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Maths Solve the Following Question.(2 Marks)

Continuity · Maharashtra Board STD 12 (MSBSHSE - Semi English Medium). 12 questions.

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Solve the Following Question.(2 Marks)

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12 questions · self-marked practice — reveal the answer and mark yourself.

Question 12 Marks
If the function $\text{f(x)}=\frac{\sin10\text{x}}{\text{x}},\text{ x}\neq0$ is continuous at x = 0, find f(0).
Answer
Since f(x) is continuous at x = 0, $\text{f}(0)=\lim\limits_{{\text{x}}\rightarrow0}\frac{\sin10\text{x}}{\text{x}}=10$
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Question 22 Marks
In the following, find the value of the constant k so that the given function is continuous at the indicated point:
$\text{f(x)}=\begin{cases}\text{k}(\text{x}^2+2),&\text{if x}\leq0\\3\text{x}+1,&\text{if x}>0\end{cases}$
Answer
$\text{f(x)}=\begin{cases}\text{k}(\text{x}^2+2),&\text{if x}\leq0\\3\text{x}+1,&\text{if x}>0\end{cases}$
We know that a function will be continuous x = 0. if
$\text{LHL}=\text{RHL}=\text{f}(0)\ ....(\text{i})$
$\text{f}(0)=\text{k}(0+2)=2\text{k}$
$\text{LHL}=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}3(\text{h})+1=1$
Thus, using (i) we get,
$2\text{k}=1$
$\text{k}=\frac{1}{2}$
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Question 32 Marks
What happens to a function f(x) at x = a, if $\lim\limits_{{\text{x}}\rightarrow\text{a}}\text{f(x})=\text{f}(\text{a})?$
Answer
If $\lim\limits_{{\text{x}}\rightarrow\text{a}}\text{f(x})=\text{f}(\text{a})$ then the function f(x) is continuse at x = a.
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Question 42 Marks
Prove that $\text{f(x)}=\begin{cases}\frac{\text{x}-|\text{x}|}{\text{x}},&\text{x}\neq0\\2,&\text{x}=0\end{cases}$ is discontinuous at x = 0.
Answer
The given function can be rewritten as
$\text{f(x)}=\begin{cases}\frac{\text{x}-\text{x}}{\text{x}},&\text{when }\text{ x}>0\\\frac{\text{x}+\text{x}}{\text{x}},&\text{when }\text{ x}<0\\2,&\text{when }\text{ x}=0\end{cases}$
$\text{f(x)}=\begin{cases}0,&\text{when }\text{ x}>0\\2,&\text{when }\text{ x}<0\\2,&\text{when }\text{ x}=0\end{cases}$
We have,
$(\text{LHL at x}= 0)=\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(-\text{h})=\lim_\limits{\text{h}\rightarrow0}2=2$
$(\text{RHL at x}= 0)=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(\text{h})=\lim_\limits{\text{h}\rightarrow0}0=0$
$\therefore\lim_\limits{\text{x}\rightarrow0^-}\text{f}(\text{x})\neq\lim_\limits{\text{x}\rightarrow0^+}\text{f}(\text{x})$
Thus, f(x) is discontinuous at x = 0
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Question 52 Marks
If $\text{f(x)}=\begin{cases}\frac{\text{x}}{\sin3\text{x}},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$ is continuous at x = 0, then write the value of k.
Answer
If f(x) is continuous at x = 0, then
$\lim\limits_{{\text{x}}\rightarrow0}\text{f(x})=\text{f}(0)$
$\Rightarrow\lim\limits_{{\text{x}}\rightarrow0}\frac{\text{x}}{\sin3\text{x}}=\text{k}$
$\Rightarrow\lim\limits_{{\text{x}}\rightarrow0}\frac{1}{\frac{\sin3\text{x}}{\text{x}}}=\text{k}$
$\Rightarrow\lim\limits_{{\text{x}}\rightarrow0}\frac{1}{\frac{3\sin3\text{x}}{3\text{x}}}=\text{k}$
$\Rightarrow\frac{1}{3}\Bigg(\frac{1}{\lim\limits_{{\text{x}}\rightarrow0}\frac{3\sin3\text{x}}{3\text{x}}}\Bigg)=\text{k}$
$\Rightarrow\text{k}=\frac{1}{3}$
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Question 62 Marks
Define continuity of a function at a point.
