Question 15 Marks
Discuss the continuity and differentiability of,$\text{f(x)}=\begin{cases}(\text{x}-\text{c})\cos\Big(\frac{1}{\text{x}-\text{c}}\Big), & \text{x}\neq 0\\0, & \text{x}= 0\end{cases}$
Answer$\text{f(x)}=\begin{cases}(\text{x}-\text{c})\cos\Big(\frac{1}{\text{x}-\text{c}}\Big), & \text{x}\neq 0\\0, & \text{x}= 0\end{cases}$
(LHL at x = c) $=\lim_\limits{\text{x}\rightarrow\text{c}^{-}}\text{f(x)}$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}(\text{c}-\text{h}-\text{c})\cos\Big(\frac{1}{\text{c}-\text{h}-\text{c}}\Big)$
$=\lim_\limits{\text{h}\rightarrow0}-\text{h}\cos\Big(-\frac{1}{\text{h}}\Big)$
$=\lim_\limits{\text{h}\rightarrow0}-\text{h}\cos\Big(\frac{1}{\text{h}}\Big)$
$=0$
(RHL at x = c) $=\lim_\limits{\text{x}\rightarrow\text{c}^{+}}\text{f(x)}$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(\text{c}+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}(\text{c}+\text{h}-\text{c})\cos\Big(\frac{1}{\text{c}+\text{h}-\text{c}}\Big)$
$=\lim_\limits{\text{h}\rightarrow0}-\text{h}\cos\Big(\frac{1}{\text{h}}\Big)$
$=0$
f(c) = 0
Since, LHL = f(x) = RHL at x = c
⇒ f(x) is continuous at x = c
(LHL at x = c) $=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{c}-\text{h})-\text{f(c)}}{-\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{(\text{c}-\text{h}-\text{c})\cos\Big(\frac{1}{\text{c}-\text{h}-\text{c}}\Big)-0}{-\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\cos\Big(-\frac{1}{\text{h}}\Big)$
$=\lim_\limits{\text{h}\rightarrow0}\cos\Big(\frac{1}{\text{h}}\Big)$
(RHL at x = c) $=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{c}+\text{h})-\text{f(c)}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{(\text{c}+\text{h}-\text{c})\cos\Big(\frac{1}{\text{c}+\text{h}-\text{c}}\Big)-0}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}\cos\Big(\frac{1}{\text{h}}\Big)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\cos\Big(\frac{1}{\text{h}}\Big)$
(LHL at x = c) = (RHL at x = c)
So,
f(x) is differentiable and continuous at x = c.
View full question & answer→Question 25 Marks
Finde the value of a and b, if the function f(x) defined by $\text{f(x)}\begin{cases}\text{x}^2+3\text{x}+\text{a}, &\text{x}\leq1\\\text{bx}+2, & \text{x}>1\end{cases}$is differentiable at x = 1.
AnswerGiven that f(x) is differentiable at x = 1, Therefore, f(x) is countinuous at x = 1.
$\lim\limits_{\text{x}\rightarrow1^{-}}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1^{+}}\text{f(x)}=\text{f(1)}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow1}(\text{x}^2+3\text{x}+\text{a})=\lim\limits_{\text{x}\rightarrow1}(\text{bx}+2)=1+3+\text{a}$
$\Rightarrow1+3+\text{a}=\text{b}+2$
$\Rightarrow\text{a}-\text{b}+2=0\dots(1)$
Again, f(x) is differentiable at x = 1. So,
(LHL at x = 1) = (RHL at x = 1)
$\Rightarrow\lim\limits_{\text{x}\rightarrow1^{-}}\frac{\text{f(z)}-\text{f}(1)}{\text{z}-1}=\lim\limits_{\text{x}\rightarrow1^{+}}\frac{\text{f(z)}-\text{f}(1)}{\text{z}-1}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow1}\frac{(\text{x}^2+3\text{x}+\text{a})-(4+\text{a})}{\text{x}-1}=\lim\limits_{\text{x}\rightarrow0}\frac{(\text{bx}+2)-(4+\text{a})}{\text{x}-1}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^2+3\text{x}-4}{\text{x}-1}=\lim\limits_{\text{x}\rightarrow1}\frac{(\text{bx}-2-\text{a})}{\text{x}-1}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow1}(\text{x}-4)=\lim\limits_{\text{x}\rightarrow1}\frac{\text{b}(\text{x}-1)}{\text{x}-1}$
$\Rightarrow5=\text{b}$
Hence, a = 3 and b = 5.
View full question & answer→Question 35 Marks
Discuss the continuity and differentiability of $\text{f(x)}=\text{e}^{|\text{x}|}.$
AnswerGiven:
$\text{f(x)}=\text{e}^{|\text{x}|}$
$\Rightarrow\text{f(x)}=\begin{cases}\text{e}^\text{x},&\text{x}\geq0\\\text{e}^{-\text{x}},&\text{x}<0\end{cases}$
f is Continuity:
(LHL at x = 0)
$\lim_\limits{\text{x}\rightarrow0{^-}}\text{f(x)}$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{e}^{-(0-\text{h})}$
$=\lim_\limits{\text{h}\rightarrow0}\text{e}^{-\text{h}}$
$=1$
(RHL at x = 0)
$\lim_\limits{\text{x}\rightarrow0{^{+}}}\text{f(x)}$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{e}^{(0+\text{h})}$
$=1$
and f(0)
$=\text{e}^0=1$
Thus, $\lim_\limits{\text{x}\rightarrow0^{-}}-\text{f(x)}=\lim_\limits{\text{h}\rightarrow0^{+}}-\text{f(x)}=\text{f(0)}$
Hence, function is continuous at x = 0.
Differentiability at x = 0.
(LHL at x = 0)
$=\lim_\limits{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(0-\text{h})-\text{f}(0)}{0-\text{h}-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^{-(0-\text{h})}-1}{-\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^\text{h}-1}{-\text{h}}=-1\ \Big[\because\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^\text{h}-1}{\text{h}}=1\Big]$
(RHL at x = 0)
$=\lim_\limits{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(0-\text{h})-\text{f}(0)}{0+\text{h}-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^{-(0-\text{h})}-1}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^\text{h}-1}{\text{h}}=1\ \Big[\because\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^\text{h}-1}{\text{h}}=1\Big]$
LHL at (x = 0) $\neq$ RHL at (x = 0)
Hence the function is not differentiable at x = 0.
