyle \textbf{Question 6. }\text{If }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,\text{ then find the values of }a\text{ and }b. \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } 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)
\displaystyle \text{Now, LHL}=\lim_{x \to 0^{-}} f(x)=\lim_{h \to 0} f(0-h)
\displaystyle \text{Put } x=0-h, \text{ when } x \to 0^{-}, \text{ then } h \to 0
\displaystyle =\lim_{h \to 0}\frac{\sin(a+1)(0-h)+2\sin(0-h)}{(0-h)}
\displaystyle =\lim_{h \to 0}\frac{-\sin(a+1)h-2\sin h}{-h}
\displaystyle =\lim_{h \to 0}\frac{\sin(a+1)h+2\sin h}{h}
\displaystyle =\lim_{h \to 0}\frac{\sin(a+1)h}{h}+\lim_{h \to 0}2\frac{\sin h}{h}
\displaystyle =\lim_{h \to 0}\frac{\sin(a+1)h}{(a+1)h}\times (a+1)+2\lim_{h \to 0}\frac{\sin h}{h}
\displaystyle =1\times (a+1)+2\times 1
\displaystyle =a+3
\displaystyle \text{and RHL}=\lim_{x \to 0^{+}} f(x)=\lim_{h \to 0} f(0+h)
\displaystyle \text{Put } x=0+h, \text{ when } x \to 0^{+}, \text{ then } h \to 0
\displaystyle =\lim_{h \to 0}\frac{\sqrt{1+b(0+h)}-1}{0+h} =\lim_{h \to 0}\frac{\sqrt{1+bh}-1}{h}
\displaystyle =\lim_{h \to 0}\frac{\sqrt{1+bh}-1}{h}\times \frac{\sqrt{1+bh}+1}{\sqrt{1+bh}+1}
\displaystyle =\lim_{h \to 0}\frac{(1+bh)-1}{h(\sqrt{1+bh}+1)}
\displaystyle =\lim_{h \to 0}\frac{bh}{h(\sqrt{1+bh}+1)}
\displaystyle =\lim_{h \to 0}\frac{b}{\sqrt{1+bh}+1}=\frac{b}{\sqrt{1+0}+1}=\frac{b}{2}
\displaystyle \text{From Eq. (i), we get } a+3=\frac{b}{2}=2
\displaystyle \Rightarrow a+3=2 \text{ and } \frac{b}{2}=2
\displaystyle \therefore a=-1 \text{ and } b=4
\\

\displaystyle \textbf{Question 7. }\text{Find the value of }k,\text{ so that the function} \ \
\displaystyle f(x)=\begin{cases}\frac{1-\cos 4x}{8x^{2}}, & x\ne 0 \\ k, & x=0\end{cases} \ \
\displaystyle \text{is continuous at }x=0. \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } f(x)=\begin{cases} \frac{1-\cos 4x}{8x^{2}}, & \text{if } x \ne 0 \\ k, & \text{if } 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{Here, LHL}=\lim_{x \to 0^{-}} f(x)=\lim_{x \to 0^{-}}\frac{1-\cos 4x}{8x^{2}}
\displaystyle =\lim_{h \to 0}\frac{1-\cos 4(0-h)}{8(0-h)^{2}}
\displaystyle \text{put } x=0-h, \text{ when } x \to 0^{-}, \text{ then } h \to 0
\displaystyle =\lim_{h \to 0}\frac{1-\cos 4h}{8h^{2}} \qquad [\because \cos(-\theta)=\cos \theta]
\displaystyle =\lim_{h \to 0}\frac{2\sin^{2}2h}{8h^{2}} \qquad [\because 1-\cos 2\theta=2\sin^{2}\theta]
\displaystyle =\lim_{h \to 0}\frac{\sin^{2}2h}{4h^{2}}=\lim_{h \to 0}\left(\frac{\sin 2h}{2h}\right)^{2}
\displaystyle =[1]^{2}=1 \qquad \left[\because \lim_{x \to 0}\frac{\sin x}{x}=1\right]
\displaystyle \text{At } x=0, f(0)=k
\displaystyle \text{Now, from Eq. (i), we have } \text{LHL}=f(0) \Rightarrow 1=k
\displaystyle \text{Hence, for } k=1, \text{ the given function } f(x) \text{ is continuous at } x=0.
