Class 12: Continuity – Exercise 9.1

$\displaystyle \textbf{Question 1: }$
$\displaystyle \text{Test the continuity of the function } f(x) \text{ at the origin:} f(x)= \begin{cases} \dfrac{x}{|x|}, & x \neq 0 \\ 1, & x = 0 \end{cases}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} \dfrac{x}{|x|}, & x \neq 0 \\ 1, & x = 0 \end{cases}$

$\displaystyle \text{We observe}$

$\displaystyle (\text{LHL at } x=0) =$

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

$\displaystyle = \lim_{h \to 0} \frac{-h}{|-h|} = \lim_{h \to 0} \frac{-h}{h} = \lim_{h \to 0} (-1) = -1$

$\displaystyle (\text{RHL at } x=0) =$

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

$\displaystyle = \lim_{h \to 0} \frac{h}{|h|} = \lim_{h \to 0} 1 = 1$

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

$\displaystyle \text{Hence, } f(x) \text{ is discontinuous at the origin.}$

$\displaystyle \textbf{Question 2: }$
$\displaystyle \text{A function } f(x) \text{ is defined as,} f(x)= \begin{cases} \dfrac{x^{2}-x-6}{x-3}, & x \neq 3 \\ 5, & x = 3 \end{cases} \quad \text{Show that } f(x) \\ \text{is continuous at } x = 3.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} \dfrac{x^{2}-x-6}{x-3}, & x \neq 3 \\ 5, & x = 3 \end{cases}$

$\displaystyle \text{We observe}$

$\displaystyle (\text{LHL at } x=3) =$

$\displaystyle \lim_{x \to 3^-} f(x) = \lim_{h \to 0} f(3-h)$

$\displaystyle \lim_{h \to 0} \frac{(3-h)^{2}-(3-h)-6}{(3-h)-3} = \lim_{h \to 0} \frac{9+h^{2}-6h-3+h-6}{-h}$

$\displaystyle = \lim_{h \to 0} \frac{h^{2}-5h}{-h} = \lim_{h \to 0} \frac{(5-h)}{-1} = \lim_{h \to 0} (5-h) = 5$

$\displaystyle \text{And, } (\text{RHL at } x=3) =$

$\displaystyle \lim_{x \to 3^+} f(x) = \lim_{h \to 0} f(3+h)$

$\displaystyle \lim_{h \to 0} \frac{(3+h)^{2}-(3+h)-6}{(3+h)-3} = \lim_{h \to 0} \frac{9+h^{2}+6h-3-h-6}{h}$

$\displaystyle = \lim_{h \to 0} \frac{h^{2}+5h}{h} = \lim_{h \to 0} (5+h) = 5$

$\displaystyle \text{Also, } f(3)=5$

$\displaystyle \therefore \lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3)$

$\displaystyle \text{Hence, } f(x) \text{ is continuous at } x=3$

$\displaystyle \textbf{Question 3: }$
$\displaystyle \text{A function } f(x) \text{ is defined as} f(x)= \begin{cases} \dfrac{x^{2}-9}{x-3}, & x \neq 3 \\ 6, & x = 3 \end{cases} \ \text{Show that } f(x) \text{ is continuous at } \\ x=3.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} \dfrac{x^{2}-9}{x-3}, & \text{if } x \neq 3 \\ 6, & \text{if } x = 3 \end{cases}$

$\displaystyle \text{We observe}$

$\displaystyle (\text{LHL at } x=3) =$

$\displaystyle \lim_{x \to 3^-} f(x) = \lim_{h \to 0} f(3-h)$

$\displaystyle \lim_{h \to 0} \frac{(3-h)^{2}-9}{(3-h)-3} = \lim_{h \to 0} \frac{3^{2}+h^{2}-6h-9}{3-h-3} = \lim_{h \to 0} \frac{h^{2}-6h}{-h}$

$\displaystyle = \lim_{h \to 0} \frac{h(h-6)}{-h} = \lim_{h \to 0} (6-h) = 6$

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

$\displaystyle \lim_{h \to 0} \frac{(3+h)^{2}-9}{(3+h)-3} = \lim_{h \to 0} \frac{3^{2}+h^{2}+6h-9}{h} = \lim_{h \to 0} \frac{h^{2}+6h}{h}$

$\displaystyle = \lim_{h \to 0} \frac{h(6+h)}{h} = \lim_{h \to 0} (6+h) = 6$

$\displaystyle \text{Given } f(3)=6$

$\displaystyle \therefore \lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3)$

$\displaystyle f(x) \text{ is continuous at } x=3$

$\displaystyle \textbf{Question 4: }$
$\displaystyle \text{If } f(x)= \begin{cases} \dfrac{x^{2}-1}{x-1}, & x \neq 1 \\ 2, & x = 1 \end{cases} \text{ find whether } f(x) \text{ is continuous at } x=1.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

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

$\displaystyle \text{We observe}$

$\displaystyle (\text{LHL at } x=1) =$

$\displaystyle \lim_{x \to 1^-} f(x) = \lim_{h \to 0} f(1-h)$

$\displaystyle \lim_{h \to 0} \frac{(1-h)^{2}-1}{(1-h)-1} = \lim_{h \to 0} \frac{1+h^{2}-2h-1}{1-h-1} = \lim_{h \to 0} \frac{h^{2}-2h}{-h}$

$\displaystyle = \lim_{h \to 0} \frac{h(h-2)}{-h} = \lim_{h \to 0} (2-h) = 2$

$\displaystyle (\text{RHL at } x=1) =$

$\displaystyle \lim_{x \to 1^+} f(x) = \lim_{h \to 0} f(1+h)$

$\displaystyle \lim_{h \to 0} \frac{(1+h)^{2}-1}{(1+h)-1} = \lim_{h \to 0} \frac{1+h^{2}+2h-1}{1+h-1} = \lim_{h \to 0} \frac{h^{2}+2h}{h}$

$\displaystyle = \lim_{h \to 0} \frac{h(h+2)}{h} = \lim_{h \to 0} (2+h) = 2$

$\displaystyle \text{Given: } f(1)=2$

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

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

$\displaystyle \textbf{Question 5: }$
$\displaystyle \text{If } f(x)= \begin{cases} \dfrac{\sin 3x}{x}, & \text{when } x \neq 0 \\ 1, & \text{when } x = 0 \end{cases} \text{ find whether } f(x) \text{ is continuous at } x=0.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given}$

$\displaystyle f(x)= \begin{cases} \dfrac{\sin 3x}{x}, & \text{when } x\neq 0 \\ 1, & \text{when } x=0 \end{cases}$

$\displaystyle \text{We observe}$

$\displaystyle (\text{LHL at } x=0)=$

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

$\displaystyle \lim_{h\to 0}\dfrac{\sin(-3h)}{-h} =\lim_{h\to 0}\dfrac{-\sin(3h)}{-h} =\lim_{h\to 0}\dfrac{3\sin(3h)}{3h} =3\lim_{h\to 0}\dfrac{\sin(3h)}{3h} =3\cdot 1=3$

$\displaystyle (\text{RHL at } x=0)=$

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

$\displaystyle \lim_{h\to 0}\dfrac{\sin 3h}{h} =\lim_{h\to 0}\dfrac{3\sin 3h}{3h} =3\lim_{h\to 0}\dfrac{\sin(3h)}{3h} =3\cdot 1=3$

$\displaystyle \text{Given:} f(0)=1$

$\displaystyle \text{It is known that for a function } f(x) \text{ to be continuous at } x=a,$

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

$\displaystyle \text{But here,}$

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

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

$\displaystyle \textbf{Question 6: }$
$\displaystyle \text{If } f(x)= \begin{cases} e^{1/x}, & \text{if } x \neq 0 \\ 1, & \text{if } x = 0 \end{cases} \text{ find whether } f \text{ is continuous at } x=0.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} e^{\frac{1}{x}}, & \text{if } x\neq 0 \\ 1, & \text{if } x=0 \end{cases}$

$\displaystyle \text{We observe}$

$\displaystyle (\text{LHL at } x=0)=$

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

$\displaystyle \lim_{h\to 0} e^{-\frac{1}{h}} =\lim_{h\to 0}\left(\dfrac{1}{e^{\frac{1}{h}}}\right) =\dfrac{1}{\lim_{h\to 0} e^{\frac{1}{h}}} =0$

$\displaystyle (\text{RHL at } x=0)=$

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

$\displaystyle \lim_{h\to 0} e^{\frac{1}{h}}=\infty$

$\displaystyle \text{Given:}$

$\displaystyle f(0)=1$

$\displaystyle \text{It is known that for a function } f(x) \text{ to be continuous at } x=a,$

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

$\displaystyle \text{But here,}$

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

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

$\displaystyle \textbf{Question 7: }$
$\displaystyle \text{Let } f(x)= \begin{cases} \dfrac{1-\cos x}{x^{2}}, & \text{when } x \neq 0 \\ 1, & \text{when } x = 0 \end{cases} \text{ show that } f(x) \text{ is discontinuous at } x=0.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} \dfrac{1-\cos x}{x^{2}}, & \text{when } x\neq 0 \\ 1, & \text{when } x=0 \end{cases}$

$\displaystyle \text{Consider:}$

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

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

$\displaystyle \Rightarrow \lim_{x\to 0} f(x) =\lim_{x\to 0}\left(\dfrac{2\sin^{2}\frac{x}{2}}{4\left(\dfrac{x^{2}}{4}\right)}\right)$

$\displaystyle \Rightarrow \lim_{x\to 0} f(x) =\lim_{x\to 0}\left(\dfrac{2\left(\sin\frac{x}{2}\right)^{2}}{4\left(\dfrac{x}{2}\right)^{2}}\right)$

$\displaystyle \Rightarrow \lim_{x\to 0} f(x) =\dfrac{2}{4}\lim_{x\to 0}\left(\dfrac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^{2}$

$\displaystyle \Rightarrow \lim_{x\to 0} f(x) =\dfrac{1}{2}\cdot 1^{2} =\dfrac{1}{2}$

$\displaystyle \text{Given:}$

$\displaystyle f(0)=1$

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

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

$\displaystyle \textbf{Question 8: }$
$\displaystyle \text{Show that } f(x)= \begin{cases} \dfrac{x-|x|}{2}, & \text{when } x \neq 0 \\ 2, & \text{when } x = 0 \end{cases} \text{ is discontinuous at } x=0.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{The given function can be rewritten as:}$

$\displaystyle f(x)= \begin{cases} \dfrac{x-x}{2}, & \text{when } x>0 \\ \dfrac{x+x}{2}, & \text{when } x<0 \\ 2, & \text{when } x=0 \end{cases} \ \ \ \ \Rightarrow\ \ \ f(x)= \begin{cases} 0, & \text{when } x>0 \\ x, & \text{when } x<0 \\ 2, & \text{when } x=0 \end{cases}$

$\displaystyle \text{We observe}$

$\displaystyle (\text{LHL at } x=0)=$

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

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

$\displaystyle (\text{RHL at } x=0)=$

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

$\displaystyle \lim_{h\to 0} 0=0$

$\displaystyle \text{And,}$

$\displaystyle f(0)=2$

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

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

$\displaystyle \textbf{Question 9: }$
$\displaystyle \text{Show that } f(x)= \begin{cases} \dfrac{|x-a|}{x-a}, & \text{when } x \neq a \\ 1, & \text{when } x = a \end{cases} \text{ is discontinuous at } x=a.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{The given function can be rewritten as:}$

