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Chapter 2: Problem 22

\(\mathrm{f}(\mathrm{x})=(\mathrm{x} \bmod 2)^{2}+(\mathrm{x} \bmod 4)\) Points of discontinuity are \(\mathrm{x}=2,4,6,8\) Hence, \(\mathrm{C}\) is correct.

Short Answer

Expert verified
Question: Identify the points of discontinuity for the function \(f(x) = (x \bmod 2)^2 + (x \bmod 4)\). Answer: The points of discontinuity for the given function are at integer multiples of 4, such as \(x=2,4,6,8,\ldots\).

Step by step solution

01

Understand the function components

We have a function \(f(x) = (x \bmod 2)^2 + (x \bmod 4)\). First, we need to analyze the components of the function: \((x \bmod 2)^2\) and \((x \bmod 4)\). The first component \((x \bmod 2)^2\) behaves like a quadratic function, and the second component \((x \bmod 4)\) behaves like a piecewise function.
02

Find where the second component is discontinuous

The second component \((x \bmod 4)\) will have different values around its integer multiples since it's a piecewise function. This means that around \(x=4k\), where \(k\) is an integer, it will cause discontinuity in the function. However, we need to consider the first component's behavior around those points as well.
03

Analyze the first component around discontinuous points

As the first component behaves like a quadratic function, it will be continuous everywhere. This means that around the discontinuity points of the second component, the only effect of this component will be that it may add or subtract a continuous function value, not affecting the discontinuity itself.
04

Combine the knowledge of both components

From steps 2 and 3, we know that the second component will be discontinuous around its integer multiples (\(x=4k\)), while the first component does not affect the overall continuity of the function. This implies that the function will be discontinuous around integer multiples of 4.
05

Find the points of discontinuity

Using the knowledge that our function has discontinuities around integer multiples of 4, we can find the first few points of discontinuity by listing those points: \(x=2,4,6,8,\ldots\). Therefore, the points of discontinuity are \(x=2,4,6,8\), and the correct answer is C.

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Most popular questions from this chapter

\(f(x)=\lim _{n \rightarrow \infty} \frac{x^{2 n-1}+a x^{2}+b x}{x^{2 n}+1}\) \(\lim _{x \rightarrow 1^{+}} f(x)=1\) \(\lim _{x \rightarrow 1^{-}} f(x)=a+b\) \(f(1)=\frac{a+b+1}{2}\) \(\Rightarrow a+b=1\)

$f(x)=[x] \& g(x)= \begin{cases}0, & x \in I \\ x^{2}, & \text { otherwise }\end{cases}$ \(g \circ f(x)=0, \quad x \in R\) $f \circ g(x)= \begin{cases}0, & x \in I \\ {\left[x^{2}\right]} & \text { otherwise }\end{cases}$

$f\left(\frac{1}{4^{n}}\right)=\left(\sin e^{n}\right) \mathrm{e}^{-n^{2}}+\frac{n^{2}}{n^{2}+1}$ $f(0)=\lim _{n \rightarrow \infty}\left(\left(\sin e^{n}\right) e^{-n^{i}}+\frac{n^{2}}{n^{2}+1}\right)$ \(=1\)

$\prod_{n=1}^{n} \cos \left(\frac{x}{2^{n}}\right)=\frac{\sin x}{2^{n} \sin \left(\frac{x}{2^{n}}\right)}$ Taking log \& differentiate w.rt, \(x\). $\sum_{n=1}^{n} \frac{1}{2^{n}} \tan \left(\frac{x}{2^{n}}\right)=\frac{1}{2^{n}} \frac{\cos \left(x / 2^{n}\right)}{\sin \left(\frac{x}{2^{n}}\right)}-\cot x .$ now, $\lim _{n \rightarrow \infty} \sum_{n=1}^{n} \frac{1}{2^{n}} \tan \left(\frac{x}{2^{n}}\right)=\frac{1}{x}-\cot x$ $\lim _{x \rightarrow \frac{\pi}{2}} \lim _{n \rightarrow \infty} \sum_{n=1}^{n} \frac{1}{2^{n}} \tan \left(\frac{x}{2^{n}}\right)=\frac{2}{\pi}$

Using AM - GM inequality A) \(\cos ^{2} A+\operatorname{Cos}^{2} B+\cos ^{2} C \leq 1+1+1\) In a triangle this is positive if $\mathrm{A} \rightarrow 0, \mathrm{~B} \rightarrow 0\( and \)\mathrm{C} \rightarrow \pi$ Hence the maximum value of 3 is not attained. Hence $\operatorname{Cos}^{2} \mathrm{~A}+\cos ^{2} \mathrm{~B}+\operatorname{Cos}^{2} \mathrm{C}<3$ So, \(P\) is correct. B) $\frac{9 \mathrm{a}+3 \mathrm{~b}+\mathrm{c}}{3} \geq \sqrt[3]{27 \mathrm{abc}}$ abc \(\leq(10)^{3}\) Now, \(\Rightarrow \log a+\log b+\log c=\log a b c\) \(\leq \log (10)^{3}=3\) So, \(R\) is correct. C) $\lim _{x \rightarrow 0} \frac{\tan x \sqrt{\tan x}-\sin x \sqrt{\sin x}}{x^{3} \sqrt{x}}$ $\lim _{x \rightarrow 0} \frac{\sin x \sqrt{\sin x}\left(1-(\cos x)^{3 / 2}\right)}{(\cos x)^{3 / 2} \cdot x^{3 / 2} \times x^{2}}$ \(=\frac{3}{4}\) So, \(P\) is correct. D) \(\lim _{x \rightarrow 0} \cos \left(x \cos \frac{1}{x}\right)=1\) $\lim _{x \rightarrow 0} \frac{\ln \left(\sec ^{2} x\right)}{x \sin x}=\lim _{x \rightarrow 0}-\frac{\ln \left(1-\sin ^{2} x\right)}{x^{2} \frac{\sin x}{x}}=1$ So, \(f(0)+g(0)=2\) So, \(Q\) is correct.
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