\displaystyle \text{Alternate Method}
\displaystyle \text{Let } f(x)=\begin{cases} \frac{1-\cos 4x}{8x^{2}}, & \text{if } x \ne 0 \\ k, & \text{if } x=0 \end{cases} \text{ is continuous at } x=0.
\displaystyle \text{Here, } f(0)=k
\displaystyle \text{and } \lim_{x \to 0} f(x)=\lim_{x \to 0}\frac{1-\cos 4x}{8x^{2}} =\lim_{x \to 0}\frac{2\sin^{2}2x}{8x^{2}}
\displaystyle =\lim_{x \to 0}\left(\frac{\sin 2x}{2x}\right)^{2}=(1)^{2}=1
\displaystyle \therefore f(x) \text{ is continuous at } x=0
\displaystyle \Rightarrow \lim_{x \to 0} f(x)=f(0)
\displaystyle \Rightarrow 1=k \Rightarrow k=1
\\

\displaystyle \textbf{Question 8. }\text{If }f(x)=\begin{cases}\frac{1-\cos 4x}{x^{2}}, & x<0 \\ a, & x=0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & x>0\end{cases} \ \
\displaystyle \text{and }f\text{ is continuous at }x=0,\text{ then find the value of }a. \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } f(x)=\begin{cases} \frac{1-\cos 4x}{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 \lim_{x \to 0^{-}} f(x)=\lim_{x \to 0^{+}} f(x)=f(0) \qquad \ldots (i)
\displaystyle \text{Here, } \lim_{x \to 0^{-}} f(x)=\lim_{x \to 0^{-}}\frac{1-\cos 4x}{x^{2}}
\displaystyle =\lim_{h \to 0}\frac{1-\cos 4(0-h)}{(0-h)^{2}}
\displaystyle \text{put } x=0-h, \text{ when } x \to 0^{-}, \text{ then } h \to 0
\displaystyle =\lim_{h \to 0}\frac{1-\cos 4h}{h^{2}} \qquad [\because \cos(-\theta)=\cos \theta]
\displaystyle =\lim_{h \to 0}\frac{2\sin^{2}2h}{h^{2}} \qquad [\because 1-\cos 2\theta=2\sin^{2}\theta]
\displaystyle =2\lim_{h \to 0}\left(\frac{\sin 2h}{2h}\right)^{2}\times 4
\displaystyle =2\times (1)^{2}\times 4 \qquad \left[\because \lim_{h \to 0}\frac{\sin h}{h}=1\right]
\displaystyle =8
\displaystyle \text{Now, from Eq. (i), we have } \lim_{x \to 0^{-}} f(x)=f(0)
\displaystyle \Rightarrow 8=a
\displaystyle \therefore a=8
\\

\displaystyle \textbf{Question 9. }\text{Find the value of }k,\text{ for which} \ \
\displaystyle f(x)=\begin{cases}\frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}, & -1\leq x<0 \\ \frac{2x+1}{x-1}, & 0\leq x<1\end{cases} \ \
\displaystyle \text{is continuous at }x=0. \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let } f(x)=\begin{cases} \frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}, & \text{if } -1 \le x<0 \\ \frac{2x+1}{x-1}, & \text{if } 0 \le x<1 \end{cases}
\displaystyle \text{is continuous at } x=0.