View full question & answer→Question 45 Marks
If $\text{f(x)}=\begin{cases}\text{ax}^2-\text{b}, & \text{if |x|}<1\\\frac{1}{|\text{x}|}, & \text{if |x|}\geq1\end{cases}$ is differentiable at x = 1, find a, b.
AnswerHere,
$\text{f(x)}=\begin{cases}\text{ax}^2-\text{b}, & \text{if |x|}<1\\\frac{1}{|\text{x}|}, & \text{if |x|}\geq1\end{cases}$
$=\begin{cases}-\frac{1}{\text{x}}, & \text{if |x|}\leq-1\\\text{ax}^2-\text{b}, & \text{if}-1<\text{x}<1\\\frac{1}{\text{x}},&\text{if x}\geq1\end{cases}$
$\text{LHL }=\lim\limits_{\text{x}\rightarrow1^{-}}\text{f(x)}$
$=\lim\limits_{\text{h}\rightarrow0}\text{f}(1-\text{h})$
$=\lim\limits_{\text{h}\rightarrow0}\text{a}(1-\text{h})^2-\text{b}$
$= \text{a}- \text{b}$
$\text{RHL }=\lim\limits_{\text{x}\rightarrow1^{+}}\text{f(x)}$
$=\lim\limits_{\text{h}\rightarrow0}\text{f}(1+\text{h})$
$=\lim\limits_{\text{h}\rightarrow0}\frac{1}{1+\text{h}}$
Since, f(x) is continuous, so
LHL = RHL
a - b = 1 .......(1)
(LHL at x = 1) $=\lim\limits_{\text{x}\rightarrow1^{-}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
$\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1-\text{h})-1}{1-\text{h}-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}(1-\text{h})^2-\text{b}-1}{-\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}(1-\text{h})^2-(\text{a}-1)-1}{-\text{h}}$
Using equation (1),
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}+\text{ah}^2-2\text{ah}-\text{a}+1-1}{-\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{ah}^2-2\text{ah}}{-\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}(2\text{a}-\text{ah})$
$=2\text{a}$
RHL at x = 1 $=\lim_\limits{\text{x}\rightarrow1^{+}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
$\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(1+\text{h})-{\text{f}(1)}}{1+\text{h}-1}$
$\lim_\limits{\text{h}\rightarrow0}\frac{\frac{1}{1+\text{h}}-1}{\text{h}}$
$\lim_\limits{\text{h}\rightarrow0}\frac{1-1-\text{h}}{(1+\text{h})\text{h}}$
$= -1$
Since f(x) is differentiable at x = 1,
(LHL at x = 1) = (RHL at x = 1)
2a = -1
$\text{a}=\frac{-1}{2}$
Put $\text{a}=\frac{-1}{2}$ in equation (1),
a - b = 1
$\Big(\frac{-1}{2}\Big)-\text{b}=1$
$\text{b}=\frac{-1}{2}-1$
$\text{b}=\frac{-3}{2}$
$\text{a}=\frac{-1}{2}$
View full question & answer→Question 55 Marks
Show that the function $\text{f(x)}\begin{cases}\text{x}^\text{m}\sin(\frac{1}{\text{x}}), &\text{x}\neq0 \\0 ,& \text{x}=0\end{cases}$
Continuous but not diffierentiable at x = 0, if 0 < m < 1
Answer$\text{LHL}=\lim_\limits{\text{x}\rightarrow0^{-}}\text{f(x)}$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}(-\text{h})^\text{m}\sin\Big(-\frac{1}{\text{h}}\Big)$
$=\lim_\limits{\text{h}\rightarrow0}(-\text{h})^\text{m}\sin\Big(\frac{1}{\text{h}}\Big)$
$=0\times\text{k}\ [\text{Where}-1\leq\text{k}\leq1]$
$=0$
$\text{RHL }=\lim_\limits{\text{x}\rightarrow0^{+}}\text{f(x)}$
$=\lim_\limits{\text{h}\rightarrow0^+}\text{f}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}(+\text{h})^\text{m}\sin\Big(\frac{1}{0+\text{h}}\Big)$
$=\lim_\limits{\text{h}\rightarrow0^-}(-\text{h})^\text{m}\sin\Big(\frac{1}{\text{h}}\Big)$
$=0\times\text{k'}\ [\text{When}-1\leq\text{k}'\leq1]$
$=0$
LHL = f(0) = RHL
$\therefore$ f(x) is continuous at x = 0
For differentiable at x = 0
(LHL at x = 0) $=\lim_\limits{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{(0-\text{h})-\text{f}(0)}{(0-\text{h})-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{(-\text{h})^\text{m}\sin\Big(-\frac{1}{\text{h}}\Big)}{-\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}(-\text{h})^\text{m-1}\sin\Big(-\frac{1}{\text{h}}\Big)$
= Not definded [Since 0 < m < 1]
(RHL at x = 0) $=\lim_\limits{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(0+\text{h})-\text{f}(0)}{(0+\text{h})-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}^\text{m}\sin\Big(\frac{1}{\text{h}}\Big)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}(\text{h})^\text{m-1}\sin\Big(\frac{1}{\text{h}}\Big)$
= Not defined [as 0, m < 1]
$\therefore$ (LHL at x = 0) and (RHL at x = 0) are not defined, so f(x) is continuous but not differentiable at x = 0, when 0 < m < 1.
View full question & answer→Question 65 Marks
Find the values of a and b so that the function $\text{f(x)}\begin{cases}\text{x}^2+3\text{x}+\text{a}, & \text{if x}\leq1\\\text{bx}+2, & \text{if x} > 1\end{cases}$ is differentiable at each $\text{x}\in\text{R}.$
AnswerIt is given that the function is differentiable at each $\text{x}\in\text{R}$ and every differentiable function is continuous.
Therefore,
Given: $\text{f(x)}=\begin{Bmatrix}\text{x}^2+3\text{x}+\text{a}, & \text{if x}\leq1\\\text{bx}+2, & \text{if x} > 1 \end{Bmatrix}$
Therefore,
f(x) is continuous at x = 1.
