3 Marks

Question 13 Marks

Write the value of the derivative of f(x) = |x − 1| + |x − 3| at x = 2.

Answer

Given: f(x) = |x - 1| + |x - 3|
$\Rightarrow\text{f(x)}=\begin{cases}-(\text{x}-1)-(\text{x}-3), & \text{x}<1\\ \text{x}-1-(\text{x}-3),& 1\leq\text{x}<3\$\text{x}-1)+(\text{x}-3),&\text{x}\geq3\end{cases}$
$\Rightarrow\text{f(x)}=\begin{cases}-2\text{x}+4, & \text{x}<1\\ 2,& 1\leq\text{x}<3\\2\text{x}-4,&\text{x}\geq3\end{cases}$
Wecheck differentiable at x = 2
(LHL at x = 2)
$\lim\limits_{\text{x}\rightarrow2^{-}}\frac{\text{f(x)}-\text{f}(2)}{\text{x}-2}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(2-\text{h})-\text{f}(2)}{2-\text{h}-2}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{2-2}{-\text{h}}$
$=0$

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Question 23 Marks

Give an example of a function which is continuos but not differentiable at at a point.

Answer

Consider a function, $\text{f(x)}=\begin{cases}\text{x}, & \text{x}> 0\\-\text{x}, & \text{x}\leq 0\end{cases}$
This mod function is continuous at x = 0 but not differentiable at x = 0.
Continuity at x - 0, We have:
(LHL at x = 0)
$\lim_\limits{\text{x}\rightarrow0^{-}}\text{f(x)}$
$=\lim_\limits{\text{x}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{x}\rightarrow0}-(0-\text{h})$
$=0$
(RHL at x = 0)
$\lim_\limits{\text{x}\rightarrow0^{+}}\text{f(x)}$
$=\lim_\limits{\text{x}\rightarrow0}\text{f}(0+\text{h})$
$=\lim_\limits{\text{x}\rightarrow0}(0+\text{h})$
$=0$
and f(0) = 0
Thus, $\lim_\limits{\text{x}\rightarrow0^{-}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0^{+}}\text{f(x)}=\text{f}(0).$
Hence, f(x) is continuous at x = 0.
Now, we will check the differentiability at x = 0, we have:
(LHL at x = 0)
$\lim_\limits{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(0-\text{h})-\text{f}(0)}{0-\text{h}-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{-(0-\text{h})-0}{-\text{h}}=-1$
(RHL at x = 0)
$\lim_\limits{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(0+\text{h})-\text{f}(0)}{0+\text{h}-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{(0+\text{h})-0}{-\text{h}}=1$
Thus, $\lim_\limits{\text{h}\rightarrow0^{-}}\text{f(x)}\neq\lim_\limits{\text{h}\rightarrow0^{+}}\text{f(x)}$
Hence f(x) is not differentiable at x = 0.

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Question 33 Marks

Find the derivative of the function f defined by f(x) = mx + c at x = 0.

Answer

Given: f(x) = mx + c
Clearly, being a polynomial function, is differentiable everywhere. Therefore the derivative of f at x is given by:
$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h}-\text{f(x)})}{\text{h}}$
$\Rightarrow\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{m}(\text{x}+\text{h})+\text{c}-\text{mx}-\text{c}}{\text{h}}$
$\Rightarrow\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{mx}+\text{mh}+\text{c}-\text{mx}-\text{c}}{\text{h}}$
$\Rightarrow\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{mh}}{\text{h}}$
$\Rightarrow\text{f}'(\text{x})=\text{m}$
Thus, $\text{f}'(0)=\text{m}$

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Question 43 Marks

If f(x) is defined by f(x) x². find f(2).

Answer

Given: f(x) = x².
We know a polynomial function is everywhere differentiable. Therefore f(x) is differentiable at x = 2.
$\text{f}'(2)=\lim_\limits{\text{k}\rightarrow0}\text{f}\frac{(2+\text{h})-\text{f}(2)}{\text{h}}$
$\Rightarrow\text{f}'(2)=\lim_\limits{\text{k}\rightarrow0}\text{f}\frac{(2+\text{h})2-22}{\text{h}}$
$\Rightarrow\text{f}'(2)=\lim_\limits{\text{k}\rightarrow0}\text{f}\frac{(4+\text{h}2-4\text{h})-4}{\text{h}}$
$\Rightarrow\text{f}'(2)=\lim_\limits{\text{k}\rightarrow0}\text{f}\frac{\text{h}(\text{h}+4)}{\text{h}}$
$\Rightarrow\text{f}'(2)=4$

