Class 12: Continuity – Exercise 9.2

$\displaystyle \textbf{Question 1: } \text{Prove that the function } f(x)= \begin{cases} \dfrac{\sin x}{x}, & x<0 \\ x+1, & x\ge 0 \end{cases} \text{ is everywhere continuous.}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When } x<0, \text{ we have}$

$\displaystyle f(x)=\frac{\sin x}{x}$

$\displaystyle \text{We know that } \sin x \text{ as well as the identity function } x \text{ are everywhere continuous.}$

$\displaystyle \text{So, the quotient function } \frac{\sin x}{x} \text{ is continuous for } x<0.$

$\displaystyle \text{When } x>0, \text{ we have}$

$\displaystyle f(x)=x+1$

$\displaystyle \text{Therefore, } f(x) \text{ is continuous at each } x>0.$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} \dfrac{\sin x}{x}, & x<0 \\ x+1, & x\ge 0 \end{cases}$

$\displaystyle \text{We have}$

$\displaystyle \text{(LHL at } x=0)=\lim_{x\to0^-} f(x) =\lim_{h\to0} f(-h) =\lim_{h\to0}\left(\frac{\sin(-h)}{-h}\right) =\lim_{h\to0}\left(\frac{\sin h}{h}\right) =1$

$\displaystyle \text{(RHL at } x=0)=\lim_{x\to0^+} f(x) =\lim_{h\to0} f(h) =\lim_{h\to0} (h+1) =1$

$\displaystyle \text{Also, } f(0)=0+1=1$

$\displaystyle \lim_{x\to0^-} f(x)=\lim_{x\to0^+} f(x)=f(0)$

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

$\displaystyle \text{Hence, } f(x) \text{ is everywhere continuous.}$

$\displaystyle \textbf{Question 2: } \text{Discuss the continuity of the function } f(x)= \begin{cases} \dfrac{x}{|x|}, & x\ne 0 \\ 0, & x=0 \end{cases}.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} \dfrac{x}{|x|}, & x\ne 0 \\ 0, & x=0 \end{cases}$

$\displaystyle |x|= \begin{cases} x, & x\ge 0 \\ -x, & x<0 \end{cases}$

$\displaystyle \Rightarrow f(x)= \begin{cases} 1, & x>0 \\ -1, & x<0 \\ 0, & x=0 \end{cases}$

$\displaystyle \text{We have}$

$\displaystyle \text{(LHL at } x=0)=\lim_{x\to0^-} f(x) =\lim_{h\to0} f(-h) =\lim_{h\to0} (-1) =-1$

$\displaystyle \text{(RHL at } x=0)=\lim_{x\to0^+} f(x) =\lim_{h\to0} f(h) =\lim_{h\to0} (1) =1$

$\displaystyle \lim_{x\to0^-} f(x)\ne \lim_{x\to0^+} f(x)$

$\displaystyle \text{Thus, } f(x) \text{ is discontinuous at } x=0.$

$\displaystyle \textbf{Question 3: } \text{Find the points of discontinuity, if any, of the following functions: }$

$\displaystyle \text{(i) } f(x)= \begin{cases} x^3-x^2+2x-2, & \text{if } x\ne 1 \\ 4, & \text{if } x=1 \end{cases}$

$\displaystyle \text{(ii) } f(x)= \begin{cases} \dfrac{x^4-16}{x-2}, & \text{if } x\ne 2 \\ 16, & \text{if } x=2 \end{cases}$

$\displaystyle \text{(iii) } f(x)= \begin{cases} \dfrac{\sin x}{x}, & \text{if } x<0 \\ 2x+3, & \text{if } x\ge 0 \end{cases}$

$\displaystyle \text{(iv) } f(x)= \begin{cases} \dfrac{\sin 3x}{x}, & \text{if } x\ne 0 \\ 4, & \text{if } x=0 \end{cases}$

$\displaystyle \text{(v) } f(x)= \begin{cases} \dfrac{\sin x}{x}+\cos x, & \text{if } x\ne 0 \\ 5, & \text{if } x=0 \end{cases}$

$\displaystyle \text{(vi) } f(x)= \begin{cases} \dfrac{x^4+x^3+2x^2}{\tan^{-1}x}, & \text{if } x\ne 0 \\ 10, & \text{if } x=0 \end{cases}$

$\displaystyle \text{(vii) } f(x)= \begin{cases} \dfrac{e^x-1}{\log_e(1+2x)}, & \text{if } x\ne 0 \\ 7, & \text{if } x=0 \end{cases}$

$\displaystyle \text{(viii) } f(x)= \begin{cases} |x-3|, & \text{if } x\ge 1 \\ \dfrac{x^2}{4}-\dfrac{3x}{2}+\dfrac{13}{4}, & \text{if } x<1 \end{cases}$

$\displaystyle \text{(ix) } f(x)= \begin{cases} |x|+3, & \text{if } x\le -3 \\ -2x, & \text{if } -3<x<3 \\ 6x+2, & \text{if } x>3 \end{cases} \hspace{5.0cm} \text{[CBSE 2010]}$

$\displaystyle \text{(x) } f(x)= \begin{cases} x^{10}-1, & \text{if } x\le 1 \\ x^2, & \text{if } x>1 \end{cases}$

$\displaystyle \text{(xi) } f(x)= \begin{cases} 2x, & \text{if } x<0 \\ 0, & \text{if } 0\le x\le 1 \\ 4x, & \text{if } x>1 \end{cases}$

$\displaystyle \text{(xii) } f(x)= \begin{cases} \sin x-\cos x, & \text{if } x\ne 0 \\ -1, & \text{if } x=0 \end{cases}$

$\displaystyle \text{(xiii) } f(x)= \begin{cases} -2, & \text{if } x\le -1 \\ 2x, & \text{if } -1<x<1 \\ 2, & \text{if } x\ge 1 \end{cases}$

$\displaystyle \text{Answer:}$

$\displaystyle \textbf{(i) }$

$\displaystyle \text{When } x\ne 1, \text{ then}$

$\displaystyle f(x)=x^3-x^2+2x-2$

$\displaystyle \text{We know that a polynomial function is everywhere continuous.}$

$\displaystyle \text{So, } f(x)=x^3-x^2+2x-2 \text{ is continuous at each } x\ne 1.$

$\displaystyle \text{At } x=1, \text{ we have}$

$\displaystyle \text{(LHL at } x=1)=\lim_{x\to1^-} f(x) =\lim_{h\to0} f(1-h) =\lim_{h\to0}\big((1-h)^3-(1-h)^2+2(1-h)-2\big) =1-1+2-2 =0$

$\displaystyle \text{(RHL at } x=1)=\lim_{x\to1^+} f(x) =\lim_{h\to0} f(1+h) =\lim_{h\to0}\big((1+h)^3-(1+h)^2+2(1+h)-2\big) =1-1+2-2 =0$

$\displaystyle \text{Also, } f(1)=4$

$\displaystyle \lim_{x\to1^-} f(x)=\lim_{x\to1^+} f(x)\ne f(1)$

$\displaystyle \text{Thus, } f(x) \text{ is discontinuous at } x=1.$

$\displaystyle \text{Hence, the only point of discontinuity of } f(x) \text{ is } x=1.$

$\displaystyle \textbf{(ii) }$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} \dfrac{x^4-16}{x-2}, & x\ne 2 \\ 16, & x=2 \end{cases}$

$\displaystyle \text{When } x\ne 2, \text{ then}$

$\displaystyle f(x)=\frac{x^4-16}{x-2} =\frac{x^4-2^4}{x-2} =\frac{(x^2+4)(x-2)(x+2)}{x-2} =(x^2+4)(x+2)$

$\displaystyle \text{We know that a polynomial function is everywhere continuous.}$

$\displaystyle \text{Therefore, the functions } (x^2+4) \text{ and } (x+2) \text{ are everywhere continuous.}$

$\displaystyle \text{So, the product function } (x^2+4)(x+2) \text{ is everywhere continuous.}$

$\displaystyle \text{Thus, } f(x) \text{ is continuous at every } x\ne 2.$

$\displaystyle \text{At } x=2, \text{ we have}$

$\displaystyle \text{(LHL at } x=2)=\lim_{x\to2^-} f(x) =\lim_{h\to0} f(2-h) =\lim_{h\to0}\big[(2-h)^2+4\big]\big[(2-h)+2\big] =8\times4 =32$

$\displaystyle \text{(RHL at } x=2)=\lim_{x\to2^+} f(x) =\lim_{h\to0} f(2+h) =\lim_{h\to0}\big[(2+h)^2+4\big]\big[(2+h)+2\big] =8\times4 =32$

$\displaystyle \text{Also, } f(2)=16$

$\displaystyle \lim_{x\to2^-} f(x)=\lim_{x\to2^+} f(x)\ne f(2)$

$\displaystyle \text{Thus, } f(x) \text{ is discontinuous at } x=2.$

$\displaystyle \text{Hence, the only point of discontinuity of } f(x) \text{ is } x=2.$

$\displaystyle \textbf{(iii) }$

$\displaystyle \text{When } x<0, \text{ then}$

$\displaystyle f(x)=\frac{\sin x}{x}$

$\displaystyle \text{We know that } \sin x \text{ as well as the identity function } x \text{ are everywhere continuous.}$

$\displaystyle \text{So, the quotient function } \frac{\sin x}{x} \text{ is continuous at each } x<0.$

$\displaystyle \text{When } x>0, \text{ then}$

$\displaystyle f(x)=2x+3, \text{ which is a polynomial function.}$

$\displaystyle \text{Therefore, } f(x) \text{ is continuous at each } x>0.$

$\displaystyle \text{Now, let us consider the point } x=0.$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} \dfrac{\sin x}{x}, & x<0 \\ 2x+3, & x\ge 0 \end{cases}$

$\displaystyle \text{We have}$

$\displaystyle \text{(LHL at } x=0)=\lim_{x\to0^-} f(x) =\lim_{h\to0} f(-h) =\lim_{h\to0}\left(\frac{\sin(-h)}{-h}\right) =\lim_{h\to0}\left(\frac{\sin h}{h}\right) =1$