Therefore,
$\lim_\limits{\text{x}\rightarrow1^{-}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^{+}}\text{f(x)}=\text{f}(1)$
Implies that $\lim_\limits{\text{x}\rightarrow1}\text{x}^2+3\text{x}+\text{a}=\lim_\limits{\text{x}\rightarrow1}\text{bx}+2=\text{a}+4$ [Using def. of f(x)]
Implies that a + 4 = b + 2 = a + 4 ......(1)
Since, f(x) is differentiable at x = 1. Therefore,
(LHL at x = 1) = (RHL at x = 1)
$\lim_\limits{\text{x}\rightarrow1^{-}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}=\lim_\limits{\text{x}\rightarrow1^{+}}\frac{\text{f(c)}-\text{f}(1)}{\text{x}-1}$
Implies that $\lim_\limits{\text{x}\rightarrow1}\frac{\text{x}^2+3\text{x}+\text{a}-\text{a}-4}{\text{x}-1}=\lim_\limits{\text{x}\rightarrow1}\frac{\text{bx}+2-4-\text{a}}{\text{x}-1}$ [Using def. of f(x)]
Implies that $\lim_\limits{\text{x}\rightarrow1}\frac{(\text{x}+4)(\text{x-1})}{\text{x}-1}=\lim_\limits{\text{x}\rightarrow1}\frac{\text{bx}-2-\text{a}}{\text{x}-1}$
Implies that $\lim_\limits{\text{x}\rightarrow1}\frac{(\text{x}+4)(\text{x}-1)}{\text{x}-1}=\lim_\limits{\text{x}\rightarrow1}\frac{\text{bx}-\text{b}}{\text{x}-1}$ [Using (1)]
Implies that $\lim_\limits{\text{x}\rightarrow1}\frac{(\text{x}+4)(\text{x}-1)}{\text{x}-1}=\lim_\limits{\text{x}\rightarrow1}\frac{\text{b}(\text{x}-1)}{\text{x}-1}$
Implies that 5 = b
From (1), we have
a + 4 = b + 2
Implies that a + 4 = 5 + 2
Implies that a = 7 - 4
Implies that a = 3
Hence, a = 3, b = 5.
View full question & answer→Question 75 Marks
Is $|\sin\text{x}|$ differentible? What about $\cos|\text{x}|?$
AnswerLet, d(x) = |sin x|
$\sin\text{x}=0,$ for $\text{x}=\text{n}\pi,$
$|\sin\text{x}|=\begin{cases}-\sin\text{x}\ (2\text{m}-1)\pi<\text{x}<2\text{mx},&\text{where m}\in\text{Z}\\\sin\text{x}\ 2\text{mx}<\text{x}<(2\text{m}+1)\pi,&\text{where m}\in\text{Z}\\-\sin\text{x}\ (2\text{m}+1)\pi<\text{x}<2(\text{m}+1)\pi,&\text{where m}\in\text{Z}\end{cases}$
$(\text{LHL at x}=2\text{mx})=\lim_\limits{\text{x}\rightarrow2\text{mx}^{-}}\frac{\text{f(x)}-\text{f}(2\text{mx})}{\text{x}-2\text{mx}}$
$=\lim_\limits{\text{x}\rightarrow2\text{mx}^{-}}\frac{-\sin(\text{x}-0)}{\text{x}-2\text{mx}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{-\sin(2\text{mx}-\text{h})}{2\text{mx}-\text{h}-2\text{mx}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{h})}{\text{h}}=-1$
$(\text{RHL at x}=2\text{mx})=\lim_\limits{\text{x}\rightarrow2\text{mx}^{+}}\frac{\text{f(x)}-\text{f}(2\text{mx})}{\text{x}-2\text{mx}}$
$=\lim_\limits{\text{x}\rightarrow2\text{mx}^{+}}\frac{\sin(\text{x)}-0}{\text{x}-2\text{mx}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(2\text{mx}+\text{h})}{2\text{mx}+\text{h}-2\text{mx}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{h})}{\text{h}}=1$
Here, $\text{LHL}\neq\text{RHL}$ So, function is not differentiable at $\text{x}=2\text{m}\pi,$ where, $\text{m}\in\text{Z}\ \dots(1)$
$[\text{LHL at x}=(2\text{m}+1)\pi]=\lim_\limits{\text{x}\rightarrow(2\text{m}+1)\pi^{-}}\frac{\text{f(x)}-\text{f}[(2\text{m}+1)\pi]}{\text{x}-(2\text{m}+1)\pi}$
$=\lim_\limits{\text{x}\rightarrow(2\text{m}+1)\pi^{-}}\frac{\sin(\text{x})-0}{\text{x}-(2\text{m}+1)\pi}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{\sin[(2\text{m}+1)\pi-\text{h}]}{(2\text{m}+1)\pi-\text{h}-(2\text{m}+1)\pi}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{\sin(\text{h})}{\text{h}}=-1$
$[\text{RHL at x}=(2\text{m}+1)\pi]=\lim_\limits{\text{x}\rightarrow(2\text{m}+1)\pi^{+}}\frac{\text{f(x)}-\text{f}[(2\text{m}+1)\pi]}{\text{x}-(2\text{m}+1)\pi}$
$=\lim_\limits{\text{x}\rightarrow(2\text{m}+1)\pi^{+}}\frac{-\sin(\text{x})-0}{\text{x}-(2\text{m}+1)\pi}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{-\sin[(2\text{m}+1)\pi+\text{h}]}{(2\text{m}+1)\pi+\text{h}-(2\text{m}+1)\pi}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{\sin(\text{h})}{\text{h}}=1$
Here, $\text{LHL}\neq\text{RHL}.$
So, function is not differentiable at $\text{x}=(2\text{m}+1)\pi,$ where, $\text{m}\in\text{Z}\ \dots(2)$
From, (1) and (2), we get
$\text{f(x)}=|\sin\text{x}|$ is not differentiable at $\text{x}=\text{n}\pi$
We know that,
$\cos|\text{x}|=\cos\text{x}$ For all $\text{x}\in\text{R}$
Also we know that cos x is differentiable at all real points.
Therefore, cos |x| is differentiable everywhere.
View full question & answer→Question 85 Marks
Show that the function $\text{f(x)}=\begin{cases}|2\text{x}-3||\text{x}|, & \text{x}\geq1\\\sin\Big(\frac{\pi\text{x}}{2}\Big),& \text{x}>1\end{cases}$ is continuous but not differentiable at x = 1.
