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Mathematics LibreTexts

Section 2.3 Answers

  • Page ID
    28697
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    1. (b) \(x_{0}\neq k\pi (l=\text{integer})\)

    2. (b) \(x_{0},y_{0})\neq (0,0)\)

    3. (b) \(x_{0}y_{0}\neq (2k+1)\frac{\pi }{2}(k=\text{integer})\)

    4. (b) \(x_{0}y_{0}>0\text{ and }x_{0}y_{0}\neq 1\)

    5. 

    1. all \((x_{0},y_{0})\)
    2. \((x_{0},y_{0})\) woth \(y_{0}\neq 0\)

    6. (b) all \((x_{0}, y_{0})\)

    7. (b) all \((x_{0}, y_{0})\)

    8. (b) \((x_{0}, y_{0})\) such that \(x_{0}\neq 4y_{0}\)

    9.

    1. all \((x_{0}, y_{0})\)
    2. all \((x_{0}, y_{0})\neq (0,0)\)

    10. 

    1. all \((x_{0}, y_{0})\)
    2. all \((x_{0}, y_{0})\) with \(y_{0}\neq\pm 1\)

    11. (b) all \((x_{0}, y_{0})\)

    12. (b) all \((x_{0}, y_{0})\) such that \((x_{0}, y_{0}) >0\)

    13. (b) all \((x_{0}, y_{0})\) with \(x_{0}\neq 1,\quad y_{0}\neq (2k+1)\frac{\pi }{2}(k=\text{integer})\)

    16. \(y=\left(\frac{3}{5}x+1 \right)^{5/3},\quad -\infty <x<\infty\), is a solution.

    Also, \[y=\left\{\begin{array}{cc}{0,}&{-\infty <x\leq -\frac{5}{3}}\\{(\frac{3}{5}x+1)^{5/3},}&{-\frac{5}{3}<x<\infty }\end{array} \right.\nonumber\] is a solution. For every \(a\geq \frac{5}{3}\), the following function is also a solution: \[y=\left\{\begin{array}{cc}{(\frac{3}{5}(x+a))^{5/3},}&{-\infty <x<-a,}\\{0,}&{-a\leq x\leq -\frac{5}{3}}\\{(\frac{3}{5}x+1)^{5/3},}&{-\frac{5}{3}<x<\infty }\end{array} \right.\nonumber\]

    17. 

    1. all  \((x_{0}, y_{0})\)
    2. all  \((x_{0}, y_{0})\) with \(y_{0}\neq 1\)

    18. \(y_{1}=1; y_{2}=1+|x|^{3};y_{3}=1-|x|^{3};y_{4}=1+x^{3};y_{5}=1-x^{3}\)

    \[y_{6}=\left\{\begin{array}{cc}{1+x^{3},}&{x\geq 0,}\\{1,}&{x<0}\end{array} \right.;\quad y_{7}=\left\{\begin{array}{cc}{1-x^{3},}&{x\geq 0,}\\{1,}&{x<0}\end{array} \right.;\nonumber\]

    \[y_{8}=\left\{\begin{array}{cc}{1,}&{x\geq 0,}\\{1+x^{3},}&{x<0}\end{array} \right.;\quad y_{9}=\left\{\begin{array}{cc}{1,}&{x\geq 0,}\\{1-x^{3},}&{x<0}\end{array} \right.\nonumber\]

    19. \(y=1+(x^{2}+4)^{3/2},\quad -\infty <x<\infty \)

    20. 

    1. The solution is unique on \((0,\infty )\). It is given by
      \[y=\left\{\begin{array}{cc}{1,}&{0<x\leq \sqrt{5}}\\{1-(x^{2}-5)^{3/2},}&{\sqrt{5}<x<\infty }\end{array} \right.\nonumber\]

    2. \[y=\left\{\begin{array}{cc}{1,}&{-\infty <x\leq\sqrt{5},}\\{1-(x^{2}-5)^{3/2},}&{\sqrt{5}<x<\infty }\end{array} \right.\nonumber\] is a solution of (A) on \((-\infty ,\infty )\). If \(\alpha\geq 0\), then \[y=\left\{\begin{array}{cc}{1+(x^{2}-\alpha ^{2})^{3/2},}&{-\infty <x<-\alpha ,}\\{1,}&{-\alpha\leq x\leq\sqrt{5},}\\{1-(x^{2}-5)^{3/2},}&{\sqrt{5}<x<\infty ,}\end{array}\right.\nonumber\] and \[y=\left\{\begin{array}{cc}{1-(x^{2}-\alpha ^{2})^{3/2},}&{-\infty <x<-\alpha ,}\\{1,}&{-\alpha\leq x\leq\sqrt{5},}\\{1-(x^{2}-5)^{3/2},}&{\sqrt{5}<x<\infty ,}\end{array} \right.\nonumber\] are also solutions of (A) on \((-\infty ,\infty)\).