$\displaystyle \text{(RHL at } x=0)=\lim_{x\to0^+} f(x) =\lim_{h\to0} f(h) =\lim_{h\to0} (2h+3) =3$

$\displaystyle \lim_{x\to0^-} f(x)\ne \lim_{x\to0^+} f(x)$

$\displaystyle \text{Thus, } f(x) \text{ is discontinuous at } x=0.$

$\displaystyle \text{Hence, the only point of discontinuity of } f(x) \text{ is } x=0.$

$\displaystyle \textbf{(iv) }$

$\displaystyle \text{When } x\ne 0, \text{ then}$

$\displaystyle f(x)=\frac{\sin 3x}{x}$

$\displaystyle \text{We know that } \sin 3x \text{ as well as the identity function } x \text{ are everywhere continuous.}$

$\displaystyle \text{So, the quotient function } \frac{\sin 3x}{x} \text{ is continuous at each } x\ne 0.$

$\displaystyle \text{Let us consider the point } x=0.$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} \dfrac{\sin 3x}{x}, & x\ne 0 \\ 4, & x=0 \end{cases}$

$\displaystyle \text{We have}$

$\displaystyle \text{(LHL at } x=0)=\lim_{x\to0^-} f(x) =\lim_{h\to0} f(-h) =\lim_{h\to0}\left(\frac{\sin(-3h)}{-h}\right) =\lim_{h\to0}\left(\frac{3\sin(3h)}{3h}\right) =3$

$\displaystyle \text{(RHL at } x=0)=\lim_{x\to0^+} f(x) =\lim_{h\to0} f(h) =\lim_{h\to0}\left(\frac{\sin(3h)}{h}\right) =\lim_{h\to0}\left(\frac{3\sin(3h)}{3h}\right) =3$

$\displaystyle \text{Also, } f(0)=4$

$\displaystyle \lim_{x\to0^-} f(x)=\lim_{x\to0^+} f(x)\ne f(0)$

$\displaystyle \text{Thus, } f(x) \text{ is discontinuous at } x=0.$

$\displaystyle \text{Hence, the only point of discontinuity of } f(x) \text{ is } x=0.$

$\displaystyle \textbf{(v) }$

$\displaystyle \text{When } x\ne 0, \text{ then}$

$\displaystyle f(x)=\frac{\sin x}{x}+\cos x$

$\displaystyle \text{We know that } \sin x \text{ as well as the identity function } x \text{ are everywhere continuous.}$

$\displaystyle \text{So, the quotient function } \frac{\sin x}{x} \text{ is continuous at each } x\ne 0.$

$\displaystyle \text{Also, } \cos x \text{ is everywhere continuous.}$

$\displaystyle \text{Therefore, } \frac{\sin x}{x}+\cos x \text{ is continuous at each } x\ne 0.$

$\displaystyle \text{Let us consider the point } x=0.$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} \dfrac{\sin x}{x}+\cos x, & x\ne 0 \\ 5, & x=0 \end{cases}$

$\displaystyle \text{We have}$

$\displaystyle \text{(LHL at } x=0)=\lim_{x\to0^-} f(x) =\lim_{h\to0} f(-h) =\lim_{h\to0}\left(\frac{\sin(-h)}{-h}+\cos(-h)\right) =\lim_{h\to0}\left(\frac{\sin h}{h}\right)+\lim_{h\to0}\cos h =1+1 =2$

$\displaystyle \text{(RHL at } x=0)=\lim_{x\to0^+} f(x) =\lim_{h\to0} f(h) =\lim_{h\to0}\left(\frac{\sin h}{h}+\cos h\right) =\lim_{h\to0}\left(\frac{\sin h}{h}\right)+\lim_{h\to0}\cos h =1+1 =2$

$\displaystyle \text{Also, } f(0)=5$

$\displaystyle \lim_{x\to0^-} f(x)=\lim_{x\to0^+} f(x)\ne f(0)$

$\displaystyle \text{Thus, } f(x) \text{ is discontinuous at } x=0.$

$\displaystyle \text{Hence, the only point of discontinuity of } f(x) \text{ is } x=0.$

$\displaystyle \textbf{(vi) }$

$\displaystyle \text{When } x\ne 0, \text{ then}$

$\displaystyle f(x)=\frac{x^4+x^3+2x^2}{\tan^{-1}x}$

$\displaystyle \text{We know that } x^4+x^3+2x^2 \text{ is a polynomial function which is everywhere continuous.}$

$\displaystyle \text{Also, } \tan^{-1}x \text{ is everywhere continuous.}$

$\displaystyle \text{So, the quotient function } \frac{x^4+x^3+2x^2}{\tan^{-1}x} \text{ is continuous at each } x\ne 0.$

$\displaystyle \text{Let us consider the point } x=0.$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} \dfrac{x^4+x^3+2x^2}{\tan^{-1}x}, & x\ne 0 \\ 10, & x=0 \end{cases}$

$\displaystyle \text{We have}$

$\displaystyle \text{(LHL at } x=0)=\lim_{x\to0^-} f(x) =\lim_{h\to0} f(-h) =\lim_{h\to0} \left( \frac{(-h)^4+(-h)^3+2(-h)^2}{\tan^{-1}(-h)} \right) =\lim_{h\to0} \left( \frac{h^4-h^3+2h^2}{-\tan^{-1}h} \right) =0$

$\displaystyle \text{(RHL at } x=0)=\lim_{x\to0^+} f(x) =\lim_{h\to0} f(h) =\lim_{h\to0} \left( \frac{h^4+h^3+2h^2}{\tan^{-1}h} \right) =0$

$\displaystyle \text{Also, } f(0)=10$

$\displaystyle \lim_{x\to0^-} f(x)=\lim_{x\to0^+} f(x)\ne f(0)$

$\displaystyle \text{Thus, } f(x) \text{ is discontinuous at } x=0.$

$\displaystyle \text{Hence, the only point of discontinuity of } f(x) \text{ is } x=0.$

$\displaystyle \textbf{(vii) }$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} \dfrac{e^x-1}{\log_e(1+2x)}, & x\ne 0 \\ 7, & x=0 \end{cases}$

$\displaystyle \text{We have}$

$\displaystyle \lim_{x\to0} f(x) =\lim_{x\to0}\frac{e^x-1}{\log_e(1+2x)} =\lim_{x\to0}\frac{\dfrac{e^x-1}{x}}{\dfrac{\log_e(1+2x)}{2x}}\times\frac{1}{2}$

$\displaystyle =\frac{1}{2}\times \lim_{x\to0}\left(\frac{e^x-1}{x}\right) \Bigg/ \lim_{x\to0}\left(\frac{\log_e(1+2x)}{2x}\right) =\frac{1}{2}$

$\displaystyle \text{It is given that } f(0)=7$

$\displaystyle \Rightarrow \lim_{x\to0} f(x)\ne f(0)$

$\displaystyle \text{Hence, the given function is discontinuous at } x=0 \text{ and continuous elsewhere.}$

$\displaystyle \textbf{(viii) }$

$\displaystyle \text{When } x>1, \text{ then}$

$\displaystyle f(x)=|x-3|$

$\displaystyle \text{Since the modulus function is continuous, } f(x) \text{ is continuous for each } x>1.$

$\displaystyle \text{When } x<1, \text{ then}$

$\displaystyle f(x)=\frac{x^2}{4}-\frac{3x}{2}+\frac{13}{4}$

$\displaystyle \text{Since } x^2 \text{ and } 3x \text{ are polynomial functions, they are continuous.}$

$\displaystyle \text{Hence, } \frac{x^2}{4} \text{ and } \frac{3x}{2} \text{ are also continuous.}$

$\displaystyle \text{Also, } \frac{13}{4} \text{ is a constant function and hence continuous.}$

$\displaystyle \Rightarrow \frac{x^2}{4}-\frac{3x}{2}+\frac{13}{4} \text{ is continuous for each } x<1.$

$\displaystyle \Rightarrow f(x) \text{ is continuous for each } x<1.$

$\displaystyle \text{At } x=1, \text{ we have}$

$\displaystyle \text{(LHL at } x=1)=\lim_{x\to1^-} f(x) =\lim_{h\to0} f(1-h) =\lim_{h\to0}\left[\frac{(1-h)^2}{4}-\frac{3(1-h)}{2}+\frac{13}{4}\right] =\frac{1}{4}-\frac{3}{2}+\frac{13}{4} =2$

$\displaystyle \text{(RHL at } x=1)=\lim_{x\to1^+} f(x) =\lim_{h\to0} f(1+h) =\lim_{h\to0}|1+h-3| =|-2| =2$

$\displaystyle f(1)=|1-3| =|-2| =2$

$\displaystyle \text{Thus, } \lim_{x\to1^-} f(x)=\lim_{x\to1^+} f(x)=f(1).$

$\displaystyle \text{Hence, } f(x) \text{ is continuous at } x=1.$

$\displaystyle \text{Thus, the given function is nowhere discontinuous.}$

$\displaystyle \textbf{(ix) }$

$\displaystyle \text{At } x\le -3, \text{ we have}$

$\displaystyle f(x)=|x|+3$

$\displaystyle \text{Since the modulus function and constant function are continuous,}$

$\displaystyle f(x)=|x|+3 \text{ is continuous for each } x\le -3.$

$\displaystyle \text{At } -3<x<3, \text{ we have}$

$\displaystyle f(x)=-2x$

$\displaystyle \text{Since a polynomial function is continuous,}$

$\displaystyle f(x)=-2x \text{ is continuous for each } -3<x<3.$

$\displaystyle \text{At } x>3, \text{ we have}$

$\displaystyle f(x)=6x+2$

$\displaystyle \text{Since a polynomial function is continuous,}$

$\displaystyle f(x)=6x+2 \text{ is continuous for each } x>3.$

$\displaystyle \text{Now, we check the continuity of the function at the point } x=3.$