$\displaystyle f(x)= \begin{cases} \dfrac{x-a}{x-a}, & \text{when } x>a \\ \dfrac{a-x}{x-a}, & \text{when } x<a \\ 1, & \text{when } x=a \end{cases}$

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

$\displaystyle \Rightarrow f(x)= \begin{cases} 1, & \text{when } x\ge a \\ -1, & \text{when } x<a \end{cases}$

$\displaystyle \text{We observe}$

$\displaystyle (\text{LHL at } x=a)=$

$\displaystyle \lim_{x\to a^-} f(x) =\lim_{h\to 0} f(a-h)$

$\displaystyle =\lim_{h\to 0} (-1)=-1$

$\displaystyle (\text{RHL at } x=a)=$

$\displaystyle \lim_{x\to a^+} f(x) =\lim_{h\to 0} f(a+h)$

$\displaystyle \lim_{h\to 0} (1)=1$

$\displaystyle \lim_{x\to a^-} f(x)\neq \lim_{x\to a^+} f(x)$

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

$\displaystyle \textbf{Question 10: } \text{Discuss the continuity of the following function at the indicated point:}$

$\displaystyle \textbf{(i) } f(x)= \begin{cases} |x|\cos\left(\dfrac{1}{x}\right), & x\neq 0 \\ 0, & x=0 \end{cases} \quad \text{at } x=0$

$\displaystyle \textbf{(ii) } f(x)= \begin{cases} x^{2}\sin\left(\dfrac{1}{x}\right), & x\neq 0 \\ 0, & x=0 \end{cases} \quad \text{at } x=0$

$\displaystyle \textbf{(iii) } f(x)= \begin{cases} (x-a)\sin\left(\dfrac{1}{x-a}\right), & x\neq a \\ 0, & x=a \end{cases} \quad \text{at } x=a$

$\displaystyle \textbf{(iv) } f(x)= \begin{cases} \dfrac{e^{x}-1}{\log(1+2x)}, & x\neq 0 \\ 7, & x=0 \end{cases} \quad \text{at } x=0$

$\displaystyle \textbf{(v) } f(x)= \begin{cases} \dfrac{1-x^{n}}{1-x}, & x\neq 1 \\ n-1, & x=1 \end{cases} \quad n\in \mathbb{N}, \ \text{at } x=1$

$\displaystyle \textbf{(vi) } f(x)= \begin{cases} \dfrac{|x^{2}-1|}{x-1}, & x\neq 1 \\ 2, & x=1 \end{cases} \quad \text{at } x=1$

$\displaystyle \textbf{(vii) } f(x)= \begin{cases} \dfrac{2|x|+x^{2}}{x}, & x\neq 0 \\ 0, & x=0 \end{cases} \quad \text{at } x=0$

$\displaystyle \textbf{(viii) } f(x)= \begin{cases} |x-a|\sin\left(\dfrac{1}{x-a}\right), & x\neq a \\ 0, & x=a \end{cases} \quad \text{at } x=a$

$\displaystyle \text{Answer:}$

$\displaystyle \textbf{(i) } \text{Given:}$

$\displaystyle f(x)= \begin{cases} |x|\cos\left(\dfrac{1}{x}\right), & x\neq 0 \\ 0, & x=0 \end{cases} \quad \text{at } x=0$

$\displaystyle \text{Consider,}$

$\displaystyle \lim_{x\to 0} f(x) =\lim_{x\to 0} |x|\cos\left(\dfrac{1}{x}\right)$

$\displaystyle =\left(\lim_{x\to 0} |x|\right)\left(\lim_{x\to 0} \cos\left(\dfrac{1}{x}\right)\right)$

$\displaystyle =0 \times \lim_{x\to 0} \cos\left(\dfrac{1}{x}\right)=0$

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

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

$\displaystyle \textbf{(ii) } \text{Given:}$

$\displaystyle f(x)= \begin{cases} x^{2}\sin\left(\dfrac{1}{x}\right), & x\neq 0 \\ 0, & x=0 \end{cases}$

$\displaystyle \text{We observe}$

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

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

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

$\displaystyle \textbf{(iii) } \text{Given:}$

$\displaystyle f(x)= \begin{cases} (x-a)\sin\left(\dfrac{1}{x-a}\right), & x\neq a \\ 0, & x=a \end{cases}$

$\displaystyle \text{Putting } x-a=y, \text{ we get}$

$\displaystyle \lim_{x\to a} (x-a)\sin\left(\dfrac{1}{x-a}\right) =\lim_{y\to 0} y\sin\left(\dfrac{1}{y}\right)$

$\displaystyle =\left(\lim_{y\to 0} y\right)\left(\lim_{y\to 0} \sin\left(\dfrac{1}{y}\right)\right) =0\times \lim_{y\to 0} \sin\left(\dfrac{1}{y}\right) =0$

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

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

$\displaystyle \textbf{(iv) } \text{Given:}$

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

$\displaystyle \text{We observe}$

$\displaystyle \lim_{x\to 0} f(x) =\lim_{x\to 0}\dfrac{e^{x}-1}{\log(1+2x)}$

$\displaystyle \Rightarrow \lim_{x\to 0} f(x) =\lim_{x\to 0}\dfrac{e^{x}-1}{2x}\cdot\dfrac{2x}{\log(1+2x)}$

$\displaystyle \Rightarrow \lim_{x\to 0} f(x) =\dfrac{1}{2}\lim_{x\to 0}\left(\dfrac{e^{x}-1}{x}\right)\Bigg/\left(\dfrac{\log(1+2x)}{2x}\right)$

$\displaystyle \Rightarrow \lim_{x\to 0} f(x) =\dfrac{1}{2}\cdot \dfrac{\lim_{x\to 0}\left(\dfrac{e^{x}-1}{x}\right)} {\lim_{x\to 0}\left(\dfrac{\log(1+2x)}{2x}\right)} =\dfrac{1}{2}\times 1\times 1 =\dfrac{1}{2}$

$\displaystyle \text{And,}$

$\displaystyle f(0)=7$

$\displaystyle \Rightarrow \lim_{x\to 0} f(x)\neq f(0)$

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

$\displaystyle \textbf{(v) } \text{Given:}$

$\displaystyle f(x)= \begin{cases} \dfrac{1-x^{n}}{1-x}, & x\neq 1 \\ n-1, & x=1 \end{cases}$

$\displaystyle \text{Here,}$

$\displaystyle f(1)=n-1$

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

$\displaystyle \Rightarrow \lim_{x\to 1} f(x) =\lim_{x\to 1}\Big[(1-x)^{n-1} + {}^{n-1}C_{1}(1-x)^{n-2}x + {}^{n-1}C_{2}(1-x)^{n-3}x^{2} +\cdots + {}^{n-1}C_{n-1}x^{\,n-1}\Big]$

$\displaystyle \Rightarrow \lim_{x\to 1} f(x) =0+0+\cdots+(1)^{\,n-1} =1\neq f(1)$

$\displaystyle \text{Thus,}$

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

$\displaystyle \textbf{(vi) } \text{Given:}$

$\displaystyle f(x)= \begin{cases} \dfrac{|x^{2}-1|}{x-1}, & x\neq 1 \\ 2, & x=1 \end{cases}$

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

$\displaystyle \text{We observe}$

$\displaystyle (\text{LHL at } x=1)=$

$\displaystyle \lim_{x\to 1^-} f(x) =\lim_{h\to 0} f(1-h)$

$\displaystyle =\lim_{h\to 0}\big[-(1-h)-1\big] =\lim_{h\to 0}(-2+h) =-2$

$\displaystyle f(1)=2$

$\displaystyle \Rightarrow \lim_{x\to 1^-} f(x)\neq f(1)$

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

$\displaystyle \textbf{(vii) } \text{Given:}$

$\displaystyle f(x)= \begin{cases} \dfrac{2|x|+x^{2}}{x}, & x\neq 0 \\ 0, & x=0 \end{cases}$

$\displaystyle \Rightarrow f(x)= \begin{cases} \dfrac{2x+x^{2}}{x}, & x>0 \\ \dfrac{-2x+x^{2}}{x}, & x<0 \\ 0, & x=0 \end{cases}$

$\displaystyle \Rightarrow f(x)= \begin{cases} x+2, & x>0 \\ x-2, & x<0 \\ 0, & x=0 \end{cases}$

$\displaystyle \text{We observe}$

$\displaystyle (\text{LHL at } x=0)=$

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

$\displaystyle (\text{RHL at } x=0)=$

$\displaystyle \lim_{x\to 0^+} f(x) =\lim_{h\to 0} f(h) =\lim_{h\to 0}(2+h) =2$

$\displaystyle \Rightarrow \lim_{x\to 0^-} f(x)\neq \lim_{x\to 0^+} f(x)$

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

$\displaystyle \textbf{(viii) } \text{Given:}$

$\displaystyle f(x)= \begin{cases} |x-a|\sin\left(\dfrac{1}{x-a}\right), & x\neq a \\ 0, & x=a \end{cases}$

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

$\displaystyle \text{We observe}$

$\displaystyle (\text{LHL at } x=a)=$

$\displaystyle \lim_{x\to a^-} f(x) =\lim_{h\to 0} (a-(a-h))\sin\left(\dfrac{1}{-h}\right) =0$

$\displaystyle (\text{RHL at } x=a)=$

$\displaystyle \lim_{x\to a^+} f(x) =\lim_{h\to 0} ((a+h)-a)\sin\left(\dfrac{1}{h}\right) =0$

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

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

$\displaystyle \textbf{Question 11: }$
$\displaystyle f(x)= \begin{cases} 1+x^{2}, & 0\le x\le 1 \\ 2-x, & x>1 \end{cases} \ \text{ is discontinuous at } x=1$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

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

$\displaystyle \text{We observe}$

$\displaystyle (\text{LHL at } x=1)=$

$\displaystyle \lim_{x\to 1^-} f(x) =\lim_{h\to 0} f(1-h)$

$\displaystyle =\lim_{h\to 0}\left[1+(1-h)^{2}\right] =\lim_{h\to 0}(2+h^{2}-2h) =2$

$\displaystyle (\text{RHL at } x=1)=$

$\displaystyle \lim_{x\to 1^+} f(x) =\lim_{h\to 0} f(1+h)$

$\displaystyle =\lim_{h\to 0}\left[2-(1+h)\right] =\lim_{h\to 0}(1-h) =1$

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

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

$\displaystyle \textbf{Question 12: }$
$\displaystyle \text{Show that} f(x)= \begin{cases} \dfrac{\sin 3x}{\tan 2x}, & x<0 \\ \dfrac{3}{2}, & x=0 \\ \dfrac{\log(1+3x)}{e^{2x}-1}, & x>0 \end{cases} \text{ is continuous at } x=0$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} \dfrac{\sin 3x}{\tan 2x}, & x<0 \\ \dfrac{3}{2}, & x=0 \\ \dfrac{\log(1+3x)}{e^{2x}-1}, & x>0 \end{cases}$

