/*! 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 67 As \(f(x)\) is increasing functi... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 67

As \(f(x)\) is increasing function \(\Rightarrow 2 a^{2}+a+1>3 a^{2}-4 a+1\) \(a^{2}-5 a<0\) \(a(a-5)<0\) as \(x \in(0, \infty), \quad a \in\\{2,3,4\\}\) Hence, \(\mathrm{B}, \mathrm{C}, \mathrm{D}\) is correct

Short Answer

Expert verified
Answer: The correct values of \(a\) are \(\{2, 3, 4\}\).

Step by step solution

01

Understand the increasing function property

An increasing function has the property that for any two values \(x_1\) and \(x_2\) in the given domain, if \(x_1 < x_2\), then \(f(x_1) < f(x_2)\).
02

Analyze the inequality

Given that \(f(x)\) is an increasing function, we have \(\Rightarrow 2a^2 + a + 1 > 3a^2 - 4a + 1\). The next step is to analyze and rearrange this inequality to find the possible values of \(a\).
03

Rearrange the inequality

We can begin by simplifying the inequality: \begin{align*} 2a^2 + a + 1 &> 3a^2 - 4a + 1 \\ (-1)a^2 + 5a &> 0 \\ a^2 - 5a &< 0 \\ a(a - 5) &< 0 \end{align*} Now, we need to determine the range of values of \(a\) for which this inequality holds true.
04

Find the values of \(a\) satisfying the inequality

To find the values of \(a\) that satisfy the inequality, first observe that \(a\) must be positive since \(x\) is in the domain \((0, \infty)\). Then identify and test the intervals between the critical points where \(a(a - 5)\) changes its sign. 1. \(0 < a < 5\): In this interval, \(a(a - 5) < 0\) since \(a > 0\) and \((a - 5) < 0\). 2. \(a > 5\): In this interval, \(a(a - 5) > 0\) since \(a > 0\) and \((a - 5) > 0\). With this information, it becomes clear that the values of \(a\) for which \(a(a - 5) < 0\) hold true are contained in the interval \((0,5)\).
05

Select the values of \(a\) from the set

Now, since we want only the integers \(\{2, 3, 4\}\) to choose from within the domain \((0, \infty)\), we can quickly see that all three values \(2, 3, 4\) lie within the valid interval \((0,5)\) of \(a\). Hence the correct values of \(a\) are \(\boxed{\{2, 3, 4\}}\).

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Most popular questions from this chapter

\(h(x)=2(1+\sin 4 x)=4\) \(\Rightarrow \sin 4 \mathrm{x}=1\) \(\Rightarrow 4 \mathrm{x}=(4 \mathrm{n}+1) \frac{\pi}{2}\) \(x=\frac{(4 n+1) \pi}{8}\) Hence, A is correct

A) \(f(x)=\frac{x}{1+x^{2}}\) $f^{\prime}(x)=\frac{\left(1+x^{2}\right)-2 x^{2}}{\left(1+x^{2}\right)^{2}}=\frac{1-x^{2}}{\left(1+x^{2}\right)^{2}}<0$ \(\Rightarrow x \in(-\infty,-1) \cup(1, \infty)\) B) \(f(x)=\tan ^{-1} x-x\) \(f^{\prime}(x)=-\frac{x^{2}}{1+x^{2}}<0\) \(\Rightarrow x \in R\) C) \(f(x)=x-e^{x}+\tan \frac{2 \pi}{7}\) \(f^{\prime}(x)=1-e^{x} \geq 0\) \(\quad x \in(-\infty, 10]\) D) \(f(x)=x^{3}-\ln \left(1+x^{3}\right)\) \(f^{\prime}(x)=3 x^{2}\left(1-\frac{1}{1+x^{3}}\right)\) \(=\frac{3 x^{5}}{1+x^{3}}\)

\(h(x)=3 f\left(\frac{x^{2}}{3}\right)+f\left(3-x^{2}\right)\) $h^{\prime}(x)=2 x\left(f^{\prime}\left(\frac{x^{2}}{3}\right)-f^{\prime}\left(3-x^{2}\right)\right)$ For \(\mathrm{h}(\mathrm{x})\) to be increasing if $x<0, \quad 3-x^{2}>\frac{x^{2}}{3}$ \(\Rightarrow 4 \mathrm{x}^{2}-9<0\) \(x \in(-3 / 2,0)\) For \(x>0, \frac{x^{2}}{3}>3-x^{2}\) \(4 x^{2}-9<0\) \(x \in(3 / 2,4)\) Hence, A, B, C, D is correct

\(f(x)=a x^{3}+b x^{2}+c x+d\) \(f^{\prime}(x)=3 a x^{2}+2 b x+c>0\) \(\Rightarrow 4 b^{2}-12 a c<0 \& a>0\) \(\Rightarrow \mathrm{b}^{2}-3 \mathrm{ac}<0\) \(b^{2}<3 a c \& a>0,3 b^{2}\left(\sqrt{3} a x+\frac{4 b}{\sqrt{3}}\right)^{2}>0\) Hence, B is correct

\(f(x)=x^{3}+2 x^{2}+5 x+2 \cos x\) \(f^{\prime}(x)=3 x^{2}+4 x+5-2 \sin x\) Min value of \(3 \mathrm{x}^{2}+4 \mathrm{x}+5\) is \(\frac{22}{3}\) \(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})>0\) \(f(0)=2 \& f(\infty)=\infty, \quad f(-\infty)-\infty\) There is no root in \([0,2 \pi]\) Hence, D is correct
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