$\displaystyle \text{We have}$

$\displaystyle \text{(LHL at } x=3)=\lim_{x\to3^-} f(x) =\lim_{h\to0} f(3-h) =\lim_{h\to0}\big[-2(3-h)\big] =-6$

$\displaystyle \text{(RHL at } x=3)=\lim_{x\to3^+} f(x) =\lim_{h\to0} f(3+h) =\lim_{h\to0}\big[6(3+h)+2\big] =20$

$\displaystyle \lim_{x\to3^-} f(x)\ne \lim_{x\to3^+} f(x)$

$\displaystyle \text{Hence, the only point of discontinuity of the given function is } x=3.$

$\displaystyle \textbf{(x) }$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} x^{10}-1, & x\le 1 \\ x^2, & x>1 \end{cases}$

$\displaystyle \text{The given function } f \text{ is defined at all points of the real line.}$

$\displaystyle \text{Let } c \text{ be a point on the real line.}$

$\displaystyle \text{Case I:}$

$\displaystyle \text{If } c<1, \text{ then } f(c)=c^{10}-1 \text{ and}$

$\displaystyle \lim_{x\to c} f(x) =\lim_{x\to c}(x^{10}-1) =c^{10}-1$

$\displaystyle \therefore \lim_{x\to c} f(x)=f(c)$

$\displaystyle \text{Therefore, } f \text{ is continuous at all points } x \text{ such that } x<1.$

$\displaystyle \text{Case II:}$

$\displaystyle \text{If } c=1, \text{ then the left hand limit of } f \text{ at } x=1 \text{ is}$

$\displaystyle \lim_{x\to1^-} f(x) =\lim_{x\to1^-}(x^{10}-1) =1^{10}-1 =0$

$\displaystyle \text{The right hand limit of } f \text{ at } x=1 \text{ is}$

$\displaystyle \lim_{x\to1^+} f(x) =\lim_{x\to1^+}(x^2) =1^2 =1$

$\displaystyle \text{It is observed that the left and right hand limits of } f \text{ at } x=1 \text{ do not coincide.}$

$\displaystyle \text{Therefore, } f \text{ is not continuous at } x=1.$

$\displaystyle \text{Case III:}$

$\displaystyle \text{If } c>1, \text{ then } f(c)=c^2 \text{ and}$

$\displaystyle \lim_{x\to c} f(x) =\lim_{x\to c}(x^2) =c^2$

$\displaystyle \therefore \lim_{x\to c} f(x)=f(c)$

$\displaystyle \text{Therefore, } f \text{ is continuous at all points } x \text{ such that } x>1.$

$\displaystyle \text{Thus, from the above observation, it can be concluded that } x=1 \text{ is the only point of discontinuity of } f.$

$\displaystyle \textbf{(xi) }$

$\displaystyle \text{The given function is }$

$\displaystyle f(x)= \begin{cases} 2x, & x<0 \\ 0, & 0\le x\le 1 \\ 4x, & x>1 \end{cases}$

$\displaystyle \text{The given function is defined at all points of the real line.}$

$\displaystyle \text{Let } c \text{ be a point on the real line.}$

$\displaystyle \text{Case I:}$

$\displaystyle \text{If } c<0, \text{ then } f(c)=2c \text{ and}$

$\displaystyle \lim_{x\to c} f(x) =\lim_{x\to c}(2x) =2c$

$\displaystyle \therefore \lim_{x\to c} f(x)=f(c)$

$\displaystyle \text{Therefore, } f \text{ is continuous at all points } x \text{ such that } x<0.$

$\displaystyle \text{Case II:}$

$\displaystyle \text{If } c=0, \text{ then } f(c)=f(0)=0.$

$\displaystyle \text{The left hand limit of } f \text{ at } x=0 \text{ is}$

$\displaystyle \lim_{x\to0^-} f(x) =\lim_{x\to0^-}(2x) =0$

$\displaystyle \text{The right hand limit of } f \text{ at } x=0 \text{ is}$

$\displaystyle \lim_{x\to0^+} f(x) =\lim_{x\to0^+}(0) =0$

$\displaystyle \therefore \lim_{x\to0} f(x)=f(0)$

$\displaystyle \text{Therefore, } f \text{ is continuous at } x=0.$

$\displaystyle \text{Case III:}$

$\displaystyle \text{If } 0<c<1, \text{ then } f(c)=0 \text{ and}$

$\displaystyle \lim_{x\to c} f(x) =\lim_{x\to c}(0) =0$

$\displaystyle \therefore \lim_{x\to c} f(x)=f(c)$

$\displaystyle \text{Therefore, } f \text{ is continuous at all points } x \text{ such that } 0<x<1.$

$\displaystyle \text{Case IV:}$

$\displaystyle \text{If } c=1, \text{ then } f(c)=f(1)=0.$

$\displaystyle \text{The left hand limit of } f \text{ at } x=1 \text{ is}$

$\displaystyle \lim_{x\to1^-} f(x) =\lim_{x\to1^-}(0) =0$

$\displaystyle \text{The right hand limit of } f \text{ at } x=1 \text{ is}$

$\displaystyle \lim_{x\to1^+} f(x) =\lim_{x\to1^+}(4x) =4$

$\displaystyle \text{Since } \lim_{x\to1^-} f(x)\ne \lim_{x\to1^+} f(x),$

$\displaystyle \text{therefore, } f \text{ is not continuous at } x=1.$

$\displaystyle \text{Case V:}$

$\displaystyle \text{If } c>1, \text{ then } f(c)=4c \text{ and}$

$\displaystyle \lim_{x\to c} f(x) =\lim_{x\to c}(4x) =4c$

$\displaystyle \therefore \lim_{x\to c} f(x)=f(c)$

$\displaystyle \text{Therefore, } f \text{ is continuous at all points } x \text{ such that } x>1.$

$\displaystyle \text{Hence, } f \text{ is not continuous only at } x=1.$

$\displaystyle \textbf{(xii) }$

$\displaystyle \text{The given function } f \text{ is}$

$\displaystyle f(x)= \begin{cases} \sin x-\cos x, & x\ne 0 \\ -1, & x=0 \end{cases}$

$\displaystyle \text{It is evident that } f \text{ is defined at all points of the real line.}$

$\displaystyle \text{Let } c \text{ be a real number.}$

$\displaystyle \text{Case I:}$

$\displaystyle \text{If } c\ne 0, \text{ then } f(c)=\sin c-\cos c.$

$\displaystyle \lim_{x\to c} f(x) =\lim_{x\to c}(\sin x-\cos x) =\sin c-\cos c$

$\displaystyle \therefore \lim_{x\to c} f(x)=f(c)$

$\displaystyle \text{Therefore, } f \text{ is continuous at all points } x \text{ such that } x\ne 0.$

$\displaystyle \text{Case II:}$

$\displaystyle \text{If } c=0, \text{ then } f(0)=-1.$

$\displaystyle \lim_{x\to0^-} f(x) =\lim_{x\to0^-}(\sin x-\cos x) =\sin 0-\cos 0 =0-1 =-1$

$\displaystyle \lim_{x\to0^+} f(x) =\lim_{x\to0^+}(\sin x-\cos x) =\sin 0-\cos 0 =0-1 =-1$

$\displaystyle \therefore \lim_{x\to0^-} f(x)=\lim_{x\to0^+} f(x)=f(0)$

$\displaystyle \text{Therefore, } f \text{ is continuous at } x=0.$

$\displaystyle \text{From the above observations, it can be concluded that } f \text{ is continuous at every point of the real line.}$

$\displaystyle \text{Thus, } f \text{ is a continuous function.}$

$\displaystyle \textbf{(xiii) }$

$\displaystyle \text{The given function } f \text{ is}$

$\displaystyle f(x)= \begin{cases} -2, & x\le -1 \\ 2x, & -1<x<1 \\ 2, & x\ge 1 \end{cases}$

$\displaystyle \text{The given function is defined at all points of the real line.}$

$\displaystyle \text{Let } c \text{ be a point on the real line.}$

$\displaystyle \text{Case I:}$

$\displaystyle \text{If } c<-1, \text{ then } f(c)=-2 \text{ and}$

$\displaystyle \lim_{x\to c} f(x) =\lim_{x\to c}(-2) =-2$

$\displaystyle \therefore \lim_{x\to c} f(x)=f(c)$

$\displaystyle \text{Therefore, } f \text{ is continuous at all points } x \text{ such that } x<-1.$

$\displaystyle \text{Case II:}$

$\displaystyle \text{If } c=-1, \text{ then } f(c)=f(-1)=-2.$

$\displaystyle \text{The left hand limit of } f \text{ at } x=-1 \text{ is}$

$\displaystyle \lim_{x\to-1^-} f(x) =\lim_{x\to-1^-}(-2) =-2$

$\displaystyle \text{The right hand limit of } f \text{ at } x=-1 \text{ is}$

$\displaystyle \lim_{x\to-1^+} f(x) =\lim_{x\to-1^+}(2x) =2(-1) =-2$

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

$\displaystyle \text{Therefore, } f \text{ is continuous at } x=-1.$

$\displaystyle \text{Case III:}$

$\displaystyle \text{If } -1<c<1, \text{ then } f(c)=2c \text{ and}$

$\displaystyle \lim_{x\to c} f(x) =\lim_{x\to c}(2x) =2c$

$\displaystyle \therefore \lim_{x\to c} f(x)=f(c)$

$\displaystyle \text{Therefore, } f \text{ is continuous at all points of the interval } (-1,1).$