$\displaystyle \text{We observe}$

$\displaystyle (\text{LHL at } x=0)=$

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

$\displaystyle =\lim_{h\to 0}\left(\dfrac{\sin 3(-h)}{\tan 2(-h)}\right) =\lim_{h\to 0}\left(\dfrac{\sin 3h}{\tan 2h}\right)$

$\displaystyle =\lim_{h\to 0} \left(\dfrac{\dfrac{3\sin 3h}{3h}}{\dfrac{2\tan 2h}{2h}}\right) =\dfrac{3\lim_{h\to 0}\left(\dfrac{\sin 3h}{3h}\right)} {2\lim_{h\to 0}\left(\dfrac{\tan 2h}{2h}\right)} =\dfrac{3\times 1}{2\times 1} =\dfrac{3}{2}$

$\displaystyle (\text{RHL at } x=0)=$

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

$\displaystyle =\lim_{h\to 0}\left(\dfrac{\log(1+3h)}{e^{2h}-1}\right)$

$\displaystyle =\lim_{h\to 0} \left(\dfrac{\dfrac{\log(1+3h)}{3h}}{\dfrac{e^{2h}-1}{2h}}\right) =\dfrac{3\lim_{h\to 0}\left(\dfrac{\log(1+3h)}{3h}\right)} {2\lim_{h\to 0}\left(\dfrac{e^{2h}-1}{2h}\right)} =\dfrac{3\times 1}{2\times 1} =\dfrac{3}{2}$

$\displaystyle \text{And,}$

$\displaystyle f(0)=\dfrac{3}{2}$

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

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

$\displaystyle \textbf{Question 13: }$
$\displaystyle \text{Find the value of 'a' for which the function } f \text{ defined by} f(x)= \begin{cases} a\sin\left(\dfrac{\pi}{2}(x+1)\right), & x\le 0 \\ \dfrac{\tan x-\sin x}{x^{3}}, & x>0 \end{cases} \\ \text{is continuous at } x=0 \hspace{7.0cm} \text{[CBSE 2011]}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} a\sin\left(\dfrac{\pi}{2}(x+1)\right), & x\le 0 \\ \dfrac{\tan x-\sin x}{x^{3}}, & x>0 \end{cases}$

$\displaystyle \text{We have}$

$\displaystyle (\text{LHL at } x=0)=$

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

$\displaystyle =\lim_{h\to 0} a\sin\left(\dfrac{\pi}{2}(-h+1)\right) =a\sin\left(\dfrac{\pi}{2}\right) =a$

$\displaystyle (\text{RHL at } x=0)=$

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

$\displaystyle =\lim_{h\to 0}\dfrac{\tan h-\sin h}{h^{3}}$

$\displaystyle =\lim_{h\to 0}\dfrac{\dfrac{\sin h}{\cos h}-\sin h}{h^{3}}$

$\displaystyle =\lim_{h\to 0}\dfrac{\sin h(1-\cos h)}{h^{3}\cos h}$

$\displaystyle =\lim_{h\to 0}\dfrac{(1-\cos h)\tan h}{h^{3}}$

$\displaystyle =\lim_{h\to 0}\dfrac{2\sin^{2}\frac{h}{2}\tan h}{4\left(\dfrac{h^{2}}{4}\right)\times h}$

$\displaystyle =\dfrac{2}{4}\lim_{h\to 0}\dfrac{\sin^{2}\frac{h}{2}\tan h}{\left(\dfrac{h^{2}}{4}\right)\times h}$

$\displaystyle =\dfrac{1}{2} \left(\lim_{h\to 0}\left(\dfrac{\sin\frac{h}{2}}{\frac{h}{2}}\right)^{2}\right) \left(\lim_{h\to 0}\dfrac{\tan h}{h}\right)$

$\displaystyle =\dfrac{1}{2}\times 1\times 1 =\dfrac{1}{2}$

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

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

$\displaystyle \Rightarrow a=\dfrac{1}{2}$

$\displaystyle \textbf{Question 14: }$
$\displaystyle \text{Examine the continuity of the function} f(x)= \begin{cases} 3x-2, & x\le 0 \\ x+1, & x>0 \end{cases} \quad \text{at } x=0$
$\displaystyle \text{Also sketch the graph of this function.}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Meaning of continuity of function}$

$\displaystyle \text{If we talk about a general meaning of continuity of a function } f(x), \text{ we can say that if we plot}$

$\displaystyle \text{the coordinates } (x,f(x)) \text{ and try to join all those points in the specified region,}$

$\displaystyle \text{we can do so without picking up our pen i.e. you will put your pen/pencil on graph paper}$

$\displaystyle \text{and you can draw the curve without any breakage.}$

$\displaystyle \text{Mathematically we define the same thing as given below:}$

$\displaystyle \text{A function } f(x) \text{ is said to be continuous at } x=c \text{ where } c \text{ is x-coordinate of the point}$

$\displaystyle \text{at which continuity is to be checked}$

$\displaystyle \text{If:} \ \lim_{h\to 0} f(c-h)=\lim_{h\to 0} f(c+h)=f(c) \text{where } h \text{ is a very small positive no (can assume } \\ h=0.0000000001 \text{ like this)}$

$\displaystyle \text{It means:-}$

$\displaystyle \text{Limiting value of the left neighbourhood of } x=c \text{ also called left hand limit LHL}$

$\displaystyle \left\{ i.e.\ \lim_{h\to 0} f(c-h) \right\} \text{ must be equal to limiting value of right neighbourhood}$

$\displaystyle \text{of } x=c \text{ called right hand limit RHL } \left\{ i.e.\ \lim_{h\to 0} f(c+h) \right\}$

$\displaystyle \text{and both must be equal to the value of } f(x) \text{ at } x=c \text{ i.e. } f(c).$

$\displaystyle \text{Thus, it is the necessary condition for a function to be continuous.}$

$\displaystyle \text{So, whenever we check continuity we try to check above equality. If it holds true,}$

$\displaystyle \text{function is continuous else it is discontinuous.}$

$\displaystyle \text{Lets solve now:}$

$\displaystyle \text{Given function is}$

$\displaystyle f(x)= \begin{cases} 3x-2, & \text{if } x\le 0 \\ x+1, & \text{if } x>0 \end{cases} \qquad \ldots(2)$

$\displaystyle \text{We need to check whether } f(x) \text{ is continuous at } x=0 \text{ or not}$

$\displaystyle \text{For this we need to check LHL, RHL and value of function at } x=0$

$\displaystyle \text{Clearly,}$

$\displaystyle f(0)=3*0-2=-2 \ \text{[from equation 2]}$

$\displaystyle \text{LHL }= \lim_{h\to 0} f(0-h) = \lim_{h\to 0} f(-h) = \lim_{h\to 0}\{3(-h)-2\} = -2$

$\displaystyle \text{RHL }= \lim_{h\to 0} f(0+h) = \lim_{h\to 0} f(h) = \lim_{h\to 0}\{h+1\} = 0+1=1$

$\displaystyle \text{As, LHL } \neq \text{ RHL}$

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

$\displaystyle \text{This can also be proved by plotting } f(x) \text{ on cartesian plane.}$

$\displaystyle \text{For } x>0, \text{ we need to plot}$

$\displaystyle y=x+1$

$\displaystyle \text{put } y=0, \text{ we get } x=-1 \text{ and for second point we put } x=0 \text{ and thus get } y=1$

$\displaystyle \text{two points are enough to plot the straight line.}$

$\displaystyle \text{Two coordinates are } (-1,0) \text{ and } (0,1)$

$\displaystyle \text{For } x\le 0, \text{ we need to plot}$

$\displaystyle y=3x-2$

$\displaystyle \text{put } x=0 \text{ then } y=-2$

$\displaystyle \text{on putting } y=0 \text{ we get } x=\dfrac{2}{3}$

$\displaystyle \text{two coordinates are } (0,-2) \text{ and } \left(\dfrac{2}{3},0\right)$

$\displaystyle \textbf{Question 15: }$
$\displaystyle \text{Discuss the continuity of the function } f(x)= \begin{cases} x, & x>0 \\ 1, & x=0 \\ -x, & x<0 \end{cases} \ \\ \text{ at the point } x=0$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

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

$\displaystyle (\text{LHL at } x=0)=$

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

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

$\displaystyle (\text{RHL at } x=0)=$

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

$\displaystyle =\lim_{h\to 0} h=0$

$\displaystyle \text{And,}$

$\displaystyle f(0)=1$

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

$\displaystyle \text{Hence,}$

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

$\displaystyle \textbf{Question 16: }$
$\displaystyle \text{Discuss the continuity of the function } f(x)= \begin{cases} x, & 0\le x<\dfrac{1}{2} \\ \dfrac{1}{2}, & x=\dfrac{1}{2} \\ 1-x, & \dfrac{1}{2}<x\le 1 \end{cases} \ \\ \text{ at the point } x=\dfrac{1}{2}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

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

$\displaystyle \text{We observe}$

$\displaystyle (\text{LHL at } x=\dfrac{1}{2})=$

$\displaystyle \lim_{x\to \left(\frac{1}{2}\right)^-} f(x) =\lim_{h\to 0} f\left(\dfrac{1}{2}-h\right)$

$\displaystyle =\lim_{h\to 0}\left(\dfrac{1}{2}-h\right) =\dfrac{1}{2}$

$\displaystyle (\text{RHL at } x=\dfrac{1}{2})=$

$\displaystyle \lim_{x\to \left(\frac{1}{2}\right)^+} f(x) =\lim_{h\to 0} f\left(\dfrac{1}{2}+h\right)$

$\displaystyle =\lim_{h\to 0}\left(1-\left(\dfrac{1}{2}+h\right)\right) =\dfrac{1}{2}$

$\displaystyle \text{Also,}$

$\displaystyle f\left(\dfrac{1}{2}\right)=\dfrac{1}{2}$

$\displaystyle \therefore\ \lim_{x\to \left(\frac{1}{2}\right)^-} f(x) =\lim_{x\to \left(\frac{1}{2}\right)^+} f(x) =f\left(\dfrac{1}{2}\right)$

$\displaystyle \text{Hence,}$

$\displaystyle f(x) \text{ is continuous at } x=\dfrac{1}{2}.$

$\displaystyle \textbf{Question 17: }$
$\displaystyle \text{Discuss the continuity of } f(x)= \begin{cases} 2x-1, & x<0 \\ 2x+1, & x\ge 0 \end{cases} \ \text{at } x=0 \hspace{7.0cm} \text{[CBSE 2002]}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

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

$\displaystyle (\text{LHL at } x=0)=$

$\displaystyle \lim_{x\to 0^-} f(x) =\lim_{x\to 0^-}(2x-1) =2(0)-1=-1$

$\displaystyle (\text{RHL at } x=0)=$

$\displaystyle \lim_{x\to 0^+} f(x) =\lim_{x\to 0^+}(2x+1) =2(0)+1=1$

$\displaystyle \Rightarrow \lim_{x\to 0^-} f(x)\neq \lim_{x\to 0^+} f(x)$

$\displaystyle \text{Hence,}$

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

$\displaystyle \textbf{Question 18: }$
$\displaystyle \text{For what value of } k \text{ is the function }f(x)= \begin{cases} \dfrac{x^{2}-1}{x-1}, & x\neq 1 \\ k, & x=1 \end{cases} \ \\ \text{continuous at } x=1?$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} \dfrac{x^{2}-1}{x-1}, & x\neq 1 \\ k, & x=1 \end{cases}$