Answer
Continuity at a point:
A function f(x) is said to be continuous at a point x = a of its domain, $\lim\limits_{{\text{x}}\rightarrow\text{a}}\text{f(x)}=\text{f(a)}$
Thus, f(x) is continuous at $\text{x}=\text{a}\Leftrightarrow\lim\limits_{{\text{x}}\rightarrow\text{a}}\text{f(x)}=\text{f}(\text{a})\Leftrightarrow\lim\limits_{{\text{x}}\rightarrow\text{a}^-}\text{f(x)}=\lim\limits_{{\text{x}}\rightarrow\text{a}^+}\text{f(x)}=\text{f}(\text{a})$
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Question 72 Marks
Determine whether $\text{f(x)}=\begin{cases}\frac{\sin\text{x}^2}{\text{x}},&\text{x}\neq0\\0,&\text{x}=0\end{cases}$ is continuous at x = 0 or not.
Answer
Given, $\text{f(x)}=\begin{cases}\frac{\sin\text{x}^2}{\text{x}},&\text{x}\neq0\\0,&\text{x}=0\end{cases}$
We have
$\lim\limits_{{\text{x}}\rightarrow0}\text{f(x})=\lim\limits_{{\text{x}}\rightarrow0}\frac{\sin\text{x}^2}{\text{x}}$
$=\lim\limits_{{\text{x}}\rightarrow0}\frac{\text{x}\sin\text{x}^2}{\text{x}^2}$
$=\lim\limits_{{\text{x}}\rightarrow0}\frac{\sin\text{x}^2}{\text{x}^2}\lim\limits_{{\text{x}}\rightarrow0}\text{x}$
$=1\times0$
$=0$
$=\text{f}(0)$
$\therefore\ \lim\limits_{{\text{x}}\rightarrow0}\text{f(x)}=\text{f}(0)$
Hence, f(x) is continuous at x = 0.
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Question 82 Marks
Discuss the continuity of the following functions at the indicated point:
$\text{f}\text{(x)}=\begin{cases}\text{(x}-\text{a})\sin\Big(\frac{1}{\text{x}-\text{a}}\Big), & \text{x} \neq 0\\\ \ 0, & \text{x} = \text{a}\end{cases}\text{at x}=\text{a}$
Answer
Given,
$\text{f}\text{(x)}=\begin{cases}\text{(x}-\text{a})\sin\Big(\frac{1}{\text{x}-\text{a}}\Big), & \text{x} \neq 0\\\ 0,\text{x} = \text{a}\end{cases}$
Putting x - a = y, we get
$\lim\limits_{\text{x} \rightarrow \text{a}}\text{(x}-\text{a})\sin\Big(\frac{1}{\text{x}-\text{a}}\Big)$
$=\lim\limits_{\text{y} \rightarrow 0}\text{y}\sin\Big(\frac{1}{\text{y}}\Big)$
$=\lim\limits_{\text{y} \rightarrow 0}\text{y}\lim\limits_{\text{y} \rightarrow 0}\sin\Big(\frac{1}{\text{y}}\Big)$
$=0 \times\lim\limits_{\text{y} \rightarrow 0}\sin\Big(\frac{1}{\text{y}}\Big)=0$
$\Rightarrow\lim\limits_{\text{x} \rightarrow \text{a}}\text{f}\text{(x)}=\text{f}\text{(a)}=0$
Hence, f(x) is continuous at x = a.