$\displaystyle \text{Case IV:}$

$\displaystyle \text{If } c=1, \text{ then } f(c)=f(1)=2.$

$\displaystyle \text{The left hand limit of } f \text{ at } x=1 \text{ is}$

$\displaystyle \lim_{x\to1^-} f(x) =\lim_{x\to1^-}(2x) =2$

$\displaystyle \text{The right hand limit of } f \text{ at } x=1 \text{ is}$

$\displaystyle \lim_{x\to1^+} f(x) =\lim_{x\to1^+}(2) =2$

$\displaystyle \therefore \lim_{x\to1^-} f(x)=\lim_{x\to1^+} f(x)=f(1)$

$\displaystyle \text{Therefore, } f \text{ is continuous at } x=1.$

$\displaystyle \text{Case V:}$

$\displaystyle \text{If } c>1, \text{ then } f(c)=2 \text{ and}$

$\displaystyle \lim_{x\to c} f(x) =\lim_{x\to c}(2) =2$

$\displaystyle \therefore \lim_{x\to c} f(x)=f(c)$

$\displaystyle \text{Therefore, } f \text{ is continuous at all points } x \text{ such that } x>1.\\ \text{Thus, from the above observations, it can be concluded that } f \\ \text{ is continuous at all points of the real line.}$

$\displaystyle \textbf{Question 4: } \text{In the following, determine the value(s) of contant(s) involved in } \\ \text{the definition so that the given function is continuous:}$

$\displaystyle \text{(i) } f(x)= \begin{cases} \dfrac{\sin 2x}{5x}, & x\ne 0 \\ 3k, & x=0 \end{cases}$

$\displaystyle \text{(ii) } f(x)= \begin{cases} kx+5, & x\le 2 \\ x-1, & x>2 \end{cases}$

$\displaystyle \text{(iii) } f(x)= \begin{cases} k(x^2+3x), & x<0 \\ \cos 2x, & x\ge 0 \end{cases}$

$\displaystyle \text{(iv) } f(x)= \begin{cases} 2, & x\le 3 \\ ax+b, & 3<x<5 \\ 9, & x\ge 5 \end{cases}$

$\displaystyle \text{(v) } f(x)= \begin{cases} 4, & x\le -1 \\ ax^2+b, & -1<x<0 \\ \cos x, & x\ge 0 \end{cases}$

$\displaystyle \text{(vi) } f(x)= \begin{cases} \dfrac{\sqrt{1+px}-\sqrt{1-px}}{x}, & -1\le x<0 \\ \dfrac{2x+1}{x-2}, & 0\le x\le 1 \end{cases}$

$\displaystyle \text{(vii) } f(x)= \begin{cases} 5, & x\le 2 \\ ax+b, & 2<x<10 \\ 21, & x\ge 10 \end{cases}$

$\displaystyle \text{(viii) } f(x)= \begin{cases} \dfrac{k\cos x}{\pi-2x}, & x<\dfrac{\pi}{2} \\ 3, & x=\dfrac{\pi}{2} \\ \dfrac{3\tan 2x}{2x-\pi}, & x>\dfrac{\pi}{2} \end{cases}$

$\displaystyle \text{Answer:}$

$\displaystyle \textbf{(i) }$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} \dfrac{\sin 2x}{5x}, & x\ne 0 \\ 3k, & x=0 \end{cases}$

$\displaystyle \text{If } f(x) \text{ is continuous at } x=0, \text{ then}$

$\displaystyle \lim_{x\to0} f(x)=f(0)$

$\displaystyle \Rightarrow \lim_{x\to0}\frac{\sin 2x}{5x}=3k$

$\displaystyle \Rightarrow \lim_{x\to0}\frac{2\sin 2x}{2\cdot5x}=3k$

$\displaystyle \Rightarrow \frac{2}{5}\lim_{x\to0}\frac{\sin 2x}{2x}=3k$

$\displaystyle \Rightarrow \frac{2}{5}\times 1=3k$

$\displaystyle \Rightarrow k=\frac{2}{15}$

$\displaystyle \textbf{(ii) }$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} kx+5, & x\le 2 \\ x-1, & x>2 \end{cases}$

$\displaystyle \text{If } f(x) \text{ is continuous at } x=2, \text{ then}$

$\displaystyle \lim_{x\to2^-} f(x)=\lim_{x\to2^+} f(x)=f(2)$

$\displaystyle \Rightarrow \lim_{h\to0} f(2-h)=\lim_{h\to0} f(2+h)$

$\displaystyle \Rightarrow \lim_{h\to0}\big[k(2-h)+5\big] =\lim_{h\to0}\big[(2+h)-1\big]$

$\displaystyle \Rightarrow 2k+5=1$

$\displaystyle \Rightarrow 2k=-4$

$\displaystyle \Rightarrow k=-2$

$\displaystyle \textbf{(iii) }$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} k(x^2+3x), & x<0 \\ \cos 2x, & x\ge 0 \end{cases}$

$\displaystyle \text{If } f(x) \text{ is continuous at } x=0, \text{ then}$

$\displaystyle \lim_{x\to0^-} f(x)=\lim_{x\to0^+} f(x)=f(0)$

$\displaystyle \Rightarrow \lim_{h\to0} f(-h)=\lim_{h\to0} f(h)$

$\displaystyle \Rightarrow \lim_{h\to0} k\big(({-h})^2+3(-h)\big) =\lim_{h\to0}\cos(2h)$

$\displaystyle \Rightarrow \lim_{h\to0} k(h^2-3h)=\lim_{h\to0}\cos(2h)$

$\displaystyle \Rightarrow 0=1$

$\displaystyle \text{This is not possible.}$

$\displaystyle \text{Hence, there does not exist any value of } k \text{ for which the given function is continuous at } x=0.$

$\displaystyle \textbf{(iv) }$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} 2, & x\le 3 \\ ax+b, & 3<x<5 \\ 9, & x\ge 5 \end{cases}$

$\displaystyle \text{If } f(x) \text{ is continuous at } x=3 \text{ and } x=5, \text{ then}$

$\displaystyle \lim_{x\to3^-} f(x)=\lim_{x\to3^+} f(x) \quad \text{and} \quad \lim_{x\to5^-} f(x)=\lim_{x\to5^+} f(x)$

$\displaystyle \Rightarrow \lim_{h\to0} f(3-h)=\lim_{h\to0} f(3+h) \quad \text{and} \quad \lim_{h\to0} f(5-h)=\lim_{h\to0} f(5+h)$

$\displaystyle \Rightarrow \lim_{h\to0}(2)=\lim_{h\to0}\big(a(3+h)+b\big) \quad \text{and} \quad \lim_{h\to0}\big(a(5-h)+b\big)=\lim_{h\to0}(9)$

$\displaystyle \Rightarrow 2=3a+b \quad \text{and} \quad 5a+b=9$

$\displaystyle \Rightarrow a=\frac{7}{2}, \quad b=-\frac{17}{2}$

$\displaystyle \textbf{(v) }$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} 4, & x<-1 \\ ax^2+b, & -1<x<0 \\ \cos x, & x\ge 0 \end{cases}$

$\displaystyle \text{If } f(x) \text{ is continuous at } x=-1 \text{ and } x=0, \text{ then}$

$\displaystyle \lim_{x\to-1^-} f(x)=\lim_{x\to-1^+} f(x) \quad \text{and} \quad \lim_{x\to0^-} f(x)=\lim_{x\to0^+} f(x)$

$\displaystyle \Rightarrow \lim_{h\to0} f(-1-h)=\lim_{h\to0} f(-1+h) \quad \text{and} \quad \lim_{h\to0} f(-h)=\lim_{h\to0} f(h)$

$\displaystyle \Rightarrow \lim_{h\to0}(4) =\lim_{h\to0}\big[a(-1+h)^2+b\big] \quad \text{and} \quad \lim_{h\to0}\big[a(-h)^2+b\big] =\lim_{h\to0}(\cos h)$

$\displaystyle \Rightarrow 4=a+b \quad \text{and} \quad b=1$

$\displaystyle \Rightarrow a=3 \text{ and } b=1$

$\displaystyle \textbf{(vi) }$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} \dfrac{\sqrt{1+px}-\sqrt{1-px}}{x}, & -1<x<0 \\ \dfrac{2x+1}{x-2}, & 0\le x\le 1 \end{cases}$

$\displaystyle \text{If } f(x) \text{ is continuous at } x=0, \text{ then}$

$\displaystyle \lim_{x\to0^-} f(x)=\lim_{x\to0^+} f(x)$

$\displaystyle \Rightarrow \lim_{h\to0} f(-h)=\lim_{h\to0} f(h)$

$\displaystyle \Rightarrow \lim_{h\to0} \left( \frac{\sqrt{1-ph}-\sqrt{1+ph}}{-h} \right) = \lim_{h\to0} \left( \frac{2h+1}{h-2} \right)$

$\displaystyle \Rightarrow \lim_{h\to0} \left( \frac{(\sqrt{1-ph}-\sqrt{1+ph})(\sqrt{1-ph}+\sqrt{1+ph})} {-h(\sqrt{1-ph}+\sqrt{1+ph})} \right) = \lim_{h\to0} \left( \frac{2h+1}{h-2} \right)$

$\displaystyle \Rightarrow \lim_{h\to0} \left( \frac{(1-ph)-(1+ph)} {-h(\sqrt{1-ph}+\sqrt{1+ph})} \right) = \lim_{h\to0} \left( \frac{2h+1}{h-2} \right)$

$\displaystyle \Rightarrow \lim_{h\to0} \left( \frac{-2ph} {-h(\sqrt{1-ph}+\sqrt{1+ph})} \right) = \lim_{h\to0} \left( \frac{2h+1}{h-2} \right)$

$\displaystyle \Rightarrow \lim_{h\to0} \left( \frac{2p} {\sqrt{1-ph}+\sqrt{1+ph}} \right) = \lim_{h\to0} \left( \frac{2h+1}{h-2} \right)$

$\displaystyle \Rightarrow \frac{2p}{2}=\frac{1}{-2}$

$\displaystyle \Rightarrow p=-\frac{1}{2}$

$\displaystyle \textbf{(vii) }$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} 5, & x\le 2 \\ ax+b, & 2<x<10 \\ 21, & x\ge 10 \end{cases}$

$\displaystyle \text{If } f(x) \text{ is continuous at } x=2 \text{ and } x=10, \text{ then}$

$\displaystyle \lim_{x\to2^-} f(x)=\lim_{x\to2^+} f(x) \quad \text{and} \quad \lim_{x\to10^-} f(x)=\lim_{x\to10^+} f(x)$