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

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

$\displaystyle \Rightarrow \lim_{x\to 1}\dfrac{x^{2}-1}{x-1}=k$

$\displaystyle \Rightarrow \lim_{x\to 1}\dfrac{(x-1)(x+1)}{x-1}=k$

$\displaystyle \Rightarrow \lim_{x\to 1}(x+1)=k$

$\displaystyle \Rightarrow k=2$

$\displaystyle \textbf{Question 19: }$
$\displaystyle \text{Determine the value of the constant } k \text{ so that the function} f(x)= \begin{cases} \dfrac{x^{2}-3x+2}{x-1}, & x\neq 1 \\ k, & x=1 \end{cases} \\ \text{is continuous at } x=1$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} \dfrac{x^{2}-3x+2}{x-1}, & x\neq 1 \\ k, & x=1 \end{cases}$

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

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

$\displaystyle \Rightarrow \lim_{x\to 1}\dfrac{x^{2}-3x+2}{x-1}=k$

$\displaystyle \Rightarrow \lim_{x\to 1}\dfrac{(x-2)(x-1)}{x-1}=k$

$\displaystyle \Rightarrow \lim_{x\to 1}(x-2)=k$

$\displaystyle \Rightarrow k=-1$

$\displaystyle \textbf{Question 20: }$
$\displaystyle \text{For what value of } k \text{ is the function} f(x)= \begin{cases} \dfrac{\sin 5x}{3x}, & x\neq 0 \\ k, & x=0 \end{cases} \\ \text{is continuous at } x=0?$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

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

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

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

$\displaystyle \Rightarrow \lim_{x\to 0}\dfrac{\sin 5x}{3x}=k$

$\displaystyle \Rightarrow \lim_{x\to 0}\dfrac{5\sin 5x}{3\times 5x}=k$

$\displaystyle \Rightarrow \dfrac{5}{3}\lim_{x\to 0}\dfrac{\sin 5x}{5x}=k$

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

$\displaystyle \Rightarrow k=\dfrac{5}{3}$

$\displaystyle \textbf{Question 21: }$
$\displaystyle \text{Determine the value of the constant } k \text{ so that the function} f(x)= \begin{cases} kx^{2}, & x\le 2 \\ 3, & x>2 \end{cases} \\ \text{is continuous at } x=2.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} kx^{2}, & x\le 2 \\ 3, & x>2 \end{cases}$

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

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

$\displaystyle \text{Now}$

$\displaystyle \lim_{x\to 2^-} f(x) =\lim_{h\to 0} f(2-h) =\lim_{h\to 0} k(2-h)^{2} =4k$

$\displaystyle f(2)=3$

$\displaystyle \text{From (1), we have}$

$\displaystyle 4k=3$

$\displaystyle \Rightarrow k=\dfrac{3}{4}$

$\displaystyle \textbf{Question 22: }$
$\displaystyle \text{Determine the value of the constant } k \text{ so that the function} f(x)= \begin{cases} \dfrac{\sin 2x}{5x}, & x\neq 0 \\ k, & x=0 \end{cases} \\ \text{is continuous at } x=0. \hspace{8.0cm} \text{[CBSE 2007]}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

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

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

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

$\displaystyle \Rightarrow \lim_{x\to 0}\dfrac{\sin 2x}{5x}=k$

$\displaystyle \Rightarrow \lim_{x\to 0}\dfrac{2\sin 2x}{5\times 2x}=k$

$\displaystyle \Rightarrow \dfrac{2}{5}\lim_{x\to 0}\dfrac{\sin 2x}{2x}=k$

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

$\displaystyle \Rightarrow k=\dfrac{2}{5}$

$\displaystyle \textbf{Question 23: }$
$\displaystyle \text{Find the values of } a \text{ so that the function} f(x)= \begin{cases} ax+5, & x\le 2 \\ x-1, & x>2 \end{cases} \\ \text{is continuous at } x=2. \hspace{8.0cm} \text{[CBSE 2002]}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

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

$\displaystyle \text{We observe}$

$\displaystyle (\text{LHL at } x=2)=$

$\displaystyle \lim_{x\to 2^-} f(x) =\lim_{h\to 0} f(2-h) =\lim_{h\to 0}\big[a(2-h)+5\big] =2a+5$

$\displaystyle (\text{RHL at } x=2)=$

$\displaystyle \lim_{x\to 2^+} f(x) =\lim_{h\to 0} f(2+h) =\lim_{h\to 0}\big[(2+h)-1\big] =1$

$\displaystyle \text{And,}$

$\displaystyle f(2)=a(2)+5=2a+5$

$\displaystyle \text{Since } f(x) \text{ is continuous at } x=2,\ \text{we have}$

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

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

$\displaystyle \Rightarrow 2a=-4$

$\displaystyle \Rightarrow a=-2$

$\displaystyle \textbf{Question 24: }$
$\displaystyle \text{Prove that the function} f(x)= \begin{cases} \dfrac{x}{|x|+2x^{2}}, & x\neq 0 \\ k, & x=0 \end{cases} \text{ remains discontinuous at } x=0,\ \text{regardless the choice of } k.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{The given function can be rewritten as:}$

$\displaystyle f(x)= \begin{cases} \dfrac{x}{|x|+2x^{2}}, & x\neq 0 \\ k, & x=0 \end{cases} \ \ \Rightarrow \ \ f(x)= \begin{cases} \dfrac{x}{x+2x^{2}}, & x>0 \\ \dfrac{x}{-x+2x^{2}}, & x<0 \\ k, & x=0 \end{cases} \\ \Rightarrow f(x)= \begin{cases} \dfrac{1}{2x+1}, & x>0 \\ \dfrac{1}{2x-1}, & x<0 \\ k, & x=0 \end{cases}$

$\displaystyle \text{We observe}$

$\displaystyle (\text{LHL at } x=0)=$

$\displaystyle \lim_{x\to 0^-} f(x) =\lim_{x\to 0^-}\dfrac{1}{2x-1} =\dfrac{1}{-1} =-1$

$\displaystyle (\text{RHL at } x=0)=$

$\displaystyle \lim_{x\to 0^+} f(x) =\lim_{x\to 0^+}\dfrac{1}{2x+1} =\dfrac{1}{1} =1$

$\displaystyle \text{So, } \lim_{x\to 0^-} f(x)\neq \lim_{x\to 0^+} f(x)$

$\displaystyle \text{Thus, } f(x) \text{ is discontinuous at } x=0,\ \text{regardless of the choice of } k.$

$\displaystyle \textbf{Question 25: }$
$\displaystyle \text{Find the value of } k \text{ if } f(x) \text{ is continuous at } x=\dfrac{\pi}{2},\ \text{where } f(x)= \begin{cases} \dfrac{k\cos x}{\pi-2x}, & x\neq \dfrac{\pi}{2} \\ 3, & x=\dfrac{\pi}{2} \end{cases}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

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

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

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

$\displaystyle \Rightarrow \lim_{x\to \frac{\pi}{2}}\dfrac{k\cos x}{\pi-2x}=3$

$\displaystyle \text{Putting } \dfrac{\pi}{2}-x=h,$

$\displaystyle \text{so that } x=\dfrac{\pi}{2}-h \text{ and } h\to 0$

$\displaystyle \Rightarrow \lim_{h\to 0}\dfrac{k\cos\left(\dfrac{\pi}{2}-h\right)}{\pi-2\left(\dfrac{\pi}{2}-h\right)}=3$

$\displaystyle \Rightarrow \lim_{h\to 0}\dfrac{k\sin h}{2h}=3$

$\displaystyle \Rightarrow \lim_{h\to 0}\dfrac{k\sin h}{h}=6$

$\displaystyle \Rightarrow k\lim_{h\to 0}\dfrac{\sin h}{h}=6$

$\displaystyle \Rightarrow k\times 1=6$

$\displaystyle \Rightarrow k=6$

$\displaystyle \text{Hence, for } k=6,\ f(x) \text{ is continuous at } x=\dfrac{\pi}{2}.$

$\displaystyle \textbf{Question 26: }$
$\displaystyle \text{Determine the values of } a,\, b,\, c \text{ for which the function } f(x)= \begin{cases} \dfrac{\sin((a+1)x)+\sin x}{x}, & x<0 \\ c, & x=0 \\ \dfrac{\sqrt{x+bx^{2}}-\sqrt{x}}{bx^{3/2}}, & x>0 \end{cases} \\ \text{is continuous at } x=0.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

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

$\displaystyle \text{For } f(x) \text{ to be continuous at } x=0,\ f(0^-)=f(0^+)=f(0)$

$\displaystyle \text{LHL }=f(0^-)=\lim_{x\to 0}\dfrac{\sin((a+1)x)+\sin x}{x}$

$\displaystyle \Rightarrow \lim_{h\to 0}\dfrac{\sin((a+1)h)+\sin h}{h}$

$\displaystyle \Rightarrow \lim_{h\to 0}\dfrac{\sin((a+1)h)}{h}+\lim_{h\to 0}\dfrac{\sin h}{h}$

$\displaystyle \Rightarrow \left(\lim_{h\to 0}\dfrac{\sin((a+1)h)}{(a+1)h}\right)(a+1)+\lim_{h\to 0}\dfrac{\sin h}{h}$

$\displaystyle \Rightarrow 1\times (a+1)+1$

$\displaystyle \Rightarrow (a+1)+1$

$\displaystyle \Rightarrow f(0^-)=a+2 \qquad \ldots(1)$

$\displaystyle \text{RHL }=f(0^+)=\lim_{x\to 0}\dfrac{\sqrt{x+bx^{2}}-\sqrt{x}}{bx^{3/2}}$

$\displaystyle \Rightarrow \lim_{x\to 0}\dfrac{\sqrt{x+bx^{2}}-\sqrt{x}}{bx^{3/2}}$

$\displaystyle \Rightarrow \lim_{h\to 0}\dfrac{\sqrt{h+bh^{2}}-\sqrt{h}}{bh^{3/2}}$

$\displaystyle \Rightarrow \lim_{h\to 0}\dfrac{\sqrt{h+bh^{2}}-\sqrt{h}}{b\,h\,h^{1/2}}$

$\displaystyle \Rightarrow \lim_{h\to 0}\dfrac{\sqrt{h+bh^{2}}-\sqrt{h}}{b\,h\sqrt{h}}$

$\displaystyle \Rightarrow \lim_{h\to 0}\dfrac{\sqrt{h(1+bh)}-\sqrt{h}}{b\,h\sqrt{h}}$

$\displaystyle \Rightarrow \lim_{h\to 0}\dfrac{\sqrt{h}\big(\sqrt{1+bh}-1\big)}{b\,h\sqrt{h}}$

$\displaystyle \Rightarrow \lim_{h\to 0}\dfrac{\sqrt{1+bh}-1}{bh}$

$\displaystyle \text{Take the complex conjugate of } \left(\sqrt{1+bh}-\sqrt{1}\right)$

$\displaystyle \text{i.e., } \left(\sqrt{1+bh}-\sqrt{1}\right) \text{ and multiply it with numerator and denominator}$

$\displaystyle \Rightarrow \lim_{h\to 0} \dfrac{\sqrt{1+bh}-\sqrt{1}}{bh} \times \dfrac{\sqrt{1+bh}+\sqrt{1}}{\sqrt{1+bh}+\sqrt{1}}$

