\displaystyle \textbf{Question 1. }\text{The graph of the function }f\text{ is shown below \hspace{0.2cm} ISC 2024} \ \ \displaystyle \text{Of the following options, at what values of }x\text{ is the function }f\text{ not differentiable?} \ \
\displaystyle \text{(a) At }x=0\text{ and }x=2 \ \
\displaystyle \text{(b) At }x=1\text{ and }x=3 \ \
\displaystyle \text{(c) At }x=-1\text{ and }x=1 \ \
\displaystyle \text{(d) At }x=-1.5\text{ and }x=1.5 \ \
\displaystyle \text{Answer:}
\displaystyle \text{(a) } \text{Given, graph of the function } f \text{ is }
\displaystyle \text{At } x=0, \text{ we cannot draw a tangent, therefore the function is not }
\displaystyle \text{differentiable at } x=0.
\displaystyle \text{Similarly, at } x=2, \text{ we cannot draw a tangent. Therefore, function }
\displaystyle \text{is not differentiable at } x=2.
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\displaystyle \textbf{Question 2. }\text{Determine the value of }k\text{ for which the following} \ \
\displaystyle \text{function is continuous at }x=3.\text{ \hspace{0.2cm} ISC 2024} \ \
\displaystyle f(x)=\begin{cases}\frac{(x+3)^{2}-36}{x-3}, & x\ne 3 \\ k, & x=3\end{cases} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } f(x)=\begin{cases} \frac{(x+3)^{2}-36}{x-3}, & x \ne 3 \\ k, & x=3 \end{cases}
\displaystyle \text{LHL }=\lim_{x \to 3^{-}} f(x)=\lim_{h \to 0} f(3-h)
\displaystyle =\lim_{h \to 0}\frac{(3-h+3)^{2}-36}{3-h-3}
\displaystyle =\lim_{h \to 0}\frac{(6-h)^{2}-36}{-h}
\displaystyle =\lim_{h \to 0}\frac{36+h^{2}-12h-36}{-h}
\displaystyle =\lim_{h \to 0}\frac{h^{2}-12h}{-h}=\lim_{h \to 0}(12-h)=12
\displaystyle \text{RHL }=\lim_{x \to 3^{+}} f(x)=\lim_{h \to 0} f(3+h)
\displaystyle =\lim_{h \to 0}\frac{(3+h+3)^{2}-36}{3+h-3}
\displaystyle \text{Since, } f \text{ is continuous at } x=3.
\displaystyle \therefore \text{LHL = RHL = } f(3)
\displaystyle \Rightarrow 12=12=k
\displaystyle \therefore k=12
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\displaystyle \textbf{Question 3. }\text{If the following function is continuous at }x=2,\text{ then} \ \
\displaystyle \text{the value of }k\text{ will be \hspace{0.2cm} ISC 2023} \ \
\displaystyle f(x)=\begin{cases}2x+1, & x<2 \\ k, & x=2 \\ 3x-1, & x>2\end{cases} \ \
\displaystyle \text{(a) }2 \ \
\displaystyle \text{(b) }3 \ \
\displaystyle \text{(c) }5 \ \
\displaystyle \text{(d) }-1 \ \
\displaystyle \text{Answer:}
\displaystyle \text{(c) } f(x) \text{ is continuous at } x=2.
\displaystyle \therefore \lim_{x \to 2}(2x+1)=k
\displaystyle \Rightarrow 2(2)+1=k \Rightarrow k=5
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\displaystyle \textbf{Question 4. }\text{If }f(x)=\frac{x^{2}-9}{x-3}\text{ is not defined at }x=3,\text{ what value} \ \
\displaystyle \text{should be assigned to }f(3)\text{ for continuity of }f(x)\text{ at }x=3?\text{ \hspace{0.2cm} ISC 2019} \ \
\displaystyle \text{Answer:}
\displaystyle \text{We have, } f(x)=\frac{x^{2}-9}{x-3}
\displaystyle \text{Now, } \text{RHL}=\lim_{x \to 3^{+}} f(x)=\lim_{x \to 3^{+}}\frac{x^{2}-9}{x-3}
\displaystyle \text{On putting } x=3+h \text{ as } x \to 3^{+} \text{ when } h \to 0
\displaystyle =\lim_{h \to 0}\frac{(3+h)^{2}-9}{3+h-3}
\displaystyle =\lim_{h \to 0}\frac{9+h^{2}+6h-9}{h}
\displaystyle =\lim_{h \to 0}(6+h)=6
\displaystyle \text{and LHL }=\lim_{x \to 3^{-}} f(x)=\lim_{x \to 3^{-}}\frac{x^{2}-9}{x-3}
\displaystyle \text{On putting } x=3-h \text{ as } x \to 3^{-} \text{ when } h \to 0
\displaystyle =\lim_{h \to 0}\frac{(3-h)^{2}-9}{3-h-3}
\displaystyle =\lim_{h \to 0}\frac{9+h^{2}-6h-9}{-h}
\displaystyle =\lim_{h \to 0}(6-h)=6
\displaystyle \text{Thus, to make } f(x) \text{ continuous, } f(3) \text{ should be equal to } 6.