$\displaystyle \Rightarrow \lim_{h\to0} f(2-h)=\lim_{h\to0} f(2+h) \quad \text{and} \quad \lim_{h\to0} f(10-h)=\lim_{h\to0} f(10+h)$

$\displaystyle \Rightarrow \lim_{h\to0}(5)=\lim_{h\to0}\big[a(2+h)+b\big] \quad \text{and} \quad \lim_{h\to0}\big[a(10-h)+b\big]=\lim_{h\to0}(21)$

$\displaystyle \Rightarrow 5=2a+b \qquad (1)$

$\displaystyle \Rightarrow 10a+b=21 \qquad (2)$

$\displaystyle \text{On solving equations } (1) \text{ and } (2), \text{ we get}$

$\displaystyle a=2 \quad \text{and} \quad b=1$

$\displaystyle \textbf{(viii) }$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} \dfrac{k\cos x}{\pi-2x}, & x<\dfrac{\pi}{2} \\ 3, & x=\dfrac{\pi}{2} \\ \dfrac{3\tan 2x}{2x-\pi}, & x>\dfrac{\pi}{2} \end{cases}$

$\displaystyle \text{If } f(x) \text{ is continuous at } x=\dfrac{\pi}{2}, \text{ then}$

$\displaystyle \lim_{x\to\frac{\pi}{2}} f(x)=f\!\left(\frac{\pi}{2}\right)$

$\displaystyle \Rightarrow \lim_{h\to0} f\!\left(\frac{\pi}{2}-h\right) =f\!\left(\frac{\pi}{2}\right)$

$\displaystyle \Rightarrow \lim_{h\to0} f\!\left(\frac{\pi}{2}-h\right)=3$

$\displaystyle \Rightarrow \lim_{h\to0} \left[ \frac{k\cos\left(\frac{\pi}{2}-h\right)} {\pi-2\left(\frac{\pi}{2}-h\right)} \right] =3$

$\displaystyle \Rightarrow \lim_{h\to0} \left[ \frac{k\sin h}{\pi-\pi+2h} \right] =3$

$\displaystyle \Rightarrow \lim_{h\to0} \left[ \frac{k\sin h}{2h} \right] =3$

$\displaystyle \Rightarrow \frac{k}{2} \lim_{h\to0}\frac{\sin h}{h} =3$

$\displaystyle \Rightarrow \frac{k}{2}\cdot 1=3$

$\displaystyle \Rightarrow k=6$

$\displaystyle \textbf{Question 5: } \text{The function } f(x)= \begin{cases} \dfrac{x^2}{a}, & \text{if } 0\le x<1 \\ a, & \text{if } 1\le x<\sqrt{2} \\ \dfrac{2b^2-4b}{x^2}, & \text{if } \sqrt{2}\le x<\infty \end{cases} \\ \text{is continuous on } [0,\infty). \text{ Find the most suitable values of } a \text{ and } b.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given: } f \text{ is continuous on } (0,\infty)$

$\displaystyle \therefore f \text{ is continuous at } x=1 \text{ and } x=\sqrt{2}.$

$\displaystyle \text{At } x=1, \text{ we have}$

$\displaystyle \lim_{x\to1^-} f(x) =\lim_{h\to0} f(1-h) =\lim_{h\to0}\left[\frac{(1-h)^2}{a}\right] =\frac{1}{a}$

$\displaystyle \lim_{x\to1^+} f(x) =\lim_{h\to0} f(1+h) =\lim_{h\to0}(a) =a$

$\displaystyle \text{At } x=\sqrt{2}, \text{ we have}$

$\displaystyle \lim_{x\to\sqrt{2}^-} f(x) =\lim_{h\to0} f(\sqrt{2}-h) =\lim_{h\to0}(a) =a$

$\displaystyle \lim_{x\to\sqrt{2}^+} f(x) =\lim_{h\to0} f(\sqrt{2}+h) =\lim_{h\to0}\left[\frac{2b^2-4b}{(\sqrt{2}+h)^2}\right] =\frac{2b^2-4b}{2} =b^2-2b$

$\displaystyle \text{Since } f \text{ is continuous at } x=1 \text{ and } x=\sqrt{2},$

$\displaystyle \lim_{x\to1^-} f(x)=\lim_{x\to1^+} f(x) \quad \text{and} \quad \lim_{x\to\sqrt{2}^-} f(x)=\lim_{x\to\sqrt{2}^+} f(x)$

$\displaystyle \Rightarrow \frac{1}{a}=a \quad \text{and} \quad b^2-2b=a$

$\displaystyle \Rightarrow a^2=1 \quad \text{and} \quad b^2-2b=a \qquad (1)$

$\displaystyle \Rightarrow a=\pm1$

$\displaystyle \text{If } a=1,$

$\displaystyle b^2-2b=1 \quad [\text{from (1)}]$

$\displaystyle \Rightarrow b^2-2b-1=0$

$\displaystyle \Rightarrow b=\frac{2\pm\sqrt{4+4}}{2} =\frac{2\pm2\sqrt{2}}{2} =1\pm\sqrt{2}$

$\displaystyle \text{If } a=-1,$

$\displaystyle b^2-2b=-1 \quad [\text{from (1)}]$

$\displaystyle \Rightarrow b^2-2b+1=0$

$\displaystyle \Rightarrow (b-1)^2=0$

$\displaystyle \Rightarrow b=1$

$\displaystyle \text{Hence, the most suitable values of } a \text{ and } b \text{ are }$

$\displaystyle (a,b)=(-1,1) \text{ or } (1,\,1\pm\sqrt{2}).$

$\displaystyle \textbf{Question 6: } \text{Find the values of } a \text{ and } b \text{ so that the function } f(x) \text{ defined by} \\ \\ f(x)= \begin{cases} x+a\sqrt{2}\,\sin x, & \text{if } 0\le x<\dfrac{\pi}{4} \\ 2x\cot x+b, & \text{if } \dfrac{\pi}{4}\le x<\dfrac{\pi}{2} \\ a\cos 2x-b\sin x, & \text{if } \dfrac{\pi}{2}\le x\le \pi \end{cases} \text{ becomes continuous on } [0,\pi].$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given: } f \text{ is continuous on } [0,\pi].$

$\displaystyle \therefore f \text{ is continuous at } x=\frac{\pi}{4} \text{ and } x=\frac{\pi}{2}.$

$\displaystyle \text{At } x=\frac{\pi}{4}, \text{ we have}$

$\displaystyle \lim_{x\to\frac{\pi}{4}^-} f(x) =\lim_{h\to0} f\!\left(\frac{\pi}{4}-h\right) =\lim_{h\to0}\left[\left(\frac{\pi}{4}-h\right) +a\sqrt{2}\sin\!\left(\frac{\pi}{4}-h\right)\right]$

$\displaystyle =\frac{\pi}{4}+a\sqrt{2}\sin\!\left(\frac{\pi}{4}\right) =\frac{\pi}{4}+a$

$\displaystyle \lim_{x\to\frac{\pi}{4}^+} f(x) =\lim_{h\to0} f\!\left(\frac{\pi}{4}+h\right) =\lim_{h\to0}\left[2\left(\frac{\pi}{4}+h\right) \cot\!\left(\frac{\pi}{4}+h\right)+b\right]$

$\displaystyle =\frac{\pi}{2}\cot\!\left(\frac{\pi}{4}\right)+b =\frac{\pi}{2}+b$

$\displaystyle \text{At } x=\frac{\pi}{2}, \text{ we have}$

$\displaystyle \lim_{x\to\frac{\pi}{2}^-} f(x) =\lim_{h\to0} f\!\left(\frac{\pi}{2}-h\right) =\lim_{h\to0}\left[2\left(\frac{\pi}{2}-h\right) \cot\!\left(\frac{\pi}{2}-h\right)+b\right] =b$

$\displaystyle \lim_{x\to\frac{\pi}{2}^+} f(x) =\lim_{h\to0} f\!\left(\frac{\pi}{2}+h\right) =\lim_{h\to0}\left[a\cos^2\!\left(\frac{\pi}{2}+h\right) -b\sin\!\left(\frac{\pi}{2}+h\right)\right] =-a-b$

$\displaystyle \text{Since } f \text{ is continuous at } x=\frac{\pi}{4} \text{ and } x=\frac{\pi}{2},$

$\displaystyle \lim_{x\to\frac{\pi}{4}^-} f(x) =\lim_{x\to\frac{\pi}{4}^+} f(x) \quad \text{and} \quad \lim_{x\to\frac{\pi}{2}^-} f(x) =\lim_{x\to\frac{\pi}{2}^+} f(x)$

$\displaystyle \Rightarrow \frac{\pi}{4}+a=\frac{\pi}{2}+b \quad \text{and} \quad b=-a-b$

$\displaystyle \Rightarrow b=-\frac{a}{2} \qquad (1)$

$\displaystyle \Rightarrow -\frac{\pi}{4}=b-a \qquad (2)$

$\displaystyle \Rightarrow -\frac{\pi}{4}=-\frac{a}{2}-a =-\frac{3a}{2}$

$\displaystyle \Rightarrow a=\frac{\pi}{6}$ $\displaystyle \Rightarrow b=-\frac{\pi}{12}$

$\displaystyle \textbf{Question 7: } \text{The function } f(x) \text{ is defined by } f(x)=\begin{cases} x^2+ax+b, & 0\le x<2 \\ 3x+2, & 2\le x\le 4 \\ 2ax+5b, & 4<x\le 8 \end{cases} \\ \text{If } f \text{ is continuous on } [0,8], \text{ find the values of } a \text{ and } b.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given: } f \text{ is continuous on } [0,8].$

$\displaystyle \therefore f \text{ is continuous at } x=2 \text{ and } x=4.$

$\displaystyle \text{At } x=2, \text{ we have}$

$\displaystyle \lim_{x\to2^-} f(x) =\lim_{h\to0} f(2-h) =\lim_{h\to0}\big[(2-h)^2+a(2-h)+b\big] =4+2a+b$