$\displaystyle \Rightarrow \lim_{h\to 0} \dfrac{(\sqrt{1+bh})^{2}-(\sqrt{1})^{2}} {bh\left(\sqrt{1+bh}+\sqrt{1}\right)}$

$\displaystyle \because (a+b)(a-b)=a^{2}-b^{2}$

$\displaystyle \Rightarrow \lim_{h\to 0} \dfrac{(1+bh)-1}{bh\left(\sqrt{1+bh}+\sqrt{1}\right)}$

$\displaystyle \Rightarrow \lim_{h\to 0} \dfrac{bh}{bh\left(\sqrt{1+bh}+\sqrt{1}\right)}$

$\displaystyle \Rightarrow \lim_{h\to 0} \dfrac{1}{\sqrt{1+bh}+\sqrt{1}}$

$\displaystyle \Rightarrow \dfrac{1}{\sqrt{1+b\times 0}+\sqrt{1}}$

$\displaystyle \Rightarrow \dfrac{1}{1+1}$

$\displaystyle \Rightarrow f(0^+)=\dfrac{1}{2} \qquad \ldots(2)$

$\displaystyle \text{Since } f(x) \text{ is continuous at } x=0,\ \text{from (1) and (2), we get}$

$\displaystyle \Rightarrow a+2=\dfrac{1}{2}$

$\displaystyle \Rightarrow a=\dfrac{1}{2}-2$

$\displaystyle \Rightarrow a=-\dfrac{3}{2}$

$\displaystyle \text{Also,}$

$\displaystyle f(0^-)=f(0^+)=f(0)$

$\displaystyle \Rightarrow f(0)=c$

$\displaystyle \Rightarrow c=a+2=\dfrac{1}{2}$

$\displaystyle \Rightarrow c=\dfrac{1}{2}$

$\displaystyle \text{So the values of } a=-\dfrac{3}{2},\ c=\dfrac{1}{2} \text{ and } b\in R-\{0\} \text{ (any real number except 0).}$

$\displaystyle \textbf{Question 27: }$
$\displaystyle \text{If }\ f(x)= \begin{cases} \dfrac{1-\cos kx}{x\sin x}, & x\neq 0 \\ \dfrac{1}{2}, & x=0 \end{cases} \text{ is continuous at } x=0,\ \text{find } k.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} \dfrac{1-\cos kx}{x\sin x}, & x\neq 0 \\ \dfrac{1}{2}, & x=0 \end{cases}$

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

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

$\displaystyle \text{Consider:}$

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

$\displaystyle \Rightarrow \lim_{x\to 0} f(x) =\lim_{x\to 0}\left(\dfrac{2\sin^{2}\frac{kx}{2}}{x^{2}\left(\dfrac{\sin x}{x}\right)}\right)$

$\displaystyle \Rightarrow \lim_{x\to 0} f(x) =\lim_{x\to 0}\left(\dfrac{\dfrac{2k^{2}}{4}\left(\sin\frac{kx}{2}\right)^{2}} {\left(\dfrac{kx}{2}\right)^{2}\left(\dfrac{\sin x}{x}\right)}\right)$

$\displaystyle \Rightarrow \lim_{x\to 0} f(x) =\dfrac{2k^{2}}{4}\lim_{x\to 0}\left(\dfrac{\left(\sin\frac{kx}{2}\right)^{2}} {\left(\dfrac{kx}{2}\right)^{2}\left(\dfrac{\sin x}{x}\right)}\right)$

$\displaystyle \Rightarrow \lim_{x\to 0} f(x) =\dfrac{2k^{2}}{4} \left(\dfrac{\lim_{x\to 0}\left(\dfrac{\sin\frac{kx}{2}}{\frac{kx}{2}}\right)^{2}} {\lim_{x\to 0}\left(\dfrac{\sin x}{x}\right)}\right)$

$\displaystyle \Rightarrow \lim_{x\to 0} f(x) =\dfrac{2k^{2}}{4}\times \dfrac{1}{1} =\dfrac{k^{2}}{2}$

$\displaystyle \text{From equation (1), we have}$

$\displaystyle \dfrac{k^{2}}{2}=f(0)$

$\displaystyle \Rightarrow \dfrac{k^{2}}{2}=\dfrac{1}{2}$

$\displaystyle \Rightarrow k^{2}=1$

$\displaystyle \Rightarrow k=\pm 1$

$\displaystyle \textbf{Question 28: }$
$\displaystyle \text{If }\ f(x)= \begin{cases} \dfrac{x-4}{|x-4|}+a, & x<4 \\ a+b, & x=4 \\ \dfrac{x-4}{|x-4|}+b, & x>4 \end{cases} \\ \text{is continuous at } x=4,\ \text{find } a,\, b.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} \dfrac{x-4}{|x-4|}+a, & x<4 \\ a+b, & x=4 \\ \dfrac{x-4}{|x-4|}+b, & x>4 \end{cases}$

$\displaystyle \text{We observe}$

$\displaystyle (\text{LHL at } x=4)=$

$\displaystyle \lim_{x\to 4^-} f(x) =\lim_{h\to 0} f(4-h)$

$\displaystyle =\lim_{h\to 0}\left(\dfrac{(4-h)-4}{|(4-h)-4|}+a\right) =\lim_{h\to 0}\left(\dfrac{-h}{|-h|}+a\right)$

$\displaystyle =\lim_{h\to 0}\left(-1+a\right)=a-1$

$\displaystyle (\text{RHL at } x=4)=$

$\displaystyle \lim_{x\to 4^+} f(x) =\lim_{h\to 0} f(4+h)$

$\displaystyle =\lim_{h\to 0}\left(\dfrac{(4+h)-4}{|(4+h)-4|}+b\right) =\lim_{h\to 0}\left(\dfrac{h}{|h|}+b\right)$

$\displaystyle =\lim_{h\to 0}\left(1+b\right)=b+1$

$\displaystyle \text{And}$

$\displaystyle f(4)=a+b$

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

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

$\displaystyle \Rightarrow a-1=b+1=a+b$

$\displaystyle \Rightarrow a-1=a+b,\quad b+1=a+b$

$\displaystyle \Rightarrow b=-1,\ a=1$

$\displaystyle \textbf{Question 29: }$
$\displaystyle \text{For what value of } k \text{ is the function} f(x)= \begin{cases} \dfrac{\sin 2x}{x}, & x\neq 0 \\ k, & x=0 \end{cases} \\ \text{ continuous at } x=0?$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

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

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

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

$\displaystyle \Rightarrow \lim_{x\to 0}\dfrac{\sin 2x}{x}=k$

$\displaystyle \Rightarrow \lim_{x\to 0}\dfrac{2\sin 2x}{2x}=k$

$\displaystyle \Rightarrow 2\lim_{x\to 0}\dfrac{\sin 2x}{2x}=k$

$\displaystyle \Rightarrow 2\times 1=k$

$\displaystyle \Rightarrow k=2$

$\displaystyle \textbf{Question 30: }$
$\displaystyle \text{Let }\ f(x)=\dfrac{\log\left(1+\dfrac{x}{a}\right)-\log\left(1-\dfrac{x}{b}\right)}{x},\ x\neq 0. \text{Find the value of } f \text{ at } x=0 \text{ so that } f \text{ becomes continuous at } x=0.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

$\displaystyle f(x)=\dfrac{\log\left(1+\dfrac{x}{a}\right)-\log\left(1-\dfrac{x}{b}\right)}{x}, \quad x\neq 0$

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

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

$\displaystyle \Rightarrow \lim_{x\to 0} \left(\dfrac{\log\left(1+\dfrac{x}{a}\right)-\log\left(1-\dfrac{x}{b}\right)}{x}\right) =f(0)$

$\displaystyle \Rightarrow \lim_{x\to 0} \left( \dfrac{\log\left(1+\dfrac{x}{a}\right)}{\dfrac{ax}{a}} - \dfrac{\log\left(1-\dfrac{x}{b}\right)}{\dfrac{bx}{b}} \right) =f(0)$

$\displaystyle \Rightarrow \dfrac{1}{a}\lim_{x\to 0} \left(\dfrac{\log\left(1+\dfrac{x}{a}\right)}{\dfrac{x}{a}}\right) -\dfrac{1}{b}\lim_{x\to 0} \left(\dfrac{\log\left(1-\dfrac{x}{b}\right)}{-\dfrac{x}{b}}\right) =f(0)$

$\displaystyle \Rightarrow \dfrac{1}{a}\times 1-\left(-\dfrac{1}{b}\right)\times 1 =f(0) \quad \left[\text{Using } \lim_{x\to 0}\dfrac{\log(1+x)}{x}=1\right]$

$\displaystyle \Rightarrow \dfrac{1}{a}+\dfrac{1}{b}=f(0)$

$\displaystyle \Rightarrow \dfrac{a+b}{ab}=f(0)$

$\displaystyle \textbf{Question 31: }$
$\displaystyle \text{If }\ f(x)= \begin{cases} \dfrac{2^{x+2}-16}{4^{x}-16}, & x\neq 2 \\ k, & x=2 \end{cases} \text{is continuous at } x=2,\ \text{find } k.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

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

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

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

$\displaystyle \Rightarrow \lim_{x\to 2}\dfrac{2^{x+2}-16}{4^{x}-16}=k$

$\displaystyle \Rightarrow \lim_{x\to 2}\dfrac{4(2^{x}-4)}{(2^{x}-4)(2^{x}+4)}=k$

$\displaystyle \Rightarrow \lim_{x\to 2}\dfrac{4}{2^{x}+4}=k$

$\displaystyle \Rightarrow \dfrac{4}{2^{2}+4}=k$

$\displaystyle \Rightarrow \dfrac{4}{8}=k$

$\displaystyle \Rightarrow k=\dfrac{1}{2}$

$\displaystyle \textbf{Question 32: }$
$\displaystyle \text{If }\ f(x)= \begin{cases} \dfrac{\cos^{2}x-\sin^{2}x-1}{\sqrt{x^{2}+1}-1}, & x\neq 0 \\ k, & x=0 \end{cases} \text{ is continuous at } x=0,\ \text{find } k.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} \dfrac{\cos^{2}x-\sin^{2}x-1}{\sqrt{x^{2}+1}-1}, & x\neq 0 \\ k, & x=0 \end{cases}$

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

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

$\displaystyle \Rightarrow \lim_{x\to 0} \dfrac{\cos^{2}x-\sin^{2}x-1}{\sqrt{x^{2}+1}-1}=k$

$\displaystyle \Rightarrow \lim_{x\to 0} \dfrac{(1-\sin^{2}x)-\sin^{2}x-1}{\sqrt{x^{2}+1}-1}=k$

$\displaystyle \Rightarrow \lim_{x\to 0} \dfrac{-2\sin^{2}x}{\sqrt{x^{2}+1}-1}=k$

$\displaystyle \Rightarrow \lim_{x\to 0} \dfrac{-2\sin^{2}x(\sqrt{x^{2}+1}+1)} {(\sqrt{x^{2}+1}-1)(\sqrt{x^{2}+1}+1)}=k$

$\displaystyle \Rightarrow \lim_{x\to 0} \dfrac{-2\sin^{2}x(\sqrt{x^{2}+1}+1)}{x^{2}}=k$