AnswerGiven: $\text{f(x)}=\begin{cases}|2\text{x}-3||\text{x}|, & \text{x}\geq1\\\sin\Big(\frac{\pi\text{x}}{2}\Big),& \text{x}>1\end{cases}$
Continuity at x = 1:
(LHL at x = 1) = $\lim_\limits{\text{x}\rightarrow1^{-}}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(1-\text{h})=\lim_\limits{\text{h}\rightarrow0}\sin\Big(\frac{\pi(1-\text{h})}{2}\Big)=\sin\frac{\pi}{2}=1$
(RHL at x = 1) $=\lim_\limits{\text{x}\rightarrow1^{+}}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(1+\text{h})$
$\Rightarrow\ \lim_\limits{\text{h}\rightarrow0}|2(1+\text{h})-3|[1+\text{h}]=\lim_\limits{\text{h}\rightarrow0}|2(1+\text{h})-3|=1$
Hence, (LHL at x = 1) = (RHL at x = 1)
Differentiable at x = 1:
(LHL at x = 1) $=\lim_\limits{\text{x}\rightarrow1^{-}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
(LHL at x = 1) $=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(1-\text{h})-\text{f}(1)}{1-\text{h}-1}$
(LHL at x = 1) $=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(1-\text{h})-\text{f}(1)}{-\text{h}}$
(LHL at x = 1) $=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\Big(\frac{\pi(1-\text{h})}{2}\Big)-1}{-\text{h}}$
(LHL at x = 1) $=\lim_\limits{\text{h}\rightarrow0}\frac{\cos\frac{\pi\text{h}}{2}-1}{-\text{h}}$
(LHL at x = 1) $=-\frac{\pi}{2}\lim_\limits{\text{h}\rightarrow0}\frac{\cos\frac{\pi\text{h}}{2}-1}{\frac{\pi}{2}\text{h}}=0$
(RHL at x = 1) $=\lim_\limits{\text{x}\rightarrow1^{+}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
(RHL at x = 1) $=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(1+\text{h})-\text{f}(1)}{1+\text{h}-1}$
(RHL at x = 1) $=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(1+\text{h})-\text{f}(1)}{\text{h}}$
(RHL at x = 1) $=\lim_\limits{\text{h}\rightarrow0}\frac{-(2(1+\text{h})-3)-1}{\text{h}}$
(RHL at x = 1) $=\lim_\limits{\text{h}\rightarrow0}\frac{-2\text{h}}{\text{h}}=-2$
$\text{LHL}\neq\text{RHL}$
Hence, the function is continuous but not differentiable at x = 1.
View full question & answer→Question 95 Marks
Discuss the continuity and differentiability of the f(x) = |x| + |x - 1| in the interval (-1, 2).
Answerf(x) = |x| + |x - 1| in the interval (-1, 2).
$\text{f(x)}=\begin{cases}\text{x}+\text{x}+1 & -1<\text{x}<0\\1 & 0\leq\text{x}\leq1\\-\text{x}-\text{x}+1&1<\text{x}<2\end{cases}$
$\text{f(x)}=\begin{cases}2\text{x}+1 & -1<\text{x}<0\\1 & 0\leq\text{x}\leq1\\-2\text{x}+1&1<\text{x}<2\end{cases}$
We know that a polynomial and a constant function is continuous and differentiable everywhere.
So, f(x) is continuous and differentiable for $\text{x}\in(-1,0),\text{x}\in(0,1)$ and (1, 2).
We need to check continuitly and differentiability at x = 0 and x = 1
Continuity at x = 0
$\lim_\limits{\text{x}\rightarrow0^{-}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0^{-}}2\text{x}+1=1$
$\lim_\limits{\text{x}\rightarrow0^{+}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0^{+}}1=1$
$\text{f(0)}=1$
$\lim_\limits{\text{x}\rightarrow0^{+}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0^{-}}\text{f(x)}=\text{f(x)}$
$\therefore$ f(x) is continuous at x = 0.
Continuity at x = 1
$\lim_\limits{\text{x}\rightarrow1^{-}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^{-}}1=1$
$\lim_\limits{\text{x}\rightarrow1^{+}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^{+}}1=1$
$\text{f(x)}=1$
$\lim_\limits{\text{x}\rightarrow1^{-}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^{+}}\text{f(x)}=\text{f}(1)$
$\therefore$ f(x) is continuous at x = 1.
View full question & answer→Question 105 Marks
Show that $\text{f(x)}=\begin{cases}12\text{x}-13, & \text{if x}\leq3\\2\text{x}^2+5, & \text{if x} > 3\end{cases}$ is differentiable at x = 3. Also, find f(3).
AnswerGiven: $\text{f(x)}=\begin{cases}12\text{x}-13, & \text{if x}\leq3\\2\text{x}^2+5, & \text{if x} > 3\end{cases}$
We have to show that the given function is differentiable at x = 3.
We have,
(LHL at x = 3) $=\lim_\limits{\text{x}\rightarrow3^{-}}\frac{\text{f(x)}-\text{f}(3)}{\text{x}-3}$
$=\lim_\limits{\text{x}\rightarrow3}\frac{12\text{(x)}-13-23}{\text{x}-3}$
$=\lim_\limits{\text{x}\rightarrow3}\frac{12\text{x}-36}{\text{x}-3}$
$=\lim_\limits{\text{x}\rightarrow3}\frac{12(\text{x}-3)}{\text{x}-3}$
$=\lim_\limits{\text{x}\rightarrow3}12$
$= 12$
(RHL at x = 3) $=\lim_\limits{\text{x}\rightarrow3^{+}}\frac{\text{f(x)}-\text{f}(3)}{\text{x}-3}$
$=\lim_\limits{\text{x}\rightarrow3}\frac{2\text{x}^2+5-23}{\text{x}-3}$
$=\lim_\limits{\text{x}\rightarrow3}\frac{2\text{x}^2-18}{\text{x}-3}$
$=\lim_\limits{\text{x}\rightarrow3}\frac{2(\text{x}^2-9)}{\text{x}-3}$
$=\lim_\limits{\text{x}\rightarrow3}2(\text{x}+3)$
$=2\times6$
$=12$
Thus, (LHL at x = 3) = (RHL at x = 3) = 12.
So, f(x) is differentiable at x = 3 and f(3) = 12.