$\displaystyle \lim_{x\to2^+} f(x) =\lim_{h\to0} f(2+h) =\lim_{h\to0}\big[3(2+h)+2\big] =8$

$\displaystyle \text{At } x=4, \text{ we have}$

$\displaystyle \lim_{x\to4^-} f(x) =\lim_{h\to0} f(4-h) =\lim_{h\to0}\big[3(4-h)+2\big] =14$

$\displaystyle \lim_{x\to4^+} f(x) =\lim_{h\to0} f(4+h) =\lim_{h\to0}\big[2a(4+h)+5b\big] =8a+5b$

$\displaystyle \text{Since } f \text{ is continuous at } x=2 \text{ and } x=4,$

$\displaystyle \lim_{x\to2^-} f(x)=\lim_{x\to2^+} f(x) \quad \text{and} \quad \lim_{x\to4^-} f(x)=\lim_{x\to4^+} f(x)$

$\displaystyle \Rightarrow 4+2a+b=8 \quad \text{and} \quad 8a+5b=14$

$\displaystyle \Rightarrow 2a+b=4 \qquad (1)$

$\displaystyle \Rightarrow 8a+5b=14 \qquad (2)$

$\displaystyle \text{On solving equations } (1) \text{ and } (2), \text{ we get}$

$\displaystyle a=3 \quad \text{and} \quad b=-2.$

$\displaystyle \textbf{Question 8: } \text{If } f(x)=\dfrac{\tan\left(\dfrac{\pi}{4}-x\right)}{\cot 2x} \text{ for } x\ne \dfrac{\pi}{4}, \text{ find the value which can be assigned} \\ \text{so that the function } f(x) \text{ at } x = \frac{\pi}{4} \text{ so that becomes continuous every where in } \left[0,\dfrac{\pi}{2}\right].$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When } x\ne \frac{\pi}{4},$

$\displaystyle \tan\!\left(\frac{\pi}{4}-x\right) \text{ and } \cot(2x) \text{ are continuous in } \left[0,\frac{\pi}{2}\right].$

$\displaystyle \text{Thus, the quotient function } \frac{\tan\!\left(\frac{\pi}{4}-x\right)}{\cot(2x)} \text{ is continuous in } \left[0,\frac{\pi}{2}\right] \text{ for each } x\ne\frac{\pi}{4}.$

$\displaystyle \text{So, if } f(x) \text{ is continuous at } x=\frac{\pi}{4}, \text{ then it will be continuous everywhere in } \left[0,\frac{\pi}{2}\right].$

$\displaystyle \text{Now, let us consider the point } x=\frac{\pi}{4}.$

$\displaystyle \text{Given } f(x)=\frac{\tan\!\left(\frac{\pi}{4}-x\right)}{\cot(2x)}, \quad x\ne\frac{\pi}{4}.$

$\displaystyle \text{We have}$

$\displaystyle \text{(LHL at } x=\tfrac{\pi}{4})= \lim_{x\to\frac{\pi}{4}^-} f(x) =\lim_{h\to0} f\!\left(\frac{\pi}{4}-h\right)$

$\displaystyle =\lim_{h\to0} \left( \frac{\tan\!\left(\frac{\pi}{4}-\frac{\pi}{4}+h\right)} {\cot\!\left(\frac{\pi}{2}-2h\right)} \right) =\lim_{h\to0}\left(\frac{\tan h}{\tan(2h)}\right)$

$\displaystyle =\lim_{h\to0} \left( \frac{\dfrac{\tan h}{h}}{\dfrac{\tan(2h)}{2h}} \right)\cdot\frac{1}{2} =\frac{1}{2}$

$\displaystyle \text{(RHL at } x=\tfrac{\pi}{4})= \lim_{x\to\frac{\pi}{4}^+} f(x) =\lim_{h\to0} f\!\left(\frac{\pi}{4}+h\right)$

$\displaystyle =\lim_{h\to0} \left( \frac{\tan\!\left(\frac{\pi}{4}-\frac{\pi}{4}-h\right)} {\cot\!\left(\frac{\pi}{2}+2h\right)} \right) =\lim_{h\to0}\left(\frac{\tan(-h)}{-\tan(2h)}\right)$

$\displaystyle =\lim_{h\to0}\left(\frac{\tan h}{\tan(2h)}\right) =\frac{1}{2}$

$\displaystyle \text{If } f(x) \text{ is continuous at } x=\frac{\pi}{4}, \text{ then}$

$\displaystyle \lim_{x\to\frac{\pi}{4}^-} f(x) =\lim_{x\to\frac{\pi}{4}^+} f(x) =f\!\left(\frac{\pi}{4}\right)$

$\displaystyle \therefore f\!\left(\frac{\pi}{4}\right)=\frac{1}{2}.$

$\displaystyle \text{Hence, for } f\!\left(\frac{\pi}{4}\right)=\frac{1}{2}, \text{ the function } f(x) \text{ is continuous everywhere in } \left[0,\frac{\pi}{2}\right].$

$\displaystyle \textbf{Question 9: } \text{Discuss the continuity of the function } f(x)= \begin{cases} 2x-1, & \text{if } x<2 \\ \dfrac{3x}{2}, & \text{if } x\ge 2 \end{cases}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When } x<2, \text{ we have}$

$\displaystyle f(x)=2x-1$

$\displaystyle \text{We know that a polynomial function is everywhere continuous.}$

$\displaystyle \therefore f(x) \text{ is continuous for each } x<2.$

$\displaystyle \text{When } x>2, \text{ we have}$

$\displaystyle f(x)=\frac{3x}{2}$

$\displaystyle \text{The functions } 3x \text{ and } 2 \text{ are continuous, being a polynomial and a constant function, respectively.}$

$\displaystyle \therefore \frac{3x}{2} \text{ is continuous for each } x>2.$

$\displaystyle \text{Now, let us consider the point } x=2.$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} 2x-1, & x<2 \\ \dfrac{3x}{2}, & x\ge 2 \end{cases}$

$\displaystyle \text{We have}$

$\displaystyle \text{(LHL at } x=2)= \lim_{x\to2^-} f(x) =\lim_{h\to0} f(2-h) =\lim_{h\to0}[2(2-h)-1] =4-1 =3$

$\displaystyle \text{(RHL at } x=2)= \lim_{x\to2^+} f(x) =\lim_{h\to0} f(2+h) =\lim_{h\to0}\left(\frac{3(2+h)}{2}\right) =3$

$\displaystyle \text{Also, } f(2)=\frac{3(2)}{2}=3$

$\displaystyle \therefore \lim_{x\to2^-} f(x) =\lim_{x\to2^+} f(x) =f(2)$

$\displaystyle \text{Thus, } f(x) \text{ is continuous at } x=2.$

$\displaystyle \text{Hence, } f(x) \text{ is continuous for all real values of } x.$

$\displaystyle \textbf{Question 10: } \text{Discuss the continuity of } f(x)=\sin |x|.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let } f(x)=\sin|x|.$

$\displaystyle \text{The function } f \text{ is defined for every real number and can be written} \\ \text{as the composition of two functions } f=h\circ g,$

$\displaystyle \text{where } g(x)=|x| \text{ and } h(x)=\sin x.$

$\displaystyle \big[\because (h\circ g)(x)=h(g(x))=h(|x|)=\sin|x|\big]$

$\displaystyle \text{First, we prove that } g(x)=|x| \text{ and } h(x)=\sin x \text{ are continuous functions.}$

$\displaystyle g(x)=|x| \text{ can be written as}$

$\displaystyle g(x)= \begin{cases} -x, & x<0 \\ x, & x\ge 0 \end{cases}$

$\displaystyle \text{Clearly, } g \text{ is defined for all real numbers.}$

$\displaystyle \text{Let } c \text{ be a real number.}$

$\displaystyle \text{Case I:}$

$\displaystyle \text{If } c<0, \text{ then } g(c)=-c \text{ and}$

$\displaystyle \lim_{x\to c} g(x) =\lim_{x\to c}(-x) =-c =g(c)$

$\displaystyle \therefore \lim_{x\to c} g(x)=g(c)$

$\displaystyle \text{Case II:}$

$\displaystyle \text{If } c>0, \text{ then } g(c)=c \text{ and}$

$\displaystyle \lim_{x\to c} g(x) =\lim_{x\to c}(x) =c =g(c)$

$\displaystyle \therefore \lim_{x\to c} g(x)=g(c)$

$\displaystyle \text{Case III:}$

$\displaystyle \text{If } c=0, \text{ then } g(0)=0.$

$\displaystyle \lim_{x\to0^-} g(x) =\lim_{x\to0^-}(-x) =0$

$\displaystyle \lim_{x\to0^+} g(x) =\lim_{x\to0^+}(x) =0$

$\displaystyle \therefore \lim_{x\to0} g(x)=g(0)$

$\displaystyle \text{From the above cases, } g(x) \text{ is continuous for all real } x.$

$\displaystyle \text{Now, let } h(x)=\sin x.$

$\displaystyle \text{Clearly, } h(x) \text{ is defined for all real numbers.}$

$\displaystyle \text{Let } c \text{ be any real number and put } x=c+k.$

$\displaystyle \text{As } x\to c, \text{ we have } k\to0.$

$\displaystyle \lim_{x\to c} h(x) =\lim_{k\to0} \sin(c+k)$

$\displaystyle =\lim_{k\to0}(\sin c\cos k+\cos c\sin k)$

$\displaystyle =\sin c\cos 0+\cos c\sin 0$

$\displaystyle =\sin c$

$\displaystyle \therefore \lim_{x\to c} h(x)=h(c)$

$\displaystyle \text{Hence, } h(x)=\sin x \text{ is a continuous function.}$

$\displaystyle \text{Since } g \text{ and } h \text{ are continuous functions, their composition } f=h\circ g \text{ is also continuous.}$

$\displaystyle \therefore f(x)=\sin|x| \text{ is continuous for all real } x.$

$\displaystyle \textbf{Question 11: } \text{Prove that} f(x)= \begin{cases} \dfrac{\sin x}{x}, & x<0 \\ x+1, & x\ge 0 \end{cases} \text{ is everywhere continuous.}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When } x<0, \text{ we have}$