\displaystyle \text{Here, } f(0)=\frac{2 \cdot 0+1}{0-1}=\frac{1}{-1}=-1
\displaystyle \text{and LHL}=\lim_{h \to 0} f(0-h)
\displaystyle =\lim_{h \to 0}\frac{\sqrt{1-kh}-\sqrt{1+kh}}{-h}
\displaystyle =\lim_{h \to 0}\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 \to 0}\frac{(1-kh)-(1+kh)}{-h\left(\sqrt{1-kh}+\sqrt{1+kh}\right)}
\displaystyle =\lim_{h \to 0}\frac{-2kh}{-h\left(\sqrt{1-kh}+\sqrt{1+kh}\right)} \qquad [\because (a+b)(a-b)=a^{2}-b^{2}]
\displaystyle =\lim_{h \to 0}\frac{2k}{\sqrt{1-kh}+\sqrt{1+kh}}=\frac{2k}{1+1}=\frac{2k}{2}=k
\displaystyle \therefore f(x) \text{ is continuous at } x=0.
\displaystyle \Rightarrow f(0)=\text{LHL} \Rightarrow -1=k
\displaystyle \therefore k=-1
\\

\displaystyle \textbf{Question 10. }\text{Find the value of }k,\text{ so that the following function is} \ \
\displaystyle \text{continuous at }x=2 \ \
\displaystyle f(x)=\begin{cases}\frac{x^{3}+x^{2}-16x+20}{(x-2)^{2}}, & x\ne 2 \\ k, & x=2\end{cases} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let } f(x)=\begin{cases} \frac{x^{3}+x^{2}-16x+20}{(x-2)^{2}}, & x \ne 2 \\ k, & x=2 \end{cases}
\displaystyle \text{is continuous at } x=2. \text{ Here, we have } f(2)=k
\displaystyle \text{and } \lim_{x \to 2} f(x)=\lim_{x \to 2}\frac{x^{3}+x^{2}-16x+20}{(x-2)^{2}}
\displaystyle =\lim_{x \to 2}\frac{(x-2)(x^{2}+3x-10)}{(x-2)^{2}}
\displaystyle =\lim_{x \to 2}\frac{(x-2)(x+5)(x-2)}{(x-2)^{2}}
\displaystyle =\lim_{x \to 2}(x+5)=2+5=7
\displaystyle \therefore f(x) \text{ is continuous at } x=2
\displaystyle \therefore \lim_{x \to 2} f(x)=f(2) \Rightarrow 7=k
\displaystyle \therefore k=7
\\

\displaystyle \textbf{Question 11. }\text{Find the value of }k,\text{ so that the function }f\text{ defined by} \ \
\displaystyle f(x)=\begin{cases}\frac{k\cos x}{\pi-2x}, & x\ne \frac{\pi}{2} \\ 3, & x=\frac{\pi}{2}\end{cases} \ \
\displaystyle \text{is continuous at }x=\frac{\pi}{2}. \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let } f(x)=\begin{cases} \frac{k\cos x}{\pi-2x}, & \text{if } x \ne \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 = RHL = } f\left(\frac{\pi}{2}\right) \qquad \ldots (i)
\displaystyle \text{Here, LHL}=\lim_{x \to \frac{\pi}{2}^{-}} f(x)=\lim_{x \to \frac{\pi}{2}^{-}}\frac{k\cos x}{\pi-2x}
\displaystyle \Rightarrow \text{LHL}=\lim_{h \to 0}\frac{k\cos \left(\frac{\pi}{2}-h\right)}{\pi-2\left(\frac{\pi}{2}-h\right)}
\displaystyle \text{put } x=\frac{\pi}{2}-h; \text{ when } x \to \frac{\pi}{2}^{-}, \text{ then } h \to 0
\displaystyle =\lim_{h \to 0}\frac{k\sin h}{\pi-\pi+2h} \qquad [\because \cos \left(\frac{\pi}{2}-\theta\right)=\sin \theta]
\displaystyle =\lim_{h \to 0}\frac{k\sin h}{2h}
\displaystyle =\frac{k}{2}\lim_{h \to 0}\frac{\sin h}{h}
\displaystyle \Rightarrow \text{LHL}=\frac{k}{2}\times 1=\frac{k}{2} \qquad \left[\because \lim_{h \to 0}\frac{\sin h}{h}=1\right]
\displaystyle \text{Also, from the given function, we get } f\left(\frac{\pi}{2}\right)=3
\displaystyle \text{From Eq. (i), we have } \text{LHL}=f\left(\frac{\pi}{2}\right) \Rightarrow \frac{k}{2}=3
\displaystyle \therefore k=6
\\

\displaystyle \textbf{Question 12. }\text{Find the value of }a\text{ for which the function }f\text{ is defined} \ \
\displaystyle \text{as} \ \
\displaystyle f(x)=\begin{cases}a\sin\frac{\pi}{2}(x+1), & x\leq 0 \\ \frac{\tan x-\sin x}{x^{3}}, & x>0\end{cases} \ \
\displaystyle \text{is continuous at }x=0. \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let } f(x)=\begin{cases} a\sin \frac{\pi}{2}(x+1), & x \le 0 \\ \frac{\tan x-\sin x}{x^{3}}, & x>0 \end{cases}
\displaystyle \text{is continuous at } x=0.