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

$\displaystyle \Rightarrow -2\times 1\times (1+1)=k$

$\displaystyle \Rightarrow k=-4$

$\displaystyle \textbf{Question 33: }$
$\displaystyle \text{Extend the definition of the following by continuity} f(x)=\dfrac{1-\cos 7(x-\pi)}{5(x-\pi)^{2}} \quad \\ \text{at the point } x=\pi.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

$\displaystyle f(x)=\dfrac{1-\cos 7(x-\pi)}{5(x-\pi)^{2}}, \quad x\neq \pi$

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

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

$\displaystyle \Rightarrow \lim_{x\to \pi}\dfrac{1-\cos 7(x-\pi)}{5(x-\pi)^{2}}=f(\pi)$

$\displaystyle \Rightarrow \dfrac{2}{5}\lim_{x\to \pi}\dfrac{\sin^{2}\!\left(\dfrac{7(x-\pi)}{2}\right)}{(x-\pi)^{2}}=f(\pi)$

$\displaystyle \Rightarrow \dfrac{2}{5}\times\dfrac{49}{4} \lim_{x\to \pi}\dfrac{\sin^{2}\!\left(\dfrac{7(x-\pi)}{2}\right)} {\dfrac{49}{4}(x-\pi)^{2}}=f(\pi)$

$\displaystyle \Rightarrow \dfrac{2}{5}\times\dfrac{49}{4} \lim_{x\to \pi}\left[\dfrac{\sin\left(\dfrac{7(x-\pi)}{2}\right)} {\dfrac{7}{2}(x-\pi)}\right]^{2}=f(\pi)$

$\displaystyle \Rightarrow \dfrac{2}{5}\times\dfrac{49}{4}\times 1=f(\pi)$

$\displaystyle \Rightarrow \dfrac{49}{10}=f(\pi)$

$\displaystyle \text{Hence, the given function will be continuous at } x=\pi.$

$\displaystyle f(\pi)=\dfrac{49}{10}.$

$\displaystyle \textbf{Question 34: }$
$\displaystyle \text{If }\ f(x)=\dfrac{2x+3\sin x}{3x+2\sin x},\ x\neq 0. \ \text{If } f(x) \text{ is continuous at } x=0,\ \text{then find } f(0).$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

$\displaystyle f(x)=\dfrac{2x+3\sin x}{3x+2\sin x}, \quad x\neq 0$

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

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

$\displaystyle \Rightarrow \lim_{x\to 0}\dfrac{2x+3\sin x}{3x+2\sin x}=f(0)$

$\displaystyle \Rightarrow \lim_{x\to 0}\dfrac{x\left(2+3\dfrac{\sin x}{x}\right)} {x\left(3+2\dfrac{\sin x}{x}\right)}=f(0)$

$\displaystyle \Rightarrow \lim_{x\to 0}\dfrac{2+3\dfrac{\sin x}{x}} {3+2\dfrac{\sin x}{x}}=f(0)$

$\displaystyle \Rightarrow \dfrac{2+3\lim_{x\to 0}\dfrac{\sin x}{x}} {3+2\lim_{x\to 0}\dfrac{\sin x}{x}}=f(0)$

$\displaystyle \Rightarrow \dfrac{2+3\times 1}{3+2\times 1}=f(0)$

$\displaystyle \Rightarrow \dfrac{5}{5}=f(0)$

$\displaystyle \Rightarrow f(0)=1$

$\displaystyle \textbf{Question 35: }$
$\displaystyle \text{Find the value of } k \text{ for which } f(x)= \begin{cases} \dfrac{1-\cos 4x}{8x^{2}}, & x\neq 0 \\ k, & x=0 \end{cases} \\ \text{is continuous at } x=0. \hspace{7.0cm} \text{[CBSE 2000, 2017]}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} \dfrac{1-\cos 4x}{8x^{2}}, & x\neq 0 \\ k, & x=0 \end{cases}$

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

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

$\displaystyle \Rightarrow \lim_{x\to 0}\dfrac{1-\cos 4x}{8x^{2}}=f(0)$

$\displaystyle \Rightarrow \lim_{x\to 0}\dfrac{2\sin^{2}2x}{8x^{2}}=f(0)$

$\displaystyle \Rightarrow \dfrac{2}{8}\lim_{x\to 0}\dfrac{\sin^{2}2x}{x^{2}}=f(0)$

$\displaystyle \Rightarrow \dfrac{1}{4}\lim_{x\to 0}\left(\dfrac{\sin 2x}{x}\right)^{2}=f(0)$

$\displaystyle \Rightarrow \dfrac{1}{4}\left(\lim_{x\to 0}\dfrac{\sin 2x}{2x}\times 2\right)^{2}=f(0)$

$\displaystyle \Rightarrow \dfrac{1}{4}\left(1\times 2\right)^{2}=f(0)$

$\displaystyle \Rightarrow 1=f(0)$

$\displaystyle \Rightarrow k=1 \quad (\because\ f(0)=k)$

$\displaystyle \textbf{Question 36: } \text{In each of the following, find the value of the constant } k \\ \text{ so that the given function is continuous at the indicated points:}$

$\displaystyle \textbf{(i) } f(x)= \begin{cases} \dfrac{1-\cos 2kx}{x^{2}}, & x\neq 0 \\ 8, & x=0 \end{cases} \quad \text{at } x=0$

$\displaystyle \textbf{(ii) } f(x)= \begin{cases} (x-1)\tan\dfrac{\pi x}{2}, & x\neq 1 \\ k, & x=1 \end{cases} \quad \text{at } x=1$

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

$\displaystyle \textbf{(iv) } f(x)= \begin{cases} kx+1, & x\le \pi \\ \cos x, & x>\pi \end{cases} \quad \text{at } x=\pi$

$\displaystyle \textbf{(v) } f(x)= \begin{cases} kx+1, & x\le 5,\\ 3x-5, & x>5 \end{cases} \qquad \text{at } x=5$

$\displaystyle \textbf{(vi) } f(x)= \begin{cases} \dfrac{x^{2}-25}{x-5}, & x\neq 5,\\ k, & x=5 \end{cases} \qquad \text{at } x=5 \hspace{2.0cm} \text{[CBSE 2007]}$

$\displaystyle \textbf{(vii) } f(x)= \begin{cases} kx^{2}, & x\ge 1,\\ 4, & x<1 \end{cases} \qquad \text{at } x=1 \hspace{3.0cm} \text{[CBSE 2007]}$

$\displaystyle \textbf{(viii) } f(x)= \begin{cases} k(x^{2}+2), & x\le 0,\\ 3x+1, & x>0 \end{cases} \qquad \text{at } x=0 \hspace{2.0cm} \text{[CBSE 2010]}$

$\displaystyle \textbf{(ix) } f(x)= \begin{cases} \dfrac{x^{3}+x^{2}-16x+20}{(x-2)^{2}}, & x\neq 2,\\ k, & x=2 \end{cases} \qquad \text{at } x=2$

$\displaystyle \text{Answer:}$

$\displaystyle \textbf{(i) }$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} \dfrac{1-\cos 2kx}{x^2}, & x\ne 0 \\ 8, & x=0 \end{cases}$

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

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

$\displaystyle \Rightarrow \lim_{x\to 0} \dfrac{1-\cos 2kx}{x^2}=8$

$\displaystyle \Rightarrow \lim_{x\to 0} \dfrac{2\sin^2(kx)}{x^2}=8$

$\displaystyle \Rightarrow \lim_{x\to 0} \dfrac{2k^2x^2\sin^2(kx)}{k^2x^2x^2}=8$

$\displaystyle \Rightarrow 2k^2 \lim_{x\to 0}\left(\dfrac{\sin kx}{kx}\right)^2=8$

$\displaystyle \Rightarrow 2k^2 \times 1=8$

$\displaystyle \Rightarrow k^2=4$

$\displaystyle \Rightarrow k=\pm 2$

$\displaystyle \textbf{(ii) }$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} (x-1)\tan\left(\dfrac{\pi x}{2}\right), & x\ne 1 \\ k, & x=1 \end{cases}$

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

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

$\displaystyle \Rightarrow \lim_{x\to 1} (x-1)\tan\left(\dfrac{\pi x}{2}\right)=k$

$\displaystyle \text{Put } x-1=y,\text{ then } x=y+1$

$\displaystyle \Rightarrow \lim_{y\to 0} y\tan\left(\dfrac{\pi(y+1)}{2}\right)=k$

$\displaystyle \Rightarrow \lim_{y\to 0} y\tan\left(\dfrac{\pi}{2}+\dfrac{\pi y}{2}\right)=k$

$\displaystyle \Rightarrow \lim_{y\to 0} \left(-y\cot\left(\dfrac{\pi y}{2}\right)\right)=k$

$\displaystyle \Rightarrow -\lim_{y\to 0} y\cdot \dfrac{\cos\left(\dfrac{\pi y}{2}\right)}{\sin\left(\dfrac{\pi y}{2}\right)}=k$

$\displaystyle \Rightarrow -\dfrac{2}{\pi}\lim_{y\to 0} \dfrac{\cos\left(\dfrac{\pi y}{2}\right)} {\dfrac{\sin\left(\dfrac{\pi y}{2}\right)}{\dfrac{\pi y}{2}}}=k$

$\displaystyle \Rightarrow -\dfrac{2}{\pi}\times \dfrac{1}{1}=k$

$\displaystyle \Rightarrow k=-\dfrac{2}{\pi}$

$\displaystyle \textbf{(iii) }$

$\displaystyle \text{Given:}$

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

$\displaystyle \text{We examine continuity at } x=0.$

$\displaystyle \text{Left-hand limit (LHL) at } x=0:$

$\displaystyle \lim_{x\to 0^-} f(x) =\lim_{h\to 0^+} f(-h) =\lim_{h\to 0^+} k(h^2+2h) =0$

$\displaystyle \text{Right-hand limit (RHL) at } x=0:$

$\displaystyle \lim_{x\to 0^+} f(x) =\lim_{h\to 0^+} f(h) =\lim_{h\to 0^+} \cos h =1$

$\displaystyle \text{Since } \lim_{x\to 0^-} f(x)\ne \lim_{x\to 0^+} f(x),$

$\displaystyle \text{the limit of } f(x) \text{ at } x=0 \text{ does not exist.}$

$\displaystyle \therefore \text{There is no value of } k \text{ for which } f(x) \text{ is continuous at } x=0.$

$\displaystyle \textbf{(iv) }$

$\displaystyle \text{Given:}$

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

$\displaystyle \text{We examine continuity at } x=\pi.$

$\displaystyle \text{Left-hand limit (LHL) at } x=\pi:$

$\displaystyle \lim_{x\to \pi^-} f(x) =\lim_{h\to 0^+} f(\pi-h) =\lim_{h\to 0^+} \bigl(k(\pi-h)+1\bigr) =k\pi+1$

$\displaystyle \text{Right-hand limit (RHL) at } x=\pi:$

$\displaystyle \lim_{x\to \pi^+} f(x) =\lim_{h\to 0^+} f(\pi+h) =\lim_{h\to 0^+} \cos(\pi+h) =\cos \pi =-1$