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Question 92 Marks
For what value of $\lambda$ is the function $\text{f(x)}=\begin{cases}\lambda(\text{x}^2-2\text{x}),&\text{if }\text{ x}\leq0\\4\text{x}+1,&\text{if }\text{ x}>0\end{cases}$ continuous at x = 0? What about continuity at $\text{x}=\pm1?$
Answer
The given function f is $\text{f(x)}=\begin{cases}\lambda(\text{x}^2-2\text{x}),&\text{if }\text{ x}\leq0\\4\text{x}+1,&\text{if }\text{ x}>0\end{cases}$
If f is continuous at x = 0, then
$\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\text{f}(0)$
$\Rightarrow\lim_\limits{\text{x}\rightarrow0^-}\lambda(\text{x}^2-2\text{x})=\lim_\limits{\text{x}\rightarrow0^+}(4\text{x}+1)=\lambda(0^2-2\times0)$
$\Rightarrow\lambda(0^2-2\times0)=4\times0+1=0$
$\Rightarrow0=1=0$ which is not possible
Therefore, there is no value of $\lambda$ for which f(x) is continuous at x = 0
At x = 1
$\Rightarrow\text{f}(1)=4\text{x}+1=4\times1+1=5$
$\Rightarrow\lim_\limits{\text{x}\rightarrow1}(4\text{x}+1)=4\times1+1=5$
$\therefore\ \lim_\limits{\text{x}\rightarrow1}\text{f(x)}=\text{f}(1)$
Therefore, for any value of $\lambda,$ f is continuous at x = 1
At x = -1 we have,
$\Rightarrow\text{f}(-1)=\lambda(1+2)=3\lambda$
$\Rightarrow\lim_\limits{\text{x}\rightarrow-1}\lambda(1+2)=3\lambda$
$\therefore\ \lim_\limits{\text{x}\rightarrow-1}\text{f(x)}=\text{f}(-1)$
Therefore, for any values of $\lambda,$ f is continuous at x = -1
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Question 102 Marks
If $\text{f(x)}=\begin{cases}\frac{\text{x}^2-16}{\text{x}-4},&\text{if }\text{ x}\neq4\\\text{k},&\text{if }\text{ x}=4\end{cases}$ is continuous at x = 4, find k.
Answer
Given, $\text{f(x)}=\begin{cases}\frac{\text{x}^2-16}{\text{x}-4},&\text{if }\text{ x}\neq4\\\text{k},&\text{if }\text{ x}=4\end{cases}$
If f(x) is continuous at x = 4, then
$\lim\limits_{{\text{x}}\rightarrow4}\text{f(x})=\text{f(4)}$
$\Rightarrow\lim\limits_{{\text{x}}\rightarrow4}\Big(\frac{\text{x}^2-16}{\text{x}-4}\Big)=\text{k}$
$\Rightarrow\lim\limits_{{\text{x}}\rightarrow4}\frac{(\text{x}+4)(\text{x}-4)}{(\text{x}-4)}=\text{k}$
$\Rightarrow\lim\limits_{{\text{x}}\rightarrow4}(\text{x}+4)=\text{k}$
$\Rightarrow\text{k}=8$
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Question 112 Marks
If $\text{f(x)}=\begin{cases}\frac{1-\cos\text{x}}{\text{x}^2},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$ is continuous at x = 0, find k.
Answer
Since f(x) is continuous at x = 0,
$\text{f}(0)=\lim\limits_{{\text{x}}\rightarrow0}\text{f(x})$
$\Rightarrow\text{k}=\lim\limits_{{\text{x}}\rightarrow0}\frac{1-\cos\text{x}}{\text{x}^2}$
$\Rightarrow\text{k}=\lim\limits_{{\text{x}}\rightarrow0}\frac{2\sin^2\frac{\text{x}}{2}}{\text{x}^2}$
$\Rightarrow\text{k}=\frac{1}{2}$
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Question 122 Marks
If $\text{f(x)}=\begin{cases}\frac{\sin^{-1}\text{x}}{\text{x}},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$ is continuous at x = 0, write the value of k.
Answer
Given, $\text{f(x)}=\begin{cases}\frac{\sin^{-1}\text{x}}{\text{x}},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$
If f(x) is continuous at x = 0, then
$\lim\limits_{{\text{x}}\rightarrow0}\text{f(x})=\text{f}(0)$
$\Rightarrow\lim\limits_{{\text{x}}\rightarrow0}\text{f(x})\Big(\frac{\sin^{-1}\text{x}}{\text{x}}\Big)=\text{f}(0)$
$\Rightarrow\lim\limits_{{\text{x}}\rightarrow0}\Big(\frac{\sin^{-1}\text{x}}{\text{x}}\Big)=\text{k}$
$\Rightarrow\text{k}=1$ $\bigg[\because\ \lim\limits_{{\text{x}}\rightarrow0}\Big(\frac{\sin^{-1}\text{x}}{\text{x}}\Big)=1\bigg]$
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