\displaystyle \therefore \text{LHL = RHL = } f(0) \qquad \ldots (i)
\displaystyle \text{Here, LHL}=\lim_{x \to 0^{-}} a\sin \frac{\pi}{2}(x+1)
\displaystyle \Rightarrow \text{LHL}=\lim_{h \to 0} a\sin \frac{\pi}{2}(0-h+1)
\displaystyle \text{put } x=0-h, \text{ when } x \to 0^{-}, \text{ then } h \to 0
\displaystyle =a\sin \frac{\pi}{2}=a \qquad \left[\because \sin \frac{\pi}{2}=1\right]
\displaystyle f(0)=a\sin \frac{\pi}{2}=a
\displaystyle \text{Now, we need to evaluate RHL at } x=0
\displaystyle \left[\because \text{LHL}=f(0)=a \text{ and from this, we can not find the value of } a\right]
\displaystyle \text{Here, RHL}=\lim_{x \to 0^{+}}\frac{\tan x-\sin x}{x^{3}}
\displaystyle \Rightarrow \text{RHL}=\lim_{h \to 0}\frac{\tan h-\sin h}{h^{3}}
\displaystyle \text{put } x=0+h, \text{ when } x \to 0^{+}, \text{ then } h \to 0
\displaystyle =\lim_{h \to 0}\frac{\frac{\sin h}{\cos h}-\sin h}{h^{3}}
\displaystyle =\lim_{h \to 0}\frac{\sin h-\sin h\cos h}{h^{3}\cos h}
\displaystyle =\lim_{h \to 0}\frac{\sin h(1-\cos h)}{h^{3}\cos h}
\displaystyle =\lim_{h \to 0}\frac{\sin h}{h}\cdot \lim_{h \to 0}\frac{1-\cos h}{h^{2}}\cdot \lim_{h \to 0}\frac{1}{\cos h}
\displaystyle =1\times \lim_{h \to 0}\frac{1-\cos h}{h^{2}}\times 1
\displaystyle \qquad \left[\because \lim_{h \to 0}\frac{\sin h}{h}=1 \text{ and } \lim_{h \to 0}\frac{1}{\cos h}=\frac{1}{\cos 0}=1\right]
\displaystyle =\lim_{h \to 0}\frac{1-\cos h}{h^{2}}
\displaystyle =\lim_{h \to 0}\frac{2\sin^{2}\frac{h}{2}}{h^{2}} \qquad [\because 1-\cos x=2\sin^{2}\frac{x}{2}]
\displaystyle =\lim_{h \to 0}2\cdot \frac{\sin^{2}\frac{h}{2}}{h^{2}}
\displaystyle =\lim_{h \to 0}2\cdot \frac{\sin^{2}\frac{h}{2}}{\left(\frac{h}{2}\right)^{2}}\cdot \frac{1}{4}
\displaystyle =\frac{1}{2}\lim_{h \to 0}\left(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right)^{2}
\displaystyle =\frac{1}{2}\times (1)^{2}=\frac{1}{2} \qquad \left[\because \lim_{x \to 0}\frac{\sin x}{x}=1\right]
\displaystyle \therefore \text{RHL}=\frac{1}{2}
\displaystyle \text{From Eq. (i), we have LHL = RHL}
\displaystyle \therefore a=\frac{1}{2}
\\

\displaystyle \textbf{Question 13. }\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. \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } f(x)=\begin{cases} 3ax+b, & x>1 \\ 11, & x=1 \\ 5ax-2b, & x<1 \end{cases}
\displaystyle \text{is continuous at } x=1.