View full question & answer→Question 115 Marks
Show that the function $\text{f(x)}\begin{cases}\text{x}^\text{m}\sin(\frac{1}{\text{x}}), &\text{x}\neq0 \\0 ,& \text{x}=0\end{cases}$
Neirher continuous but not diffierentiable, if $\text{m}\leq0$
Answer$\text{LHL }=\lim_\limits{\text{x}\rightarrow0^{-}}\text{f(x)}$ $=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$ $=\lim_\limits{\text{h}\rightarrow0}(-\text{h})^\text{m}\sin\Big(-\frac{1}{\text{h}}\Big)$ = Not defined as $\text{m}\leq0$ $\text{RHL }=\lim_\limits{\text{x}\rightarrow0^{+}}\text{f(x)}$ $=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})$ $=\lim_\limits{\text{h}\rightarrow0}(+\text{h})^\text{m}\sin\Big(\frac{1}{0+\text{h}}\Big)$ =Not defined, as $\text{m}\leq0$ Since RHL and LHL are not difined, so f(x) is not continuous Let x = 0 for $\text{m}\leq0.$ (LHL at x = 0) $=\lim_\limits{\text{x}\rightarrow0^{-1}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$ $=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(0-\text{h})-(0)}{0-\text{h}-0}$ $=\lim_\limits{\text{h}\rightarrow0}\frac{(-\text{h})^\text{m}\sin\Big(-\frac{1}{\text{h}}\Big)}{-\text{h}}$ $=\lim_\limits{\text{h}\rightarrow0}(-\text{h})^\text{m-1}\sin\Big(-\frac{1}{\text{h}}\Big)$ = Not definded, as $\text{m}\leq0$ (RHL at x = 0) $=\lim_\limits{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$ $=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(0+\text{h})-\text{f}(0)}{0+\text{h}-0}$ $=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}^\text{m}\sin\Big(\frac{1}{\text{h}}\Big)}{\text{h}}$ $=\lim_\limits{\text{h}\rightarrow0}(\text{h})^{\text{m}^{-1}}\sin\Big(\frac{1}{\text{h}}\Big)$ = Not defined $\text{m}\leq0$ Thus,f(x) is neither continuous not differentiable at x = 0 for $\text{m}\leq0.$
View full question & answer→Question 125 Marks
Show that the function f defined as follows,
$\text{f(x)}=\begin{cases}3\text{x}-2, & 0<\text{x}\leq1\\2\text{x}^2-\text{x,} & 1<\text{x}\leq2\\5\text{x}-4,&\text{x}>2\end{cases}$
is countinuous at x = 2, but not differentiable there at x = 2.
AnswerGiven:
$\text{f(x)}=\begin{cases}3\text{x}-2, & 0<\text{x}\leq1\\2\text{x}^2-\text{x,} & 1<\text{x}\leq2\\5\text{x}-4,&\text{x}>2\end{cases}$
First, we will show that f(x) is continuos at x = 2.
We have,
(LHL at x = 2)
$=\lim_\limits{\text{x}\rightarrow2^{-}}\text{f(x)}$
$=\lim_\limits{\text{x}\rightarrow0}\text{f}(2-\text{h)}$
$=\lim_\limits{\text{x}\rightarrow0}2(2-\text{h)}^2-(2-\text{h})$
$=\lim_\limits{\text{x}\rightarrow0}(8+2\text{h}^2-8\text{h}-2+\text{h})$
$=6$
(RHL at x = 2)
$=\lim_\limits{\text{x}\rightarrow2^{+}}\text{f(x)}$
$=\lim_\limits{\text{x}\rightarrow0}\text{f}(2+\text{h)}$
$=\lim_\limits{\text{x}\rightarrow0}5(2+\text{h)}-4$
$=\lim_\limits{\text{x}\rightarrow0}(10+5\text{h}-4)$
$=6$
and $\text{f}(2)=2\times4-2=6$
Thus, $=\lim_\limits{\text{x}\rightarrow2^{-}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow2^{+}}\text{f(x)}=\text{f}(2)$
Hence the function is continuous at x = 2.
Now, we will check whether the given function is differerentiable at x = 2.
We have,
(LHL at x = 2)
$\lim_\limits{\text{x}\rightarrow2^{-}}\frac{\text{f(x)}-\text{f}(2)}{\text{x}-2}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{\text{f}(2-\text{h})-\text{f}(2)}{-\text{h}}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{2\text{h}^2-7\text{h}+6-6}{-\text{h}}$
$=\lim_\limits{\text{x}\rightarrow0}-2\text{h}+7$
$=7$
(RHL at x = 2)
$\lim_\limits{\text{x}\rightarrow2^{+}}\frac{\text{f(x)}-\text{f}(2)}{\text{x}-2}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{\text{f}(2+\text{h})-\text{f}(2)}{\text{h}}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{10+5\text{h}-4-6}{\text{h}}$
$=5$
Thus, LHL at x = 2 $\neq$ RHL at x = 2.
Hence, function is no differentiable at x = 2.
View full question & answer→Question 135 Marks
If for function $\phi(\text{x})=\lambda\text{x}^2+7\text{x}-4, \phi(5)=97,$ find $\lambda.$
AnswerGiven: $\phi(\text{x})=\lambda\text{x}^2+7\text{x}-4$
Clearly, being a polynomial function, is differentiable everywhere. Therefore the derivative of $\phi$ at x is given by:
$\phi'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\phi(\text{x}+\text{h})-\phi(\text{x})}{\text{h}}$
$\Rightarrow\phi'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\lambda(\text{x}+\text{h})^2+7(\text{x}+\text{h})-4-\lambda\text{x}^2-7\text{x}+4}{\text{h}}$
$\Rightarrow\phi'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\lambda(\text{x}+\text{h})^2+7(\text{x}+\text{h})-4-\lambda\text{x}^2-7\text{x}+4}{\text{h}}$
$\Rightarrow\phi'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\lambda\text{h}^2+2\lambda\text{xh}+7\text{h}}{\text{h}}$
$\Rightarrow\phi'(\text{x})=\lim_\limits{\text{h}\rightarrow\infty}\frac{\text{h}(\lambda\text{h}+2\lambda\text{x}+7)}{\text{h}}$
$\Rightarrow\phi'(\text{x})=2\lambda\text{x}+7$
It is given $\phi'(5)=97$
Thus, $\phi'(5)=10\lambda+7=97$
$\Rightarrow10\lambda+7=97$
$\Rightarrow10\lambda=90$
$\Rightarrow\lambda=9$
View full question & answer→Question 145 Marks
Find whether the function is differentiable at x = 1 and x = 2
$\text{f(x)}=\begin{cases}\text{x} & \text{x}\leq1\\2-\text{x} & 1\leq\text{x}\leq2\\-2+3\text{x}&\text{x}>2\end{cases}$
Answer$\text{f(x)}=\begin{cases}\text{x} & \text{x}\leq1\\2-\text{x} & 1\leq\text{x}\leq2\\-2+3\text{x}&\text{x}>2\end{cases}$
$\text{f(x)}=\begin{cases}\text{x} & \text{x}\leq1\\-1 & 1\leq\text{x}\leq2\\3-2\text{x}&\text{x}>2\end{cases}$
Now,
LHL $=\lim_\limits{\text{x}\rightarrow1^{-}}\text{f}'(\text{x})=\lim_\limits{\text{x}\rightarrow1^{-}}1=1$
RHL $=\lim_\limits{\text{x}\rightarrow1^{+}}\text{f}'(\text{x})=\lim_\limits{\text{x}\rightarrow1^{+}}-1=-1$
Since, at $\text{x}=1,\text{LHL}\neq\text{RHL}$
Hence, f(x) is not differentiable at x = 1
Again,
LHL $=\lim_\limits{\text{x}\rightarrow2^{-}}\text{f}'(\text{x})=\lim_\limits{\text{x}\rightarrow2^{-}}-1=-1$
RHL $=\lim_\limits{\text{x}\rightarrow2^{+}}\text{f}'(\text{x})=\lim_\limits{\text{x}\rightarrow2^{+}}3-2\text{x}=3-4=-1$
Since, at x = 2, LHL = RHL
Hence, f(x) is differentiable at x = 2.