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

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

$\displaystyle \Rightarrow k\pi+1=-1$

$\displaystyle \Rightarrow k=-\dfrac{2}{\pi}$

$\displaystyle \textbf{(v) }$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} kx+1, & x\le 5 \\ 3x-5, & x>5 \end{cases}$

$\displaystyle \text{We examine continuity at } x=5.$

$\displaystyle \text{Left-hand limit (LHL) at } x=5:$

$\displaystyle \lim_{x\to 5^-} f(x) =\lim_{h\to 0^+} f(5-h) =\lim_{h\to 0^+} \bigl(k(5-h)+1\bigr) =5k+1$

$\displaystyle \text{Right-hand limit (RHL) at } x=5:$

$\displaystyle \lim_{x\to 5^+} f(x) =\lim_{h\to 0^+} f(5+h) =\lim_{h\to 0^+} \bigl(3(5+h)-5\bigr) =10$

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

$\displaystyle \lim_{x\to 5^-} f(x)=\lim_{x\to 5^+} f(x)$

$\displaystyle \Rightarrow 5k+1=10$

$\displaystyle \Rightarrow k=\dfrac{9}{5}$

$\displaystyle \textbf{(vi) }$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} \dfrac{x^2-25}{x-5}, & x\ne 5 \\ k, & x=5 \end{cases}$

$\displaystyle \text{For } x\ne 5,$

$\displaystyle f(x)=\dfrac{(x-5)(x+5)}{x-5}$

$\displaystyle \Rightarrow f(x)=x+5,\quad x\ne 5$

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

$\displaystyle \lim_{x\to 5} f(x)=f(5)$

$\displaystyle \Rightarrow \lim_{x\to 5} (x+5)=k$

$\displaystyle \Rightarrow k=5+5=10$

$\displaystyle \textbf{(vii) }$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} kx^2, & x\ge 1 \\ 4, & x<1 \end{cases}$

$\displaystyle \text{We examine continuity at } x=1.$

$\displaystyle \text{Left-hand limit (LHL) at } x=1:$

$\displaystyle \lim_{x\to 1^-} f(x) =\lim_{h\to 0^+} f(1-h) =\lim_{h\to 0^+} 4 =4$

$\displaystyle \text{Right-hand limit (RHL) at } x=1:$

$\displaystyle \lim_{x\to 1^+} f(x) =\lim_{h\to 0^+} f(1+h) =\lim_{h\to 0^+} k(1+h)^2 =k$

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

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

$\displaystyle \Rightarrow 4=k$

$\displaystyle \Rightarrow k=4$

$\displaystyle \textbf{(viii) }$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} k(x^2+2), & x\le 0 \\ 3x+1, & x>0 \end{cases}$

$\displaystyle \text{We examine continuity at } x=0.$

$\displaystyle \text{Left-hand limit (LHL) at } x=0:$

$\displaystyle \lim_{x\to 0^-} f(x) =\lim_{h\to 0^+} f(0-h) =\lim_{h\to 0^+} k\bigl(( -h)^2+2\bigr) =2k$

$\displaystyle \text{Right-hand limit (RHL) at } x=0:$

$\displaystyle \lim_{x\to 0^+} f(x) =\lim_{h\to 0^+} f(0+h) =\lim_{h\to 0^+} (3h+1) =1$

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

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

$\displaystyle \Rightarrow 2k=1$

$\displaystyle \Rightarrow k=\dfrac{1}{2}$

$\displaystyle \textbf{(ix) }$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} \dfrac{x^3+x^2-16x+20}{(x-2)^2}, & x\ne 2 \\ k, & x=2 \end{cases}$

$\displaystyle \text{For } x\ne 2,$

$\displaystyle f(x)=\dfrac{x^3+x^2-16x+20}{x^2-4x+4}$

$\displaystyle =\dfrac{(x-2)^2(x+5)}{(x-2)^2}$

$\displaystyle \Rightarrow f(x)=x+5,\quad x\ne 2$

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

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

$\displaystyle \Rightarrow \lim_{x\to 2}(x+5)=k$

$\displaystyle \Rightarrow k=2+5=7$

$\displaystyle \textbf{Question 37: }$

$\displaystyle \text{Find the values of } a \text{ and } b \text{ so that the function } f \text{ given by}$

$\displaystyle f(x)= \begin{cases} 1, & x<3 \\ ax+b, & 3\le x<5 \\ 7, & x\ge 5 \end{cases} \text{ is continuous at } x=3 \text{ and } x=5. \hspace{1.0cm} \text{[CBSE 2013]}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} 1, & x\le 3 \\ ax+b, & 3<x<5 \\ 7, & x\ge 5 \end{cases}$

$\displaystyle \text{We have}$

$\displaystyle \text{(LHL at } x=3):$

$\displaystyle \lim_{x\to 3^-} f(x) =\lim_{h\to 0^+} f(3-h) =\lim_{h\to 0^+} 1 =1$

$\displaystyle \text{(RHL at } x=3):$

$\displaystyle \lim_{x\to 3^+} f(x) =\lim_{h\to 0^+} f(3+h) =\lim_{h\to 0^+} \bigl(a(3+h)+b\bigr) =3a+b$

$\displaystyle \text{(LHL at } x=5):$

$\displaystyle \lim_{x\to 5^-} f(x) =\lim_{h\to 0^+} f(5-h) =\lim_{h\to 0^+} \bigl(a(5-h)+b\bigr) =5a+b$

$\displaystyle \text{(RHL at } x=5):$

$\displaystyle \lim_{x\to 5^+} f(x) =\lim_{h\to 0^+} f(5+h) =\lim_{h\to 0^+} 7 =7$

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

$\displaystyle \lim_{x\to 3^-} f(x)=\lim_{x\to 3^+} f(x) \quad \text{and} \quad \lim_{x\to 5^-} f(x)=\lim_{x\to 5^+} f(x)$

$\displaystyle \Rightarrow 1=3a+b \qquad \ldots (1)$

$\displaystyle \text{and } 5a+b=7 \qquad \ldots (2)$

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

$\displaystyle a=3 \text{ and } b=-8$

$\displaystyle \textbf{Question 38: }$

$\displaystyle \text{If } f(x)= \begin{cases} \dfrac{x^2}{2}, & \text{if } 0\le x\le 1 \\ 2x^2-3x+\dfrac{3}{2}, & \text{if } 1<x\le 2 \end{cases}. \\ \text{Show that } f \text{ is continuous at } x=1.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} \dfrac{x^2}{2}, & 0\le x\le 1 \\ 2x^2-3x+\dfrac{3}{2}, & 1<x\le 2 \end{cases}$

$\displaystyle \text{We have}$

$\displaystyle \text{(LHL at } x=1):$

$\displaystyle \lim_{x\to 1^-} f(x) =\lim_{h\to 0^+} f(1-h) =\lim_{h\to 0^+} \dfrac{(1-h)^2}{2} =\dfrac{1}{2}$

$\displaystyle \text{(RHL at } x=1):$

$\displaystyle \lim_{x\to 1^+} f(x) =\lim_{h\to 0^+} f(1+h) =\lim_{h\to 0^+} \left[2(1+h)^2-3(1+h)+\dfrac{3}{2}\right] =\dfrac{1}{2}$

$\displaystyle \text{Also,}$

$\displaystyle f(1)=\dfrac{(1)^2}{2}=\dfrac{1}{2}$

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

$\displaystyle \text{Hence, the given function is continuous at } x=1.$

$\displaystyle \textbf{Question 39: }$

$\displaystyle \text{Discuss the continuity of the } f(x) \text{ at the indicated points:}$

$\displaystyle \text{(i) } f(x)=|x|+|x-1| \text{ at } x=0,1.$

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

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }$

$\displaystyle \text{Given:}$

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

$\displaystyle \text{We have}$

$\displaystyle \text{(LHL at } x=0):$

$\displaystyle \lim_{x\to 0^-} f(x) =\lim_{h\to 0^+} f(0-h) =\lim_{h\to 0^+}\bigl(|0-h|+|0-h-1|\bigr) =1$

$\displaystyle \text{(RHL at } x=0):$

$\displaystyle \lim_{x\to 0^+} f(x) =\lim_{h\to 0^+} f(0+h) =\lim_{h\to 0^+}\bigl(|0+h|+|0+h-1|\bigr) =1$

$\displaystyle \text{Also,}$

$\displaystyle f(0)=|0|+|0-1|=0+1=1$

$\displaystyle \text{Now,}$

$\displaystyle \text{(LHL at } x=1):$

$\displaystyle \lim_{x\to 1^-} f(x) =\lim_{h\to 0^+} f(1-h) =\lim_{h\to 0^+}\bigl(|1-h|+|1-h-1|\bigr) =1$

$\displaystyle \text{(RHL at } x=1):$

$\displaystyle \lim_{x\to 1^+} f(x) =\lim_{h\to 0^+} f(1+h) =\lim_{h\to 0^+}\bigl(|1+h|+|1+h-1|\bigr) =1$

$\displaystyle \text{Also,}$

$\displaystyle f(1)=|1|+|1-1|=1+0=1$

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

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

$\displaystyle \text{Hence, } f(x) \text{ is continuous at } x=0,1.$

$\displaystyle \text{(ii) }$

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

$\displaystyle \text{We have}$

$\displaystyle \text{(LHL at } x=-1):$

$\displaystyle \lim_{x\to -1^-} f(x) =\lim_{h\to 0^+} f(-1-h) =\lim_{h\to 0^+} \bigl(|-1-h-1|+|-1-h+1|\bigr) =2+0=2$

$\displaystyle \text{(RHL at } x=-1):$

$\displaystyle \lim_{x\to -1^+} f(x) =\lim_{h\to 0^+} f(-1+h) =\lim_{h\to 0^+} \bigl(|-1+h-1|+|-1+h+1|\bigr) =2+0=2$

$\displaystyle \text{Also,}$

$\displaystyle f(-1)=|-1-1|+|-1+1|=|{-2}|+0=2$

$\displaystyle \text{Now,}$

$\displaystyle \text{(LHL at } x=1):$

$\displaystyle \lim_{x\to 1^-} f(x) =\lim_{h\to 0^+} f(1-h) =\lim_{h\to 0^+} \bigl(|1-h-1|+|1-h+1|\bigr) =0+2=2$

$\displaystyle \text{(RHL at } x=1):$

$\displaystyle \lim_{x\to 1^+} f(x) =\lim_{h\to 0^+} f(1+h) =\lim_{h\to 0^+} \bigl(|1+h-1|+|1+h+1|\bigr) =0+2=2$

$\displaystyle \text{Also,}$

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

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

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

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

$\displaystyle \textbf{Question 40: }$

$\displaystyle \text{Prove that } f(x)= \begin{cases} \dfrac{x-|x|}{x}, & x\ne 0 \\ 2, & x=0 \end{cases} \text{ is discontinuous at } x=0$

$\displaystyle \text{Answer:}$

$\displaystyle \text{The given function can be rewritten as}$

$\displaystyle f(x)= \begin{cases} \dfrac{x-x}{x}, & x>0 \\ \dfrac{x+x}{x}, & x<0 \\ 2, & x=0 \end{cases} \ \ \ \Rightarrow \ \ \ \ f(x)= \begin{cases} 0, & x>0 \\ 2, & x<0 \\ 2, & x=0 \end{cases}$

