/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 42 \(f(x)=\frac{\tan x \log x}{1-\c... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 42

\(f(x)=\frac{\tan x \log x}{1-\cos 4 x}\) Firstly, \(x \neq(-\infty, 0]\) \(\& 1-\cos 4 x \neq 0\) \(\cos 4 x \neq 1\) \(\Rightarrow 4 \mathrm{x} \neq 2 \mathrm{n} \pi\) \(x \neq \frac{n \pi}{2}\)

Short Answer

Expert verified
Based on the given function, \(f(x) = \frac{\tan x \log x}{1-\cos 4x}\), with the restrictions \(x \neq(-\infty, 0]\) and \(x \neq \frac{n \pi}{2}\), explain the reason for these restrictions and how they affect the domain of the function.

Step by step solution

01

Understanding the function

We are given the function \(f(x) = \frac{\tan x \log x}{1-\cos 4x}\). Notice that this function consists of a rational function, with a trigonometric function \(\tan x\) and a logarithmic function \(\log x\) in the numerator, and another trigonometric function, \(\cos 4x\), in the denominator.
02

Determine the restrictions

First, we must examine the restrictions placed on \(x\). For the function \(f(x)\) to be defined, the denominator must not be equal to 0. In our function, the denominator is given by \(1-\cos 4x\), so we require that \(1-\cos 4x\neq 0\).
03

Find the range of the cosine function

The range of the cosine function is \([-1,1]\). Therefore, \(\cos 4x\) can take values between -1 and 1, inclusive. If we want the denominator to be non-zero, we need to determine when \(\cos 4x = 1\) because this is the only value in the range of cosine that makes the denominator zero.
04

Determine when \(\cos 4x = 1\)

Cosine equals 1 at even multiples of \(\pi\), that is, \(\cos 4x = 1 \Rightarrow 4x = 2n\pi\), where \(n\) is an integer. This means that \(x = \frac{n \pi}{2}\), for any integer \(n\).
05

Taking into account the domain of the logarithm

Since \(\log x\) is only defined for \((0, \infty)\), and we are given the restriction \(x \neq (-\infty,0]\), we don't need to worry about the domain of logarithm in this case, as it is already covered by the given restriction.
06

Final restrictions on \(x\)

Taking all restrictions into account, we have that \(x \neq (-\infty, 0]\) and \(x \neq \frac{n \pi}{2}\) for any integer \(n\). Now the function \(f(x) = \frac{\tan x \log x}{1-\cos 4x}\) is well-defined for all values of \(x\) that satisfy these conditions.

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Most popular questions from this chapter

\(f(x)=x^{4}-14 x^{3}+p x^{2}+q x-105\) Sum of roots are 14 \& product is \(-105\). Hence, roots are \(-1,3,5,7\) \(g(x)=(x-1)(x-3)(x-5)(x-7)\) \(F(x)=\frac{(x+1)(x-3)(x-5)(x-7)}{(x-1)(x-3)(x-5)(x-7)}\) \(=\frac{x+1}{x-1} \forall x \neq 1,3,5,7\)

\(\lim _{x \rightarrow 0^{\prime}} \frac{b e^{x}-\cos x-x}{x^{2}}\) For limit to exist, \(b=1\) \(\lim _{x \rightarrow 0^{+}} \frac{e^{x}-\cos x-x}{x^{2}}=1=a\) $\lim _{x \rightarrow 0^{-}} \frac{2\left(\tan ^{+} e^{x}-\frac{\pi}{4}\right)}{x}=1$

$\lim _{x \rightarrow a}(\cos (x-a))^{\frac{6}{(x-a) 2 \sin (x-a) \cos (x-a)}}$ $\Rightarrow e^{-\lim 2 \sin ^{2} \frac{(x-a)}{2} \times \frac{6}{(x-a) \sin (x-a) \cos (x-a)}}$ \(\Rightarrow e^{-3 / 2}\) Now, $h(x)=\lim _{n \rightarrow \infty} \frac{\sin \frac{x}{2^{n}}}{\frac{x}{2^{n}}} \times x=x$ \(h(a)=a\) \(\Rightarrow \mathrm{k}=\frac{3}{2}\) Hence, \(\mathrm{C}\) is correct.

Correction $\rightarrow f(x)=\frac{1-\cos x(\cos 2 x)^{1 / 2} \cos (3 x)^{1 / 3}}{x^{2}}$ Sol using L-Hospital's Rule $\begin{aligned} & \lim _{x \rightarrow 0}(\sin x)(\cos 2 x)^{1 / 2}(\cos 3 x)^{1 / 3}+\cos x(\cos 3 x)^{1 / 3} \\ \lim _{x \rightarrow 0} f(x)=& \frac{\sin 2 x}{\sqrt{\cos 2 x}+\cos x(\cos 2 x)^{1 / 3} \frac{\sin 3 x}{(\cos 3 x)^{2 / 3}}}{2} \\\=& \lim _{x \rightarrow 0}\left[\frac{1}{2}+1+\frac{3}{2}\right]=3 \end{aligned}$

\(\lim _{x \rightarrow 0} \frac{\sin a x}{b x}=\frac{a}{b}\) \(\lim _{x \rightarrow 0^{\circ}}(a x+1)=1\) $\lim _{x \rightarrow 1}(a x+1)=a+1 \quad \lim _{x \rightarrow 2}\left(c x^{2}-2\right)=4 c-2$ $\lim _{x \rightarrow 1^{\prime}}\left(c x^{2}-2\right)=c-2 \quad \lim _{x \rightarrow 2^{\prime}} \frac{d\left(x^{2}-4\right)}{\sqrt{x}}=0$
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