\displaystyle \therefore \text{LHL}=\text{RHL}=f(1) \qquad \ldots (i)
\displaystyle \text{Here, LHL}=\lim_{x \to 1^{-}} f(x)=\lim_{x \to 1^{-}}(5ax-2b)
\displaystyle =\lim_{h \to 0}[5a(1-h)-2b] \qquad \text{put } x=1-h
\displaystyle =\lim_{h \to 0}(5a-5ah-2b)=5a-2b
\displaystyle \text{and RHL}=\lim_{x \to 1^{+}} f(x)=\lim_{x \to 1^{+}}(3ax+b)
\displaystyle =\lim_{h \to 0}[3a(1+h)+b] \qquad \text{put } x=1+h
\displaystyle =\lim_{h \to 0}(3a+3ah+b)=3a+b
\displaystyle \text{Also, given that } f(1)=11
\displaystyle \text{Now, from Eq. (i), we have } \text{RHL}=f(1) \Rightarrow 3a+b=11 \qquad \ldots (ii)
\displaystyle \text{and } \text{LHL}=f(1) \Rightarrow 5a-2b=11 \qquad \ldots (iii)
\displaystyle \text{On multiplying Eq. (ii) by 5 and Eq. (iii) by 3 and subtracting, we get}
\displaystyle 15a+5b=55
\displaystyle 15a-6b=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 \Rightarrow 3a=9 \Rightarrow a=3
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\displaystyle \textbf{Question 14. }\text{Find the values of }a\text{ and }b\text{ such that the following} \ \
\displaystyle \text{function }f(x)\text{ is a continuous function.} \ \
\displaystyle f(x)=\begin{cases}5, & x\leq 2 \\ ax+b, & 2<x<10 \\ 21, & x\geq 10\end{cases} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } f(x)=\begin{cases} 5, & x \le 2 \\ ax+b, & 2<x<10 \\ 21, & x \ge 10 \end{cases}
\displaystyle \text{is a continuous function.}
\displaystyle \text{So, it is continuous at } x=2 \text{ and } x=10.
\displaystyle \text{By definition, } (\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 \to 2^{-}} f(x)=\lim_{x \to 2^{-}}5=5
\displaystyle \text{and RHL}=\lim_{x \to 2^{+}} f(x)=\lim_{x \to 2^{+}}(ax+b)
\displaystyle =\lim_{h \to 0}[a(2+h)+b] \qquad \text{put } x=2+h
\displaystyle =\lim_{h \to 0}(2a+ah+b)=2a+b
\displaystyle \text{From Eq. (i), we have LHL = 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 \to 10^{-}} f(x)=\lim_{x \to 10^{-}}(ax+b)
\displaystyle =\lim_{h \to 0}[a(10-h)+b] \qquad \text{put } x=10-h
\displaystyle =\lim_{h \to 0}(10a-ah+b)=10a+b
\displaystyle \text{and RHL}=\lim_{x \to 10^{+}} f(x)=\lim_{x \to 10^{+}}21=21
\displaystyle \text{Now, from Eq. (ii), we have LHL = 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. (iii), we get}
\displaystyle 4+b=5 \Rightarrow b=1
\displaystyle \text{Hence, } a=2 \text{ and } b=1.
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