$\displaystyle f(x)=\frac{\sin x}{x}$

$\displaystyle \text{We know that } \sin x \text{ as well as the identity function } x \text{ are everywhere continuous.}$

$\displaystyle \text{So, the quotient function } \frac{\sin x}{x} \text{ is continuous at each } x<0.$

$\displaystyle \text{When } x>0, \text{ we have}$

$\displaystyle f(x)=x+1,$

$\displaystyle \text{which is a polynomial function.}$

$\displaystyle \therefore f(x) \text{ is continuous at each } x>0.$

$\displaystyle \text{Now, let us consider the point } x=0.$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} \dfrac{\sin x}{x}, & x<0 \\ x+1, & x\ge 0 \end{cases}$

$\displaystyle \text{We have}$

$\displaystyle \text{(LHL at } x=0)= \lim_{x\to0^-} f(x) =\lim_{h\to0} f(0-h) =\lim_{h\to0} f(-h) =\lim_{h\to0}\left(\frac{\sin(-h)}{-h}\right)$

$\displaystyle =\lim_{h\to0}\left(\frac{\sin h}{h}\right) =1$

$\displaystyle \text{(RHL at } x=0)= \lim_{x\to0^+} f(x) =\lim_{h\to0} f(0+h) =\lim_{h\to0} f(h) =\lim_{h\to0}(h+1) =1$

$\displaystyle \text{Also, } f(0)=0+1=1$

$\displaystyle \therefore \lim_{x\to0^-} f(x) =\lim_{x\to0^+} f(x) =f(0)$

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

$\displaystyle \text{Hence, } f(x) \text{ is continuous everywhere.}$

$\displaystyle \textbf{Question 12: } \text{Show that the function } g(x)=x-[x] \text{ is discontinuous at all integral points.} \\ \text{Here } [x] \text{ denotes the greatest integer function.}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given: } g(x)=x-[x].$

$\displaystyle \text{It is evident that } g \text{ is defined at all real numbers.}$

$\displaystyle \text{Let } n\in\mathbb{Z}.$

$\displaystyle \text{Then, } g(n)=n-[n]=n-n=0.$

$\displaystyle \text{The left hand limit of } g \text{ at } x=n \text{ is}$

$\displaystyle \lim_{x\to n^-} g(x) =\lim_{x\to n^-}(x-[x]) =\lim_{x\to n^-}x-\lim_{x\to n^-}[x] =n-(n-1) =1.$

$\displaystyle \text{The right hand limit of } g \text{ at } x=n \text{ is}$

$\displaystyle \lim_{x\to n^+} g(x) =\lim_{x\to n^+}(x-[x]) =\lim_{x\to n^+}x-\lim_{x\to n^+}[x] =n-n =0.$

$\displaystyle \text{It is observed that } \lim_{x\to n^-} g(x)\ne \lim_{x\to n^+} g(x).$

$\displaystyle \text{Therefore, } g \text{ is not continuous at } x=n,\; n\in\mathbb{Z}.$

$\displaystyle \text{Hence, } g \text{ is discontinuous at all integral points.}$

$\displaystyle \textbf{Question 13: } \text{Discuss the continuity of the following functions:}$

$\displaystyle \text{(i) } f(x)=\sin x+\cos x \\ \text{(ii) } f(x)=\sin x-\cos x \\ \text{(iii) } f(x)=\sin x\cos x$

$\displaystyle \text{Answer:}$

$\displaystyle \text{It is known that if } g \text{ and } h \text{ are two continuous functions, then } g+h,\; g-h \text{ and } g\times h \text{ are also continuous.}$

$\displaystyle \text{It has to be proved first that } g(x)=\sin x \text{ and } h(x)=\cos x \text{ are continuous functions.}$

$\displaystyle \text{Let } g(x)=\sin x.$

$\displaystyle \text{It is evident that } g(x)=\sin x \text{ is defined for every real number.}$

$\displaystyle \text{Let } c \text{ be a real number. Put } x=c+h.$

$\displaystyle \text{If } x\to c, \text{ then } h\to 0.$

$\displaystyle g(c)=\sin c.$

$\displaystyle \lim_{x\to c} g(x)=\lim_{x\to c}\sin x =\lim_{h\to 0}\sin(c+h)$

$\displaystyle =\lim_{h\to 0}[\sin c\cos h+\cos c\sin h]$

$\displaystyle =\lim_{h\to 0}(\sin c\cos h)+\lim_{h\to 0}(\cos c\sin h)$

$\displaystyle =\sin c\cos 0+\cos c\sin 0$

$\displaystyle =\sin c+0$

$\displaystyle =\sin c$

$\displaystyle \therefore\ \lim_{x\to c} g(x)=g(c).$

$\displaystyle \text{So, } g \text{ is a continuous function.}$

$\displaystyle \text{Let } h(x)=\cos x.$

$\displaystyle \text{It is evident that } h(x)=\cos x \text{ is defined for every real number.}$

$\displaystyle \text{Let } c \text{ be a real number. Put } x=c+h.$

$\displaystyle \text{If } x\to c, \text{ then } h\to 0.$

$\displaystyle h(c)=\cos c.$

$\displaystyle \lim_{x\to c} h(x)=\lim_{x\to c}\cos x =\lim_{h\to 0}\cos(c+h)$

$\displaystyle =\lim_{h\to 0}[\cos c\cos h-\sin c\sin h]$

$\displaystyle =\lim_{h\to 0}(\cos c\cos h)-\lim_{h\to 0}(\sin c\sin h)$

$\displaystyle =\cos c\cos 0-\sin c\sin 0$

$\displaystyle =\cos c\times 1-\sin c\times 0$

$\displaystyle =\cos c$

$\displaystyle \therefore\ \lim_{x\to c} h(x)=h(c).$

$\displaystyle \text{So, } h \text{ is a continuous function.}$

$\displaystyle \text{Therefore, it can be concluded that}$

$\displaystyle \text{(i) } f(x)=g(x)+h(x)=\sin x+\cos x \text{ is a continuous function.}$

$\displaystyle \text{(ii) } f(x)=g(x)-h(x)=\sin x-\cos x \text{ is a continuous function.}$

$\displaystyle \text{(iii) } f(x)=g(x)\times h(x)=\sin x\cos x \text{ is a continuous function.}$

$\displaystyle \textbf{Question 14: } \text{Show that } f(x)=\cos x^2 \text{ is a continuous function.}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given: } f(x)=\cos(x^2)$

$\displaystyle \text{This function } f \text{ is defined for every real number and } f \text{ can be written as the composition of two functions as } f=g\circ h, \text{ where } g(x)=\cos x \text{ and } h(x)=x^2.$

$\displaystyle [:\ (g\circ h)(x)=g(h(x))=g(x^2)=\cos(x^2)=f(x)\ :]$

$\displaystyle \text{It has to be first proved that } g(x)=\cos x \text{ and } h(x)=x^2 \text{ are continuous functions.}$

$\displaystyle \text{It is evident that } g \text{ is defined for every real number.}$

$\displaystyle \text{Let } c \text{ be a real number.}$

$\displaystyle \text{Then, } g(c)=\cos c.$

$\displaystyle \text{If } x\to c, \text{ then } h\to 0.$

$\displaystyle \lim_{x\to c} g(x)=\lim_{x\to c}\cos x =\lim_{h\to 0}\cos(c+h) =\lim_{h\to 0}[\cos c\cos h-\sin c\sin h]$

$\displaystyle =\lim_{h\to 0}\cos c\cos h-\lim_{h\to 0}\sin c\sin h =\cos c\cos 0-\sin c\sin 0$

$\displaystyle =\cos c\times1-\sin c\times0 =\cos c$

$\displaystyle \therefore\ \lim_{x\to c} g(x)=g(c).$

$\displaystyle \text{So, } g(x)=\cos x \text{ is a continuous function.}$

$\displaystyle \text{Now, } h(x)=x^2.$

$\displaystyle \text{Clearly, } h \text{ is defined for every real number.}$

$\displaystyle \text{Let } k \text{ be a real number, then } h(k)=k^2.$

$\displaystyle \lim_{x\to k} h(x)=\lim_{x\to k} x^2=k^2$

$\displaystyle \therefore\ \lim_{x\to k} h(x)=h(k).$

$\displaystyle \text{So, } h \text{ is a continuous function.}$

$\displaystyle \text{It is known that for real valued functions } g \text{ and } h, \text{ such that } (g\circ h) \text{ is defined at } x=c,$

$\displaystyle \text{if } g \text{ is continuous at } x=c \text{ and } h \text{ is continuous at } g(c), \text{ then } (f\circ g) \text{ is continuous at } x=c.$

$\displaystyle \text{Therefore, } f(x)=(g\circ h)(x)=\cos(x^2) \text{ is a continuous function.}$

$\displaystyle \textbf{Question 15: } \text{Show that } f(x)=|cos x^2| \text{ is a continuous function.}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{The given function is } f(x)=|\cos x|.$

$\displaystyle \text{This function } f \text{ is defined for every real number and can be written as the composition of two functions } f=g\circ h,$
$\displaystyle \text{where } g(x)=|x| \text{ and } h(x)=\cos x.$

$\displaystyle [\because (g\circ h)(x)=g(h(x))=g(\cos x)=|\cos x|=f(x)].$

$\displaystyle \text{First, we prove that } g(x)=|x| \text{ and } h(x)=\cos x \text{ are continuous functions.}$

$\displaystyle g(x)=|x| \text{ can be written as }$
$\displaystyle g(x)= \begin{cases} -x, & x<0,\\ x, & x\ge 0. \end{cases}$

$\displaystyle \text{Clearly, } g \text{ is defined for all real numbers.}$

$\displaystyle \text{Let } c \text{ be a real number.}$

$\displaystyle \text{Case I: If } c<0,$
$\displaystyle g(c)=-c \text{ and } \lim_{x\to c}g(x)=\lim_{x\to c}(-x)=-c.$