View full question & answer→Question 155 Marks
If $f(x) = x^3 + 7x^2 + 8x - 9,$ find $f(4).$
Answer$f(x) = x^3 + 7x^2 + 8x - 9$ is a polynomial function. So, it is differentiable every.
$\text{f}'(4)=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(4+\text{h})-\text{h}(4)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{[(4+\text{h})^3+7(4+\text{h})^2+8(4+\text{h})-9]-[64+112+32-9]}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{[64+\text{h}^3+48\text{h}+12\text{h}^2+112+7\text{h}^2+56\text{h}+32+8\text{h}-9]-[210-9]}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}^3+19\text{h}^2+112\text{h}+210-9-210+9}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}^3+19\text{h}^2+112\text{h}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}(\text{h}^2+19\text{h}+112)}{\text{h}}$
$\text{f}'(4)=112$
View full question & answer→Question 165 Marks
Show that the function $\text{f(x)}\begin{cases}\text{x}^\text{m}\sin(\frac{1}{\text{x}}), &\text{x}\neq0 \\0 ,& \text{x}=0\end{cases}$
Differential at x = 0, if m > 1
AnswerLet m = 2, then the function $\text{f(x)}=\begin{Bmatrix}\text{x}^2\sin(\frac{1}{\text{x}}) &\text{x}\neq0 \\0 & \text{x}=0 \end{Bmatrix}$
Differentiability at x = 0:
$\lim_\limits{\text{x}\rightarrow0}\frac{\text{f(x)}-\text{f(0)}}{\text{x}-0}=\lim_\limits{\text{x}\rightarrow0}\frac{\text{f(x)}}{\text{x}}=\lim_\limits{\text{x}\rightarrow0}\text{x}\sin\Big(\frac{1}{\text{x}}\Big)=0.$
$\big[\therefore\lim_\limits{\text{x}\rightarrow0}\text{x}\sin\Big(\frac{1}{\text{x}}\Big)\text{x}=0,$ as
$\begin{vmatrix}\text{x}\sin\frac{1}{\text{x}}-0\end{vmatrix}=\begin{vmatrix}\text{x}\sin\frac{1}{\text{x}} \end{vmatrix}=\begin{vmatrix}\text{x} \end{vmatrix}\begin{vmatrix}\sin\frac{1}{\text{x}} \end{vmatrix} \leq\begin{vmatrix}\text{x}\end{vmatrix}$
$\because\ |\sin\theta|\leq1\ \text{for all }\theta$
Hence, $\begin{vmatrix}\text{x}\sin\frac{1}{\text{x}} \end{vmatrix}<0\ \text{when}\ |\text{x-0}|<\epsilon|\text{x}-0|<\epsilon\big]$
Therefore, f'(x) = 0, which means f is differentiable at x = 0.
Hence the given function is differentiable at x = 0.
View full question & answer→Question 175 Marks
Write the points where $f(x) = |\log_e x|$ is not differentiable.
AnswerGiven: $\text{f(x)}=|\log_\text{e}\text{x}|=\begin{cases}-\log_\text{e}\text{x}, & 0<\text{x}<1\\\log_\text{e}\text{x}, & \text{x}\geq1\end{cases}$
Clearly $f(x)$ is differentiable for all $x > 1$ and for all $x < 1$.
So, we have to check the differentiability at $x = 1$.
$(\text{LHL}$ at $x = 1)$
$\lim_\limits{\text{x}\rightarrow1^{-}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
$=\lim_\limits{\text{x}\rightarrow1^{-}}\frac{-\log\text{x}-\log1}{\text{x}-1}$
$=-\lim_\limits{\text{x}\rightarrow1^{-}}\frac{\log\text{x}}{\text{x}-1}$
$=-\lim_\limits{\text{h}\rightarrow0^{-}}\frac{\log(1-\text{h})}{1-\text{h}-1}$
$=-\lim_\limits{\text{h}\rightarrow0^{-}}\frac{\log(1-\text{h})}{-\text{h}}$
$=-1$
$(\text{RHL}$ at $x = 1)$
$\lim_\limits{\text{x}\rightarrow1^{+}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
$=\lim_\limits{\text{x}\rightarrow1^{+}}\frac{-\log\text{x}-\log1}{\text{x}-1}$
$=\lim_\limits{\text{x}\rightarrow1^{+}}\frac{\log\text{x}}{\text{x}-1}$
$=-\lim_\limits{\text{h}\rightarrow0^{+}}\frac{\log(1+\text{h})}{1+\text{h}-1}$
$=\lim_\limits{\text{h}\rightarrow0^{-}}\frac{\log(1+\text{h})}{\text{h}}$
$=1$
Thus$, (\text{LHL}$ at $x = 1) \neq (\text{RHL}$ at $x = 1)$
So$, f(x)$ is not differentiable at $x = 1$.
