/*! 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 47 \(|\sin 2 \mathrm{x}|=|\mathrm{x... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 47

\(|\sin 2 \mathrm{x}|=|\mathrm{x}|-\mathrm{a}\) \(|\mathbf{x}|-\mathrm{a}>1 \quad\) for no solution \(|x|>1+a\)

Short Answer

Expert verified
1. When \(x \geq 0\) 2. When \(x < 0\)

Step by step solution

01

Break down the absolute values

Start by examining the two absolute value functions separately. We know that \(|\sin 2x| = \sin 2x\) when \(2x\) is in the first or second quadrants, and \(|\sin 2x| = -\sin 2x\) when \(2x\) is in the third or fourth quadrants. The same logic can be applied to \(|x|-a\): Case 1 (When \(x \geq 0\)): \(|x|-a = x - a\) Case 2 (When \(x < 0\)): \(|x|-a = -x-a\)
02

Solve for each case

For each case, we will have two subcases (when \(\sin 2x\) is positive and when it is negative), giving us a total of four equations: Subcase 1.1 (\(x \geq 0\) and \(2x\) is in the first or second quadrant): \(x - a = \sin 2x\) Subcase 1.2 (\(x \geq 0\) and \(2x\) is in the third or fourth quadrant): \(x - a = -\sin 2x\) Subcase 2.1 (\(x < 0\) and \(2x\) is in the first or second quadrant): \(-x - a = \sin 2x\) Subcase 2.2 (\(x < 0\) and \(2x\) is in the third or fourth quadrant): \(-x - a = -\sin 2x\)
03

Solve for the conditions when there are no solutions

We are given that the equation has no solutions when \(|x|>1+a\). We need to consider this condition for each case: For \(x \geq 0\): \(x > 1 + a\) For \(x < 0\): \(-x > 1 + a\)
04

Solve the equations and analyze the conditions

Now solve each equation in Step 2, and compare the results to the conditions found in Step 3: Subcase 1.1: Solve for \(x\), \(x - a = \sin 2x\), and compare it to \(x > 1 + a\) Solution: Graphically or algebraically (using numerical methods) find the values of \(x\) that satisfy this equation and do not satisfy the condition \(x > 1 + a\). Subcase 1.2: Solve for \(x\), \(x - a = -\sin 2x\), and compare it to \(x > 1 + a\) Solution: Graphically or algebraically (using numerical methods) find the values of \(x\) that satisfy this equation and do not satisfy the condition \(x > 1 + a\). Subcase 2.1: Solve for \(x\), \(-x - a = \sin 2x\), and compare it to \(-x > 1 + a\) Solution: Graphically or algebraically (using numerical methods) find the values of \(x\) that satisfy this equation and do not satisfy the condition \(-x > 1 + a\). Subcase 2.2: Solve for \(x\), \(-x - a = -\sin 2x\), and compare it to \(-x > 1 + a\) Solution: Graphically or algebraically (using numerical methods) find the values of \(x\) that satisfy this equation and do not satisfy the condition \(-x > 1 + a\). By solving these equations and analyzing the conditions, we will find the values of \(x\) that satisfy the given equation, \(|\sin 2x| = |x| - a\), and its given condition, \(|x| - a > 1\).

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Most popular questions from this chapter

\(y=\frac{x^{2}}{4}-2\) \(\frac{d y}{d x}=\frac{x}{2}\) \(\left(y-\frac{x_{\perp}^{2}}{4}+2\right)=\frac{-2}{x}\left(x-x_{1}\right)\) \(-1-\frac{x_{1}^{2}}{4}+2=\frac{-2}{x_{1}}\left(1-x_{1}\right)\) \(\Rightarrow 2 x_{1}-\frac{x_{1} 3}{4}=-2+2 x_{1}\) \(x_{1}^{3}=8\) \(x_{1}=2, \quad y=-1\)