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\displaystyle \textbf{Question 5. }\text{Show that the function }f(x)=|x-4|,\ x\in R\text{ is} \ \
\displaystyle \text{continuous but not differentiable at }x=4.\text{ \hspace{0.2cm} ISC 2019} \ \
\displaystyle \text{Answer:}
\displaystyle \text{We have, } f(x)=|x-4|=\begin{cases} x-4, & x \ge 4 \\ 4-x, & x<4 \end{cases}
\displaystyle \text{Check continuity at } x=4
\displaystyle \text{LHL}=\lim_{x \to 4^{-}} f(x)=\lim_{h \to 0} f(4-h)
\displaystyle =\lim_{h \to 0}[4-(4-h)]=\lim_{h \to 0}h=0
\displaystyle \text{RHL}=\lim_{x \to 4^{+}} f(x)=\lim_{h \to 0} f(4+h)
\displaystyle =\lim_{h \to 0}[(4+h)-4]=\lim_{h \to 0}h=0
\displaystyle \text{and } f(4)=4-4=0
\displaystyle \text{Here, LHL = RHL = } f(4)=0
\displaystyle \text{Hence, } f(x) \text{ is continuous at } x=4
\displaystyle \text{Check differentiability at } x=4
\displaystyle \text{LHD}=\lim_{h \to 0}\frac{f(4-h)-f(4)}{-h}
\displaystyle =\lim_{h \to 0}\frac{4-(4-h)-0}{-h}=\lim_{h \to 0}\frac{h}{-h}=-1
\displaystyle \text{RHD}=\lim_{h \to 0}\frac{f(4+h)-f(4)}{h}
\displaystyle =\lim_{h \to 0}\frac{(4+h)-4-0}{h}=\lim_{h \to 0}\frac{h}{h}=1
\displaystyle \text{Here, LHD } \ne \text{ RHD}
\displaystyle \text{Hence, } f(x) \text{ is not differentiable at } x=4
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\displaystyle \textbf{Question 6. }\text{Find the value of constant }k\text{ so that the function }f(x) \ \
\displaystyle \text{defined as} \ \
\displaystyle f(x)=\begin{cases}\frac{x^{2}-2x-3}{x+1}, & x\ne -1 \\ k, & x=-1\end{cases} \ \
\displaystyle \text{is continuous at }x=-1.\text{ \hspace{0.2cm} ISC 2018} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } f(x)=\begin{cases} \frac{x^{2}-2x-3}{x+1}, & x \ne -1 \\ k, & x=-1 \end{cases}
\displaystyle \text{is continuous at } x=-1.
\displaystyle \therefore \lim_{x \to -1} f(x)=f(-1)
\displaystyle \Rightarrow \lim_{x \to -1}\frac{x^{2}-2x-3}{x+1}=k
\displaystyle =\lim_{x \to -1}\frac{(x+1)(x-3)}{x+1}
\displaystyle =\lim_{x \to -1}(x-3)
\displaystyle =-1-3=-4
\displaystyle \therefore k=-4
\displaystyle \text{Hence, the value of } k \text{ is } -4.
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\displaystyle \textbf{Question 7. }\text{Show that the function }f(x)=\begin{cases}x^{2}, & x\leq 1 \\ \frac{1}{x}, & x>1\end{cases}\text{ is} \ \
\displaystyle \text{continuous at }x=1\text{ but not differentiable. \hspace{0.2cm} ISC 2018} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } f(x)=\begin{cases} x^{2}, & x \le 1 \\ \frac{1}{x}, & x>1 \end{cases}
\displaystyle \text{Continuity at } x=1
\displaystyle \text{LHL}=\lim_{x \to 1^{-}} f(x)=\lim_{h \to 0} f(1-h)
\displaystyle =\lim_{h \to 0}(1-h)^{2}=\lim_{h \to 0}(1-2h+h^{2})=1
\displaystyle \text{RHL}=\lim_{x \to 1^{+}} f(x)=\lim_{h \to 0} f(1+h)
\displaystyle =\lim_{h \to 0}\frac{1}{1+h}=1
\displaystyle \text{and } f(1)=1
\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 \text{Differentiability at } x=1
\displaystyle \text{LHD}=\lim_{h \to 0}\frac{f(1-h)-f(1)}{-h}
\displaystyle =\lim_{h \to 0}\frac{(1-h)^{2}-1}{-h}
\displaystyle =\lim_{h \to 0}\frac{1-2h+h^{2}-1}{-h}
\displaystyle =\lim_{h \to 0}\frac{-2h+h^{2}}{-h}=\lim_{h \to 0}(2-h)=2
\displaystyle \text{RHD}=\lim_{h \to 0}\frac{f(1+h)-f(1)}{h}
\displaystyle =\lim_{h \to 0}\frac{\frac{1}{1+h}-1}{h}
\displaystyle =\lim_{h \to 0}\frac{1-(1+h)}{h(1+h)}
\displaystyle =\lim_{h \to 0}\frac{-h}{h(1+h)}=\lim_{h \to 0}\frac{-1}{1+h}=-1
\displaystyle \text{Since, LHD } \ne \text{ RHD}
\displaystyle \text{Hence, } f \text{ is not differentiable at } x=1
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