View full question & answer→Question 185 Marks
If $\lim\limits_{\text{x}\rightarrow{\text{c}}}\frac{\text{f(x)}-\text{f(c)}}{\text{x}-\text{c}}$ exists finitely, write the value of $\lim\limits_{\text{x}\rightarrow{\text{c}}}\text{f(x)}.$
AnswerLHL = f(1) = RHL
So, f(x) is continuous at x = 1
Now,
(LHL at x = 1) $=\lim\limits_{\text{x}\rightarrow1^{-}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1-\text{h})-\text{f}-(1)}{(1-\text{h})-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{1-1}{-\text{h}}$
= Not defined
(RHL at x = 1) $=\lim\limits_{\text{x}\rightarrow1^{+}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1+\text{h})-\text{f}-(1)}{(1+\text{h})-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{2(1+\text{h})-1-1}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{2\text{h}}{\text{h}}$
$=1$
(LHL at x = 1) $\neq$ (RHL at x = 1)
$\therefore$ f(x) is continuous but not differentiable at x = 0 and 1.
$\lim\limits_{\text{x}\rightarrow{\text{c}}}\frac{\text{f(x)}-\text{f(c)}}{\text{x}-\text{c}}$ exists finitely
So,
$\text{f}'(\text{c})=\lim\limits_{\text{x}\rightarrow{\text{c}}}\frac{\text{f(x)}-\text{f(c)}}{\text{x}-\text{c}}$
$\text{f}'(\text{c})\lim\limits_{\text{x}\rightarrow{\text{c}}}(\text{x}-\text{c})=\lim\limits_{\text{x}\rightarrow{\text{c}}}(\text{x})-\text{f(c)}$
$\text{f}'(\text{c})(\text{c}-\text{c})=\lim\limits_{\text{x}\rightarrow{\text{c}}}(\text{x})-\text{f(c)}$
$0=\lim\limits_{\text{x}\rightarrow{\text{c}}}(\text{x})-\text{f(c)}$
$\lim\limits_{\text{x}\rightarrow{\text{c}}}\text{f(x})=\text{f(c)}$
View full question & answer→Question 195 Marks
If $\text{f(x)}=\sqrt{\text{x}^2+9},$ Write the value of $\lim\limits_{\text{x}\rightarrow4}\frac{\text{f(x)}-\text{f}(4)}{\text{x}-4}.$
AnswerGiven: $\text{f(x)}=\text{x}^2+9$
Now,
$\text{f}(4)=\sqrt{16+9}$
$=\sqrt{25}$
$=5$
So,
$\frac{\text{f(x)}-\text{f}(4)}{\text{x}-4}=\frac{\sqrt{\text{x}^2+9-5}}{\text{x}-4}$
On rationalising the numeratore, we get
$\frac{\text{f(x)}-\text{f}(4)}{\text{x}-4}$
$=\frac{\sqrt{\text{x}^2+9-5}}{\text{x}-4}\times\frac{\sqrt{\text{x}^2+9}+5}{\sqrt{\text{x}^2+9}+5}$
$=\frac{\text{x}^2+9-25}{(\text{x}-4)(\sqrt{\text{x}^2+9+5})}$
$=\frac{\text{x}^2-16}{(\text{x}-4)(\sqrt{\text{x}^2+9+5})}$
$=\frac{\text{x}+4}{\sqrt{\text{x}^2+9+5}}$
Taking limit x → 4, we have
$\lim\limits_{\text{x}\rightarrow4}\frac{\text{f(x)}-\text{f}(4)}{\text{x}-4}=\lim\limits_{\text{x}\rightarrow4}\frac{\text{x}+4}{\sqrt{\text{x}^2+9+5}}$
$=\frac{8}{10}$
$=\frac{4}{5}$
View full question & answer→Question 205 Marks
Write the number of points where f(x) = |x| + |x − 1| is continuous but not differentiable.
AnswerGiven:
f(x) = |x| + |x - 1|
$\Rightarrow\text{f(x)}=\begin{cases}-\text{x}-(\text{x}-1),&\text{x}<0\\\text{x}-(\text{x}-1),&0\leq\text{x}<1\\\text{x}+(\text{x}-1),&\text{x}\geq1\end{cases}$
$\Rightarrow\text{f(x)}=\begin{cases}-2\text{x}+1,&\text{x}<0\\1,&0\leq\text{x}<12\\\text{x}-1,&\text{x}\geq1\end{cases}$
When x < 0, we have:
f(x) = -2x + 1 which, being a polynomial function is continuous and differentiable.
When $0\leq\text{x}<1,$ we have:
f(x) = 1 which, being a polynimial function is continuous and differentiable on (0, 1).
When $\text{x}\leq1,$ we have:
f(x) = 2x - 1 which, being a polynimial function is continuous and differentiable on x > 2.
Thus, the possible points of non differentiability of f(x) are 0 and 1.
Now,
(LHL at x = 0)
$\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$\lim\limits_{\text{x}\rightarrow0^{-}}\frac{2\text{x}+1-1}{\text{x}-0}\ [\because\text{f(x)}=-2\text{x}+1,\text{x}<0]$
$\lim\limits_{\text{x}\rightarrow0}\frac{-2\text{x}}{\text{x}}=-2$
(RHL at x = 0)
$=\lim\limits_{\text{x}\rightarrow0^{+}}\text{f(x)}-\text{f}(0)\text{x}-0$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{1-1}{\text{x}-1}$
$=0\ [\because\text{f(x)}=1,0\leq\text{x}<1]$
Thus, (LHL at x = 1) $\neq$ (RHL at x = 1)
Hence f(x) is not differentiable at x = 1.
Therefore, 0, 1 are the points where f(x) is continuous but not differentiable.
View full question & answer→Question 215 Marks
Show that f(x) = |x − 2| is continuous but not differentiable at x = 2.
