/*! 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 2 \(f(x)=\sin ^{2} x-3 \cos ^{2} x... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 2

\(f(x)=\sin ^{2} x-3 \cos ^{2} x+2 a x-4\) \(f^{\prime}(x)=4(\sin 2 x)+2 a \geq 0\) \(\Rightarrow \mathrm{a} \in[2, \infty)\) Hence, \(\mathrm{C}\) is correct

Short Answer

Expert verified
To ensure that the function's derivative, \(4(\sin 2x) + 2a\), remains nonnegative, the parameter "a" must fall within the range of \(a \in [2, \infty)\). Therefore, the correct answer is \(\boxed{\mathrm{C}}\).

Step by step solution

01

Analyze the given constraint on the function's derivative

The constraint on the derivative of the function \(f^{\prime}(x)\) is that it must be nonnegative: \(4(\sin 2x) + 2a \geq 0\).
02

Solve the inequality

In order to solve the inequality, we need to find the minimum value of the term \((\sin 2x)\), since this will allow us to find the range of values for "a" leading to a nonnegative \(f^{\prime}(x)\). The term \(\sin(2x)\) can take values in the range of \([-1, 1]\). Therefore, the minimum value of \(\sin(2x)\) is -1. When \(\sin(2x) = -1\), the inequality becomes: \(4(-1) + 2a \geq 0 \) \(-4 + 2a \geq 0\) \(2a \geq 4\) \(a \geq 2\)
03

Provide the correct answer

Based on the solution of the inequality, we have determined that the parameter "a" must be greater than or equal to 2; that is, \(a \in [2, \infty)\). Thus, the correct answer is \(\boxed{\mathrm{C}}\).

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Most popular questions from this chapter

Let \(f(x)=x^{1 / x}\) $\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x}^{1 / \mathrm{x}} \frac{(1-\ln \mathrm{x})}{\mathrm{x}^{2}}$ $\mathrm{f}^{\prime}(\mathrm{x})<0, \forall \mathrm{x} \in(\mathrm{e}, \infty)$ \(\Rightarrow \mathrm{f}(\mathrm{x})\) is decreasing As \(1999<2000\) \(f(1999)>\mathrm{f}(2000)\) \((1999)^{x-m}>(2000)^{\%}\) \((1999)^{2000}>(2000)^{1999} .\) Hence, A is correct

A: $\begin{aligned} f(x) &=x^{4}-8 x^{3}+22 x^{2}-24 x+21 \\ f^{\prime}(x) &=4 x^{3}-24 x^{2}+44 x-24 \\ &=4\left(x^{3}-6 x^{2}+11 x-6\right) \\\ &=4(x-1)\left(x^{2}-5 x+6\right) \\ &=4(x-1)(x-2)(x-3) \end{aligned}$

A) \(\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}+\frac{4}{\mathrm{x}^{2}}+7\) \(f^{\prime}(x)=3 x^{2}-\frac{8}{x^{3}}=\frac{3 x^{5}-8}{x^{3}}\) \(\mathrm{f}(\mathrm{x})\) is \(\uparrow\) in \((-\infty, \infty)\) B) \(g(x)=\sqrt{t}+\sqrt{1+t}-4\) \(g^{\prime}(x)=\frac{1}{2 \sqrt{t}}+\frac{1}{2 \sqrt{1+t}}>0\) \(\mathrm{g}(\mathrm{x})\) is increasing in \((0, \infty)\) C) \(\mathrm{r}(\theta)=\theta+\sin ^{2}\left(\frac{\theta}{3}\right)-8\) \(r^{\prime}(\theta)=1+\frac{1}{3} \sin \left(\frac{2 \theta}{3}\right)>0\) \(\mathrm{r}(\theta)\) is an increasing function D) \(r(\theta)=\tan \theta-\cot \theta-\theta\) \(r^{\prime}(\theta)=\sec ^{2} \theta+\operatorname{cosec}^{2} \theta-1\) \(=\tan ^{2} \theta+\operatorname{cosec}^{2} \theta\) \(\mathrm{r}(\theta)\) is an increasing function Hence, $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$ is correct

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

\(\mathrm{A}: \mathrm{f}(\mathrm{x})=2 \cos \mathrm{x}+3 \sin \mathrm{x}\) \(\mathrm{f}^{\prime}(\mathrm{x})=3 \cos \mathrm{x}-2 \sin \mathrm{x}\) \(=\sqrt{13} \cos (x+\alpha)\) where \(\cos \alpha=\frac{3}{\sqrt{13}}\) $\mathrm{f}^{-1}(\mathrm{x})=\sin ^{-1} \frac{\mathrm{x}}{\sqrt{13}}-\tan ^{-1} \frac{3}{2}$ Hence, A is correct
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