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Question 53 Marks

Examine the differentialiblilty of the function f defined by $\text{f(x)}=\begin{cases}2\text{x}+3 & \text{if}-3\leq\text{x}\leq-2\\\text{x}+1 & \text{if} -2\leq\text{x}\leq0\\\text{x}+2&\text{if}\ 0\leq\text{x}\leq1\end{cases}$

Answer

$\text{f(x)}=\begin{cases}2\text{x}+3 & \text{if}-3\leq\text{x}\leq-2\\\text{x}+1 & \text{if} -2\leq\text{x}\leq0\\\text{x}+2&\text{if}\ 0\leq\text{x}\leq1\end{cases}$
$\text{f}'(\text{x})=\begin{cases}2 & \text{if}-3\leq\text{x}\leq-2,\\1 & \text{if} -2\leq\text{x}\leq0\\1&\text{if}\ 0\leq\text{x}\leq1\end{cases}$
Now,
$\text{LHL}=\lim_\limits{\text{x}\rightarrow2^{-}}\text{f}'(\text{x})=\lim_\limits{\text{x}\rightarrow2^{-}}2=2$
$\text{RHL}=\lim_\limits{\text{x}\rightarrow2^{+}}\text{f}'(\text{x})=\lim_\limits{\text{x}\rightarrow2^{+}}1=1$
Since, at $\text{x}=-2,\text{LHL}\neq\text{RHL}$
Hence, f(x) is not differentiable at x = -2
Again,
$\text{LHL}=\lim_\limits{\text{x}\rightarrow0^{-}}\text{f}'(\text{x})=\lim_\limits{\text{x}\rightarrow0^{-}}1=1$
$\text{RHL}=\lim_\limits{\text{x}\rightarrow0^{+}}\text{f}'(\text{x})=\lim_\limits{\text{x}\rightarrow0^{+}}1=1$
Since, at $\text{x}=0,$
$\text{LHL}=\text{RHL}$
Hence, f(x) is differentiable at x = 0

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Question 63 Marks

Write the derivative of f(x) = |x|³ at x = 0.

Answer

Given: $\text{f(x)}=|\text{x}^3|=\begin{cases}\text{x}^3,&\text{x}\geq0\\-\text{x}^3,&\text{x}<0\end{cases}$
(LHL at x = 0)
$\lim_\limits{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{\text{f}(0-\text{h})-\text{f}(0)}{\text{x}}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{\text{h}^3}{-\text{h}}$
$=0$
(RHL at x = 0)
$\lim_\limits{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{\text{f}(0+\text{h})-\text{f}(0)}{\text{x}}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{\text{h}^3-0}{-\text{h}}$
$=0$
And f(0) = 0.
Thus, (LHL at x = 0) = (RHL at x = 0) = f(0)
Hence, $\lim_\limits{\text{x}\rightarrow0}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}=\text{f}'(0)=0$.

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Question 73 Marks

If f(x) = |x - 2| write whether f(2) exists or not.

Answer

Given: $\text{f(x)}=|\text{x}-2|=\begin{cases}\text{x}-2, & \text{x}> 2\\-\text{x}+2, & \text{x}\leq 2\end{cases}$
Now,
(LHL at x = 2)
$\lim_\limits{\text{x}\rightarrow2^{-}}\frac{\text{f(x)}-\text{f}(2)}{\text{x}-2}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(2-\text{h})-\text{f}(2)}{2-\text{h}-2}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{(-2+\text{h}+2)-0}{-\text{h}}$
$=-1$
(RHL at x = 2)
$\lim_\limits{\text{x}\rightarrow2^{+}}\frac{\text{f(x)}-\text{f}(2)}{\text{x}-2}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(2+\text{h})-\text{f}(2)}{2+\text{h}-2}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{2+\text{h}+2-0}{\text{h}}$
$=1$
Thus, (LHL at x = 2) $\neq$ (RHL at x = 2)
Hence, $\lim_\limits{\text{x}\rightarrow2}\frac{\text{f(x)}-\text{f}(2)}{\text{x}-2}=\text{f'}(2)$ does not exist.

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Maths 3 Marks

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