AnswerGiven: $\text{f}(\text{x})=|\text{x}-3|=\begin{vmatrix}\text{x}-3,&\text{x}\geq3 \\-\text{x}+3, & \text{x}<3 \end{vmatrix}$
Continuity at X = 2: We have,
(LHL at x = 3)
$=\lim_\limits{\text{x} \rightarrow 2}\text{f}(\text{x})$
$=\lim_\limits{\text{x} \rightarrow 0}\text{f}(3-\text{h})$
$=\lim_\limits{\text{x} \rightarrow 0}0(-3+\text{h})+3$
$=0.$
(RHL at x = 3)
$=\lim_\limits{\text{x} \rightarrow 2^{-}}\text{f}(\text{x})$
$=\lim_\limits{\text{x} \rightarrow 0}\text{f}(3+\text{h})$
$=\lim\limits_{\text{x} \rightarrow 0}3+\text{h}-3$
$=0.$
and f(2) = 0
Thus, $\lim_\limits{\text{x} \rightarrow 2^{-}}\text{f}(\text{x})=\text{f}(2).$
Hence,
f(x) is continuous at x = 3
We have,
(LHL at x = 3)
$=\lim_\limits{\text{x} \rightarrow 2^{-}}-\frac{\text{f}(\text{x})-\text{f}(2)}{\text{x}-3}$
$=\lim_\limits{\text{x} \rightarrow 2}\frac{(-\text{x}+3)-0}{\text{x}-3}$
$=\lim_\limits{\text{x} \rightarrow 2}\text{f}(\text{x}-3)\text{x}-2$
$\lim_\limits{\text{x} \rightarrow 2}(-1)$
$=-1$
(RHL at x = 3)
$=\lim_\limits{\text{x} \rightarrow 2^{-}}\frac{\text{f}(\text{x})-\text{f}(2)}{\text{x}-3}$
$=\lim_\limits{\text{x} \rightarrow 2}\frac{(\text{x}-3)-0}{\text{x}-3}$
$=\lim_\limits{\text{x}\rightarrow2}1=1$
Thus, $=\lim_\limits{\text{x}\rightarrow2^{-}}\text{f}(\text{x})\neq\lim_\limits{\text{x}\rightarrow2^{-}}\text{f}(\text{x}).$
Hence, f(x) is not differentiable at x = 3.
View full question & answer→Question 225 Marks
Show that the derivative of the function $f $ given by $f(x) = 2x^3 - 9x^2 + 12x + 9,$ at $x = 1$ and $x = 2$ are equal.
AnswerGiven: $f(x) = 2x^3 - 9x^2 + 12x + 9$
Clearly, being a polynomial function, is differentiable everywhere.
Therefore the derivative of $f$ at $x$ is given by:
$\text{f}\ '(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f(x)}}{\text{h}}$
$\Rightarrow\text{f}\ '(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{2(\text{x}+\text{h})^3-9(\text{x}+\text{h})^2+12(\text{x}+\text{h})+9-2\text{x}^3+9\text{x}^2-12\text{x}-9}{\text{h}}$
$\Rightarrow\text{f}\ '(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{2\text{x}^3+2\text{h}^3+6\text{x}^2\text{h}+6\text{xh}^2-9\text{x}^2-9\text{h}^2-18\text{xh}+12\text{x}+12\text{h}+9-2\text{x}^3+9\text{x}^2-12\text{x}-9}{\text{h}}$
$\Rightarrow\text{f}\ '(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{2\text{h}^3+6\text{x}^2\text{h}+6\text{xh}^2-9\text{h}^2-18\text{xh}+12\text{h}}{\text{h}}$
$\Rightarrow\text{f}\ '(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}(\text{h}^2+6\text{x}^2+6\text{xh}-9\text{h}-18\text{x}+12)}{\text{h}}$
$=6\text{x}^2-18\text{x}+12$
So,
$\text{f}\ '(1)=6(\text{x}^2-3\text{x}+2)$
$=6\times(1-3+2)$
$=0$
$\text{f}\ '(2)=6(\text{x}^2-3\text{x}+2)$
$=6\times(4-6+2)$
$=0$
Hence, the derivative at $x = 1$ and $x = 2$ are equal.
View full question & answer→Question 235 Marks
If $f$ is defined by $\text{f(x)}=\text{x}^2-4\text{x}+7,$ show that $\text{f}\ '(5)=2\text{f}\ '\Big(\frac{7}{2}\Big).$
Answer$f(x) = x^2 - 4x + 7$ is a polynomial function, So it is differentiable everywhere.
$\text{f}\ '(5)=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(5+\text{h})-\text{f}(5)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\big\{(5+\text{h})^2-4(5+\text{h})+7\big\}-[25-20+7]}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}^2+25+10\text{h}-20-4\text{h}+7-12}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}^2+6\text{h}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}(\text{h}+6)$
$=6$
$\text{f}\ '\Big(\frac{7}{2}\Big)=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}\Big(\frac{7}{2}+\text{h}\Big)^2-\text{f}\Big(\frac{7}{2}\Big)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\Big[\Big(\frac{7}{2}+\text{h}\Big)^2-4\Big(\frac{7}{2}+\text{h}\Big)+7\Big]-\Big[\Big(\frac{7}{2}\Big)^2-4\Big(\frac{7}{2}\Big)+7\Big]}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\Big[\frac{49}{2}+\text{h}^2+7\text{h}-14-4\text{h}+7\Big]-\Big[\frac{49}{2}-14+7\Big]}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\frac{49}{2}+\text{h}^2+7\text{h}-14-4\text{h}+7-\frac{49}{2}-14+7}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}^2+3\text{h}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}(\text{h}+3)$
$=3$
$\text{f}\ '(5)=6$
$=2(3)$
$=\text{f}\ '(5)$
$=2\text{f}\ '\Big(\frac{7}{2}\Big)$
View full question & answer→Question 245 Marks
Show that $\text{f}(\text{x})=\text{x}^\frac{1}{3}$ is not differentible at x = 0.
Answer$\text{f}(\text{x})=\text{x}^\frac{1}{3}$
(LHL at x = 0) $=\lim_\limits{\text{x}\rightarrow0^{-}}\frac{\text{f}(\text{x})-\text{f}(0)}{\text{x}-0}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{\text{f}(0-\text{h})-\text{f}(0)}{0-\text{h}-0}$
$=\lim_\limits{\text{x}\rightarrow0^{-}}\frac{\text{f}(\text{x})-\text{f}(0)}{\text{x}-0}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{(-\text{h})^\frac{1}{3}}{-\text{h}}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{(-1)^\frac{1}{3}\text{h}^\frac{1}{3}}{(-1)-\text{h}}$
$=\lim_\limits{\text{x}\rightarrow0}(-1)^\frac{-2}{3}\text{h}^\frac{-2}{3}$
= Not defined
(RHL at x = 0) $=\lim_\limits{\text{x}\rightarrow0^{+}}\frac{\text{f}(\text{x})-\text{f}(0)}{\text{x}-0}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{\text{f}(0+\text{h})-\text{f}(0)}{0+\text{h}-0}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{(\text{h})^\frac{1}{3}-0}{\text{h}}$
$=\lim_\limits{\text{x}\rightarrow0}\text{h}^\frac{-2}{3}$
= Not defined
Since,
LHL and RHL does not exists at x = 0
$\therefore$ f(x) is not differentiable at x = 0
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