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Chapter 6: Problem 30

$\left.\mathrm{f}^{\prime}(\mathrm{x})=\log _{1 / 3} \log _{3}(\sin (\mathrm{x})+\mathrm{a})\right)<0$ \(\left.\log _{3}(\sin (x)+a)\right)>1\) \(\sin (x)+a>3\) \(a>3-\sin x\) \(\mathrm{a}>4\) Hence B is correct

Short Answer

Expert verified
Given the inequality: \(\left.f'(\mathrm{x})=\log_{1/3}\log_{3}(\sin(\mathrm{x})+\mathrm{a})\right)< 0\), and the information provided, we applied the properties of logarithms, substituted the given inequality, and solved the inequality related to the trigonometric function. We found the value of 'a' that satisfies both inequalities, which is when \(a > 4\).

Step by step solution

01

Apply the properties of logarithms

The given inequality is: \(\left.\mathrm{f}^{\prime}(\mathrm{x})=\log _{1 / 3} \log _{3}(\sin(x)+\mathrm{a})\right)<0\). Since the base of the logarithm is a fraction (1/3), we can rewrite it using the change of base formula: $$\mathrm{f}^{\prime}(\mathrm{x})=\frac{\log _{e}\log _{3}(\sin(x)+\mathrm{a})}{\log _{e}(1 / 3)} < 0$$
02

Substitute the given inequality

Now, we are given that: \(\left.\log _{3}(\sin (x)+a)\right)>1\). We can substitute this inequality into the inequality we derived in Step 1: $$\frac{\log _{e} (1)}{\log _{e}(1 / 3)} < 0$$ $$0 < 0$$ This inequality is invalid.
03

Solve the inequality related to the trigonometric function

Since the inequality in Step 2 is invalid, we need to go back to the inequality given: $$\sin(x) + a > 3$$ Now, we need to solve this inequality for 'a'. We can do this by subtracting \(\sin(x)\) from both sides: $$a > 3 - \sin(x)$$
04

Find the value of 'a'

We are also given that \(\mathrm{a}>4\). Now, we need to find the value of 'a' that satisfies both inequalities: $$a > 3 - \sin(x)$$ $$a > 4$$ If we consider the first inequality \(a > 3 - \sin(x)\), the maximum value for \(\sin(x)\) is 1, so \(a > 3 - 1 = 2\). But in the second inequality, we have that \(a > 4\). The value that satisfies both inequalities is when \(a > 4\). So, the correct answer is B.

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Most popular questions from this chapter

\(f^{\prime}(x)=g(x)(x-a)^{2}\) \(f(x)\) is increasing if \(g(a)>0\) \& \(\mathrm{f}(\mathrm{x})\) is decreasing if \(\mathrm{g}(\mathrm{a})<0\) Hence, A, D is correct

\(f(x)=3 \cos ^{4} x+10 \cos ^{3} x+6 \cos ^{2} x-3\) \(f^{\prime}(x)=-6 \sin x\left(2 \cos ^{3} x+5 \cos ^{2} x+2 \cos x\right)\) \(=-3 \sin 2 x\left(2 \cos ^{2} x+5 \cos x+2\right)\) \(=-3 \sin 2 x(\cos x+2)(2 \cos x+1)\) Hence, \(\mathrm{C}\) is correct

$\begin{aligned} \lim _{x \rightarrow \infty} f(x) &=\lim _{x \rightarrow \infty}\left(\frac{2}{x}+2+\frac{\sqrt{x^{2}+4}}{x}\right) \\ &=3 \end{aligned}$

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

\(f^{\prime}(x)=\frac{-2}{x^{2}}\left(\frac{\sqrt{x^{2}+4}+2}{\sqrt{x^{2}+4}}+2\right)<0\) Hence, A is correct
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