$\displaystyle \text{We have}$

$\displaystyle \text{(LHL at } x=0):$

$\displaystyle \lim_{x\to 0^-} f(x) =\lim_{h\to 0^+} f(0-h) =\lim_{h\to 0^+} f(-h) =\lim_{h\to 0^+} 2 =2$

$\displaystyle \text{(RHL at } x=0):$

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

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

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

$\displaystyle \textbf{Question 41: }$

$\displaystyle \text{If } f(x)= \begin{cases} 2x^2+k, & x\ge 0 \\ -2x^2+k, & x<0 \end{cases} \text{ then what should be the value of } k \text{ so that } f(x) \\ \text{is continuous at } x=0.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{The given function can be rewritten as}$

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

$\displaystyle \text{We have}$

$\displaystyle \text{(LHL at } x=0):$

$\displaystyle \lim_{x\to 0^-} f(x) =\lim_{h\to 0^+} f(0-h) =\lim_{h\to 0^+} \bigl(-2(-h)^2+k\bigr) =k$

$\displaystyle \text{(RHL at } x=0):$

$\displaystyle \lim_{x\to 0^+} f(x) =\lim_{h\to 0^+} f(0+h) =\lim_{h\to 0^+} \bigl(2h^2+k\bigr) =k$

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

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

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

$\displaystyle \therefore \; k \text{ can be any real number.}$

$\displaystyle \textbf{Question 42: }$

$\displaystyle \text{For what value of } \lambda \text{ is the function}$

$\displaystyle f(x)= \begin{cases} \lambda(x^2-2x), & \text{if } x\le 0 \\ 4x+1, & \text{if } x>0 \end{cases} \text{ continuous at } x=0\text{? What about continuity at } x=\pm 1\text{?}$

$\displaystyle \text{Answer:}$

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

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

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

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

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

$\displaystyle \lim_{x\to 0^+} f(x) =\lim_{x\to 0^+} (4x+1) =1$

$\displaystyle \text{Since } \lim_{x\to 0^-} f(x)\ne \lim_{x\to 0^+} f(x),$

$\displaystyle \text{there is no value of } \lambda \text{ for which } f(x) \text{ is continuous at } x=0.$

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

$\displaystyle f(1)=4(1)+1=5$

$\displaystyle \text{Hence, for any value of } \lambda,\ f(x) \text{ is continuous at } x=1.$

$\displaystyle \text{At } x=-1,$

$\displaystyle f(-1)=\lambda\bigl((-1)^2-2(-1)\bigr)=\lambda(1+2)=3\lambda$

$\displaystyle \lim_{x\to -1} f(x)=\lambda(1+2)=3\lambda$

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

$\displaystyle \text{Hence, for any value of } \lambda,\ f(x) \text{ is continuous at } x=-1.$

$\displaystyle \textbf{Question 43: }$

$\displaystyle \text{For what value of } k \text{ is the following function continuous at } x=2?$

$\displaystyle f(x)= \begin{cases} 2x+1, & x<2 \\ k, & x=2 \\ 3x-1, & x>2 \end{cases} \hspace{7.0cm} \text{[CBSE 2008]}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

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

$\displaystyle \text{We have}$

$\displaystyle \text{(LHL at } x=2):$

$\displaystyle \lim_{x\to 2^-} f(x) =\lim_{h\to 0^+} f(2-h) =\lim_{h\to 0^+} \bigl(2(2-h)+1\bigr) =5$

$\displaystyle \text{(RHL at } x=2):$

$\displaystyle \lim_{x\to 2^+} f(x) =\lim_{h\to 0^+} f(2+h) =\lim_{h\to 0^+} \bigl(3(2+h)-1\bigr) =5$

$\displaystyle \text{Also,}$

$\displaystyle f(2)=k$

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

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

$\displaystyle \Rightarrow 5=5=k$

$\displaystyle \text{Hence, for } k=5,\ f(x) \text{ is continuous at } x=2.$

$\displaystyle \textbf{Question 44: }$

$\displaystyle \text{Let } f(x)= \begin{cases} \dfrac{1-\sin^3 x}{3\cos^2 x}, & \text{if } x<\dfrac{\pi}{2} \\ a, & \text{if } x=\dfrac{\pi}{2} \\ \dfrac{b(1-\sin x)}{(\pi-2x)^2}, & \text{if } x>\dfrac{\pi}{2} \end{cases} \text{. If } f(x) \text{ is continuous at } x=\dfrac{\pi}{2}, \text{ find } a \text{ and } b. \hspace{0.1cm} \text{[CBSE 2008, 2016]}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given } f(x)= \begin{cases} \dfrac{1-\sin^3 x}{3\cos^2 x}, & \text{if } x<\dfrac{\pi}{2} \\ a, & \text{if } x=\dfrac{\pi}{2} \\ \dfrac{b(1-\sin x)}{(\pi-2x)^2}, & \text{if } x>\dfrac{\pi}{2} \end{cases}$

$\displaystyle \text{We have}$

$\displaystyle \text{(LHL at } x=\dfrac{\pi}{2}\text{):}$

$\displaystyle \lim_{x\to \frac{\pi}{2}^-} f(x) =\lim_{h\to 0^+} f\left(\dfrac{\pi}{2}-h\right)$

$\displaystyle =\lim_{h\to 0^+}\left(\dfrac{1-\sin^3\left(\dfrac{\pi}{2}-h\right)}{3\cos^2\left(\dfrac{\pi}{2}-h\right)}\right)$

$\displaystyle =\lim_{h\to 0^+}\left(\dfrac{1-\cos^3 h}{3\sin^2 h}\right)$

$\displaystyle =\dfrac{1}{3}\lim_{h\to 0^+}\left(\dfrac{(1-\cos h)(1+\cos h+\cos^2 h)}{(1-\cos h)(1+\cos h)}\right)$

$\displaystyle =\dfrac{1}{3}\lim_{h\to 0^+}\left(\dfrac{1+\cos h+\cos^2 h}{1+\cos h}\right)$

$\displaystyle =\dfrac{1}{3}\left(\dfrac{1+1+1}{1+1}\right)=\dfrac{1}{2}$

$\displaystyle \text{(RHL at } x=\dfrac{\pi}{2}\text{):}$

$\displaystyle \lim_{x\to \frac{\pi}{2}^+} f(x) =\lim_{h\to 0^+} f\left(\dfrac{\pi}{2}+h\right)$

$\displaystyle =\lim_{h\to 0^+}\left(\dfrac{b\left[1-\sin\left(\dfrac{\pi}{2}+h\right)\right]}{\left[\pi-2\left(\dfrac{\pi}{2}+h\right)\right]^2}\right)$

$\displaystyle =\lim_{h\to 0^+}\left(\dfrac{b(1-\cos h)}{(-2h)^2}\right)$

$\displaystyle =\lim_{h\to 0^+}\left(\dfrac{b\cdot 2\sin^2\left(\dfrac{h}{2}\right)}{4h^2}\right)$

$\displaystyle =\dfrac{b}{8}\lim_{h\to 0^+}\left(\dfrac{\sin\left(\dfrac{h}{2}\right)}{\dfrac{h}{2}}\right)^2$

$\displaystyle =\dfrac{b}{8}\times 1=\dfrac{b}{8}$

$\displaystyle \text{Also, } f\left(\dfrac{\pi}{2}\right)=a$

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

$\displaystyle \lim_{x\to \frac{\pi}{2}^-} f(x)=\lim_{x\to \frac{\pi}{2}^+} f(x)=f\left(\dfrac{\pi}{2}\right)$

$\displaystyle \Rightarrow \dfrac{1}{2}=\dfrac{b}{8}=a$

$\displaystyle \Rightarrow a=\dfrac{1}{2} \text{ and } b=4$

$\displaystyle \textbf{Question 45: }$

$\displaystyle \text{If the function } f(x), \text{ defined below is continuous at } x=0, \text{ find the value of } k:$

$\displaystyle f(x)= \begin{cases} \dfrac{1-\cos 2x}{2x^2}, & x<0 \\ k, & x=0 \\ \dfrac{x}{|x|}, & x>0 \end{cases} \hspace{7.0cm} \text{[CBSE 2010]}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} \dfrac{1-\cos 2x}{2x^2}, & x<0 \\ k, & x=0 \\ \dfrac{x}{|x|}, & x>0 \end{cases}$

$\displaystyle \Rightarrow f(x)= \begin{cases} \dfrac{1-\cos 2x}{2x^2}, & x<0 \\ k, & x=0 \\ 1, & x>0 \end{cases}$

$\displaystyle \text{We have}$

$\displaystyle \text{(LHL at } x=0):$

$\displaystyle \lim_{x\to 0^-} f(x) =\lim_{h\to 0^+} f(0-h) =\lim_{h\to 0^+}\left(\dfrac{1-\cos 2(-h)}{2(-h)^2}\right)$

$\displaystyle =\lim_{h\to 0^+}\left(\dfrac{1-\cos 2h}{2h^2}\right)$

$\displaystyle =\dfrac{1}{2}\lim_{h\to 0^+}\left(\dfrac{2\sin^2 h}{h^2}\right)$

$\displaystyle =\lim_{h\to 0^+}\left(\dfrac{\sin h}{h}\right)^2 =1$

$\displaystyle \text{(RHL at } x=0):$

$\displaystyle \lim_{x\to 0^+} f(x) =\lim_{h\to 0^+} f(h) =\lim_{h\to 0^+} 1 =1$

$\displaystyle \text{Also,}$

$\displaystyle f(0)=k$

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

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

$\displaystyle \Rightarrow 1=1=k$

$\displaystyle \text{Hence, the required value of } k \text{ is } 1.$

$\displaystyle \textbf{Question 46: }$

$\displaystyle \text{Find the relationship between } a \text{ and } b \text{ so that the function } f \text{ defined by}$

$\displaystyle f(x)= \begin{cases} ax+1, & x<3 \\ bx+3, & x>3 \end{cases} \text{ is continuous at } x=3. \hspace{2.0cm} \text{[CBSE 2011]}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given:}$

$\displaystyle f(x)= \begin{cases} ax+1, & x<3 \\ bx+3, & x>3 \end{cases}$

$\displaystyle \text{We have}$

$\displaystyle \text{(LHL at } x=3):$

$\displaystyle \lim_{x\to 3^-} f(x) =\lim_{h\to 0^+} f(3-h) =\lim_{h\to 0^+} \bigl(a(3-h)+1\bigr) =3a+1$

$\displaystyle \text{(RHL at } x=3):$

$\displaystyle \lim_{x\to 3^+} f(x) =\lim_{h\to 0^+} f(3+h) =\lim_{h\to 0^+} \bigl(b(3+h)+3\bigr) =3b+3$

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

$\displaystyle \lim_{x\to 3^-} f(x)=\lim_{x\to 3^+} f(x)$

$\displaystyle \Rightarrow 3a+1=3b+3$

$\displaystyle \Rightarrow 3a-3b=2$

$\displaystyle \text{Hence, the required relationship between } a \text{ and } b \text{ is } 3a-3b=2.$

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

Subscribe to get the latest posts sent to your email.

ICSE / ISC / CBSE Mathematics Portal for K12 Students

Ultimate Math Study Guide for Students

Class 12: Continuity – Exercise 9.1

Like this:

Related

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

Leave a ReplyCancel reply

Share this:

Like this:

Related

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

Leave a ReplyCancel reply

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