$\displaystyle \therefore \lim_{x\to c} g(x)=g(c).$

$\displaystyle \text{So, } g \text{ is continuous at all points } x<0.$

$\displaystyle \text{Case II: If } c>0,$
$\displaystyle g(c)=c \text{ and } \lim_{x\to c}g(x)=\lim_{x\to c}x=c.$

$\displaystyle \therefore \lim_{x\to c} g(x)=g(c).$

$\displaystyle \text{So, } g \text{ is continuous at all points } x>0.$

$\displaystyle \text{Case III: If } c=0,$
$\displaystyle g(0)=0,$

$\displaystyle \lim_{x\to0^-} g(x)=\lim_{x\to0^-}(-x)=0 \quad \text{and} \quad \lim_{x\to0^+} g(x)=\lim_{x\to0^+}x=0.$

$\displaystyle \therefore \lim_{x\to0} g(x)=g(0).$

$\displaystyle \text{Hence, } g(x)=|x| \text{ is continuous at all real numbers.}$

$\displaystyle \text{Now, let } h(x)=\cos x.$

$\displaystyle \text{It is evident that } h(x)=\cos x \text{ is defined for every real number.}$

$\displaystyle \text{Let } c \text{ be a real number and put } x=c+k.$

$\displaystyle \text{If } x\to c, \text{ then } k\to0.$

$\displaystyle h(c)=\cos c.$

$\displaystyle \lim_{x\to c} h(x)=\lim_{k\to0}\cos(c+k)$
$\displaystyle =\lim_{k\to0}[\cos c\cos k-\sin c\sin k]$
$\displaystyle =\cos c\cdot1-\sin c\cdot0$
$\displaystyle =\cos c.$

$\displaystyle \therefore \lim_{x\to c} h(x)=h(c).$

$\displaystyle \text{So, } h(x)=\cos x \text{ is a continuous function.}$

$\displaystyle \text{It is known that if } g \text{ and } h \text{ are continuous at } x=c,$
$\displaystyle \text{then } (g\circ h) \text{ is continuous at } x=c.$

$\displaystyle \text{Therefore, } f(x)=(g\circ h)(x)=g(h(x))=|\cos x| \text{ is a continuous function.}$

$\displaystyle \textbf{Question 16: } \text{Find all the points of discontinuity of } f \text{ defined by } f(x)=|x|-|x+1|.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given } f(x)=|x|-|x+1|.$

$\displaystyle \text{Define the functions } g \text{ and } h \text{ as } g(x)=|x| \text{ and } h(x)=|x+1|.$

$\displaystyle \text{Then, } f=g-h.$

$\displaystyle \text{We first examine the continuity of } g \text{ and } h.$

$\displaystyle g(x)=|x| \text{ can be written as }$
$\displaystyle g(x)= \begin{cases} -x, & x<0,\\ x, & x\ge 0. \end{cases}$

$\displaystyle \text{Clearly, } g \text{ is defined for all real numbers.}$

$\displaystyle \text{Let } c \in \mathbb{R}.$

$\displaystyle \text{Case I: If } c<0,$
$\displaystyle g(c)=-c \text{ and } \lim_{x\to c} g(x)=\lim_{x\to c}(-x)=-c.$
$\displaystyle \therefore \lim_{x\to c} g(x)=g(c).$

$\displaystyle \text{Hence, } g \text{ is continuous for all } x<0.$

$\displaystyle \text{Case II: If } c>0,$
$\displaystyle g(c)=c \text{ and } \lim_{x\to c} g(x)=\lim_{x\to c}(x)=c.$
$\displaystyle \therefore \lim_{x\to c} g(x)=g(c).$

$\displaystyle \text{Hence, } g \text{ is continuous for all } x>0.$

$\displaystyle \text{Case III: If } c=0,$
$\displaystyle g(0)=0,$
$\displaystyle \lim_{x\to 0^-} g(x)=\lim_{x\to 0^-}(-x)=0,$
$\displaystyle \lim_{x\to 0^+} g(x)=\lim_{x\to 0^+}(x)=0.$

$\displaystyle \therefore \lim_{x\to 0} g(x)=g(0).$

$\displaystyle \text{Thus, } g \text{ is continuous at } x=0.$

$\displaystyle \text{Hence, } g \text{ is continuous for all real } x.$

$\displaystyle h(x)=|x+1| \text{ can be written as }$
$\displaystyle h(x)= \begin{cases} -(x+1), & x<-1,\\ x+1, & x\ge -1. \end{cases}$

$\displaystyle \text{Clearly, } h \text{ is defined for all real numbers.}$

$\displaystyle \text{Let } c \in \mathbb{R}.$

$\displaystyle \text{Case I: If } c<-1,$
$\displaystyle h(c)=-(c+1) \text{ and } \lim_{x\to c} h(x)=\lim_{x\to c}[-(x+1)]=-(c+1).$
$\displaystyle \therefore \lim_{x\to c} h(x)=h(c).$

$\displaystyle \text{Hence, } h \text{ is continuous for all } x<-1.$

$\displaystyle \text{Case II: If } c>-1,$
$\displaystyle h(c)=c+1 \text{ and } \lim_{x\to c} h(x)=\lim_{x\to c}(x+1)=c+1.$
$\displaystyle \therefore \lim_{x\to c} h(x)=h(c).$

$\displaystyle \text{Hence, } h \text{ is continuous for all } x>-1.$

$\displaystyle \text{Case III: If } c=-1,$
$\displaystyle h(-1)=0,$
$\displaystyle \lim_{x\to -1^-} h(x)=\lim_{x\to -1^-}[-(x+1)]=0,$
$\displaystyle \lim_{x\to -1^+} h(x)=\lim_{x\to -1^+}(x+1)=0.$

$\displaystyle \therefore \lim_{x\to -1} h(x)=h(-1).$

$\displaystyle \text{Thus, } h \text{ is continuous at } x=-1.$

$\displaystyle \text{Hence, } h \text{ is continuous for all real } x.$

$\displaystyle \text{Since } g \text{ and } h \text{ are continuous everywhere, their difference } f=g-h \text{ is also continuous.}$

$\displaystyle \text{Therefore, } f(x)=|x|-|x+1| \text{ is continuous for all real } x.$

$\displaystyle \textbf{Question 17: } \text{Is } f(x)= \begin{cases} x^2\sin\left(\dfrac{1}{x}\right), & x\ne 0 \\ 0, & x=0 \end{cases} \text{ is a continuous function?}$

$\displaystyle \text{The given function } f \text{ is } f(x)= \begin{cases} x^{2}\sin\left(\dfrac{1}{x}\right), & x\ne 0,\\[6pt] 0, & x=0. \end{cases}$

$\displaystyle \text{It is evident that } f \text{ is defined at all points of the real line.}$

$\displaystyle \text{Let } c \text{ be a real number.}$

$\displaystyle \text{Case I:}$

$\displaystyle \text{If } c\ne 0,$
$\displaystyle f(c)=c^{2}\sin\left(\dfrac{1}{c}\right).$

$\displaystyle \lim_{x\to c} f(x) = \lim_{x\to c}\left(x^{2}\sin\left(\dfrac{1}{x}\right)\right) = \left(\lim_{x\to c} x^{2}\right) \left(\lim_{x\to c} \sin\left(\dfrac{1}{x}\right)\right) = c^{2}\sin\left(\dfrac{1}{c}\right).$

$\displaystyle \therefore \lim_{x\to c} f(x)=f(c).$

$\displaystyle \text{Hence, } f \text{ is continuous at all points } x\ne 0.$

$\displaystyle \text{Case II:}$

$\displaystyle \text{If } c=0,$
$\displaystyle f(0)=0.$

$\displaystyle \lim_{x\to 0} f(x) = \lim_{x\to 0}\left(x^{2}\sin\left(\dfrac{1}{x}\right)\right).$

$\displaystyle \text{It is known that } -1 \le \sin\left(\dfrac{1}{x}\right) \le 1 \text{ for } x\ne 0.$

$\displaystyle \Rightarrow -x^{2} \le x^{2}\sin\left(\dfrac{1}{x}\right) \le x^{2}.$

$\displaystyle \Rightarrow \lim_{x\to 0}(-x^{2}) \le \lim_{x\to 0}\left(x^{2}\sin\left(\dfrac{1}{x}\right)\right) \le \lim_{x\to 0}(x^{2}).$

$\displaystyle \Rightarrow 0 \le \lim_{x\to 0}\left(x^{2}\sin\left(\dfrac{1}{x}\right)\right) \le 0.$

$\displaystyle \Rightarrow \lim_{x\to 0}\left(x^{2}\sin\left(\dfrac{1}{x}\right)\right)=0.$

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

$\displaystyle \text{Hence, } f \text{ is continuous at } x=0.$

$\displaystyle \text{From the above observations, } f \text{ is continuous at every point of the real line.}$

$\displaystyle \text{Thus, } f \text{ is a continuous function.}$

$\displaystyle \text{Answer:}$

$\displaystyle \textbf{Question 18: } \text{Given the function} f(x)=\dfrac{1}{x+2}. \\ \text{Find the points of discontinuity of the function } f(f(x)).$

$\displaystyle \text{Answer:}$

$\displaystyle f(f(x))=\frac{1}{\dfrac{1}{x+2}+2} =\frac{x+2}{2x+5}.$

$\displaystyle \text{So, } f(f(x)) \text{ is not defined when the denominator is zero.}$

$\displaystyle \text{That is, when } x+2=0 \text{ or } 2x+5=0.$

$\displaystyle \text{If } x+2=0,\text{ then } x=-2.$

$\displaystyle \text{If } 2x+5=0,\text{ then } x=-\dfrac{5}{2}.$

$\displaystyle \text{Hence, the function } f(f(x)) \text{ is discontinuous at } x=-2 \text{ and } x=-\dfrac{5}{2}.$

$\displaystyle \textbf{Question 19: } \text{Find all point of discontinuity of the function} \\ f(t)=\dfrac{1}{t^2+t-2}, \text{ where } t=\dfrac{1}{x-1}$