\(y=6-x-x^{2}\) \(\frac{d y}{d x}=-1-2 x\) eqn of tangent $y-6+x_{1}+x_{1}^{2}=-\left(1+2 x_{1}\right)\left(x-x_{1}\right)$ \(x y=x+3\) \(x \frac{d y}{d x}+y=1\) \(\frac{d y}{d x}=\frac{1-y_{2}}{x}\) eqn of tangent \(\rightarrow\) \(y-\frac{x_{2}+3}{x_{2}}=\frac{1-\frac{x_{2}+3}{x_{2}}}{x_{2}}\left(x-x_{2}\right)\) \(x_{2} y-\left(x_{2}+3\right)=\frac{-3}{x_{2}}\left(x-x_{2}\right)\) Comparing the two eqns $\frac{1}{x_{2}}=\frac{-\left(1+2 x_{1}\right)}{-\frac{3}{x_{2}}}=\frac{x_{1}\left(1+2 x_{1}\right)+6-x_{1}-x_{1}^{2}}{3+x_{2}+3}$ $\frac{1}{x_{2}}=\frac{x_{2}\left(1+2 x_{1}\right)}{3} \& \frac{1}{x_{2}}=\frac{x_{1}^{2}+6}{x_{2}+6}$ \(3=x_{2}^{2}+2 x_{1} x_{2}^{2} \quad\) \& \(x_{2}+6=x_{1}^{2} x_{2}+6 x_{2}\) $3=\left(1+2 \mathrm{x}_{1}\right) \mathrm{x}_{2}^{2} \quad \& \quad \mathrm{x}_{2}=\frac{6}{5-\mathrm{x}_{1}^{2}}$ \(\Rightarrow 3=\frac{\left(1+2 x_{1}\right) 36}{\left(5-x_{1}^{2}\right)^{2}}\) $\Rightarrow\left(5-\mathrm{x}_{1}^{2}\right)^{2}=12\left(1+2 \mathrm{x}_{1}\right)$ Let \(\mathrm{y}=\mathrm{mx}+\mathrm{c}\) be the common tangent \(m x+c=6-x-x^{2}\) \(\Rightarrow \mathrm{x}^{2}+(\mathrm{m}+1) \mathrm{x}+(\mathrm{c}-6)=0\) \(\mathrm{D}=0\) \((m+1)^{2}-4(c-6)=0\) For \(x y=x+3\) \(x(m x+c)=x+3\) \(m x^{2}+(c-1) x-3=0\) \(\mathrm{D}=0\) \((c-1)^{2}+12 m=0\) Using (I) \& (II) \((m+1)^{2}+24=4 c\) \(\mathrm{c}-1=\frac{(\mathrm{m}+1)^{2}+20}{4}\) $\Rightarrow\left(\frac{(\mathrm{m}+\mathrm{l})^{2}+20}{4}\right)^{2}+12 \mathrm{~m}=0$ \(\left((m+1)^{2}+20\right)^{2}+192 m=0\) Upon solving, \(\mathrm{m}=-3\) \(\Rightarrow \mathrm{c}=7\) eqn of tangent \(=y=-3 x+7\)

Ellipse \(\rightarrow \frac{x^{2}}{16}+\frac{y^{2}}{9}=1\) \(\Rightarrow \frac{x}{8} \frac{d x}{d t}+\frac{2 y}{9} \frac{d y}{d t}=1\) $\Rightarrow \frac{4 \sqrt{1-y^{2} / 9}}{8} \frac{d x}{d t}+\frac{2 y}{9} \times \frac{d y}{d x} \times \frac{d x}{d t}=1$ Now put \(\frac{d y}{d x}=1 \quad \& \quad y=1\) $\Rightarrow \frac{4}{6 \sqrt{2}} \frac{d x}{d t}+\frac{2}{9} \frac{d x}{d t}=1$ \(\Rightarrow \frac{d x}{d t}\left(\frac{\sqrt{2}}{3}+\frac{2}{9}\right)=1\) \(\Rightarrow \frac{d x}{d t}=\frac{9}{3 \sqrt{2}+2}\)

\(y=\frac{x^{2}}{4}-3 x+10\) For \(y=2-\frac{x^{2}}{4}\) eqn of tangent \(y-2=m x+m^{2}\) Solving (II) \& (I) \(m x+2+m^{2}=\frac{x^{2}}{4}-3 x+10\) \(\Rightarrow \frac{x^{2}}{4}-(m+3) x+\left(8-m^{2}\right)=0\) \(\mathrm{D}=0\) \((m+3)^{2}-\left(8-m^{2}\right)=0\) \(2 m^{2}+6 m+1=0\) \(\mathrm{m}_{1}+\mathrm{m}_{2}=-3\)

\(x^{4}+y^{4}=a^{4}\) \(x^{3}+y^{3} \frac{d y}{d x}=0\) \(\frac{d y}{d x}=\frac{-x^{3}}{y^{3}}\) eqn of tangent \(\rightarrow\) \(\mathrm{y}-\mathrm{y}_{1}=-\frac{\mathrm{x}_{1}^{3}}{\mathrm{y}_{1}^{3}}\left(\mathrm{x}-\mathrm{x}_{1}\right)\) \(\mathrm{p}=\frac{\mathrm{y}_{1}^{4}}{\mathrm{x}_{1}^{3}}+\mathrm{x}_{1}=\frac{\mathrm{x}_{1}^{4}+\mathrm{y}_{1}^{4}}{\mathrm{x}_{1}^{3}}=\frac{\mathrm{a}^{4}}{\mathrm{x}_{1}^{3}}\) \(\mathrm{q}=\mathrm{y}_{1}+\frac{\mathrm{x}_{1}^{4}}{\mathrm{y}_{1}^{3}}=\frac{\mathrm{a}^{4}}{\mathrm{y}_{1}^{3}}\) Now, \(\mathrm{p}^{-4 / 3}+\mathrm{q}^{-4 / 3}=\mathrm{a}^{-4 / 3}\)
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