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## Algebra: Polynomials Test-1

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Question 1 |

$ \displaystyle f\,\,\,x=7-4\sqrt{3},\,then\,\,\,the\,\,\,value\,\,\,of\,\,\left( x+\frac{1}{x} \right)\,\,is:$

$ \displaystyle 3\sqrt{3}$ | |

$ \displaystyle 8\sqrt{3}$ | |

$ \displaystyle 14+8\sqrt{3}$ | |

14 |

Question 1 Explanation:

Substituting values of x we have

$ \displaystyle \begin{array}{l}x+~\frac{1}{x}=7-4\sqrt{3}~+~~\frac{1}{7-4\sqrt{3}}\\7-4\sqrt{3}+\frac{7+4\sqrt{3}~}{\left( 7-4\sqrt{3}~ \right)\left( 7+4\sqrt{3}~ \right)}\\7-4\sqrt{3}~+\frac{7+4\sqrt{3}}{49-48}\\7-4\sqrt{3}+7+4\sqrt{3}\,\,=14\end{array}$

$ \displaystyle \begin{array}{l}x+~\frac{1}{x}=7-4\sqrt{3}~+~~\frac{1}{7-4\sqrt{3}}\\7-4\sqrt{3}+\frac{7+4\sqrt{3}~}{\left( 7-4\sqrt{3}~ \right)\left( 7+4\sqrt{3}~ \right)}\\7-4\sqrt{3}~+\frac{7+4\sqrt{3}}{49-48}\\7-4\sqrt{3}+7+4\sqrt{3}\,\,=14\end{array}$

Question 2 |

If $ \displaystyle p=\frac{5}{8},\,q=\frac{7}{12},r=\frac{13}{16}\,\,and\,\,\,s=\frac{16}{29}\,\,then$

$ \displaystyle p | |

$ \displaystyle s | |

$ \displaystyle p | |

$ \displaystyle s |

Question 2 Explanation:

LCM (8,12,16,29)=1392

$ \displaystyle \begin{array}{l}p=\frac{5}{8}=0.625\\q=\frac{7}{12}=0.583\\r=\frac{13}{16}=0.8125\\s=\frac{16}{29}=0.552\\s

$ \displaystyle \begin{array}{l}p=\frac{5}{8}=0.625\\q=\frac{7}{12}=0.583\\r=\frac{13}{16}=0.8125\\s=\frac{16}{29}=0.552\\s

Question 3 |

Given that $ \displaystyle {{10}^{0.48}}=x,\,{{10}^{0.70\,}}=y,\,\,\,and\,\,\,{{x}^{z}}={{y}^{2}},\,\,\,then\,\,\,the\,\,\,value\,\,\,of\,\,\,z\,\,\,is\,\,\,close\,\,to$

1.45 | |

1.88 | |

2.9 | |

3.7 |

Question 3 Explanation:

$ {{x}^{2}}={{y}^{2}}$ is

$ {{10}^{0.48\times z}}={{10}^{0.70\times 2}}$

Taking log to the base 10 both sides: $ \displaystyle lo{{g}_{10}}{{10}^{(0.48\,\text{ }\!\!\times\!\!\text{ }\,z)}}=lo{{g}_{10}}{{10}^{(0.70\text{ }\!\!\times\!\!\text{ }2)}}$

we get

0.48Ă—z=0.70Ă—2

z = 2.9 approx.

$ {{10}^{0.48\times z}}={{10}^{0.70\times 2}}$

Taking log to the base 10 both sides: $ \displaystyle lo{{g}_{10}}{{10}^{(0.48\,\text{ }\!\!\times\!\!\text{ }\,z)}}=lo{{g}_{10}}{{10}^{(0.70\text{ }\!\!\times\!\!\text{ }2)}}$

we get

0.48Ă—z=0.70Ă—2

z = 2.9 approx.

Question 4 |

$ \displaystyle f\,\,\,x=1+2\sqrt{2},\,\,then\,\,\,the\,\,\,value\,\,\,of\,\,\,\sqrt{x}-\frac{1}{\sqrt{x}}\,is$

1 | |

$ \displaystyle \frac{14\sqrt{2}+9}{8}$ | |

$ \displaystyle \frac{14\sqrt{2}-9}{8}$ | |

$ \displaystyle 3\sqrt{3}$ |

Question 4 Explanation:

$ \displaystyle \begin{array}{l}{{\left( \sqrt{x}-\frac{1}{\sqrt{x}} \right)}^{2}}=x+\frac{1}{x}-2\\=\left( 1+3\sqrt{2} \right)+\frac{1}{1+2\sqrt{2}}-2\\=\left( 1+2\sqrt{2} \right)+\frac{\left( 1-2\sqrt{2} \right)}{\left( 1+2\sqrt{2} \right)\left( 1-2\sqrt{2} \right)}-2\\=\left( 1+2\sqrt{2} \right)+\frac{\left( 1-2\sqrt{2} \right)}{-8}-2\\=\,\frac{-8-16\sqrt{2}+1-2\sqrt{2}+16}{-8}\\\sqrt{x}-\frac{1}{\sqrt{x}}=\frac{14\sqrt{2}-9}{8}\end{array}$

Question 5 |

$ \displaystyle If\,\,\,x=\frac{\sqrt{3}+1}{\sqrt{3}-1}\,\,and\,\,y=\frac{\sqrt{3}-1}{\sqrt{3}+1}\,\,the\,\,\,\,value\,\,\,of\,\,{{x}^{2}}+{{y}^{2}}\,\,is:$

14 | |

13 | |

15 | |

10 |

Question 5 Explanation:

$ \displaystyle \begin{array}{l}x=\frac{\sqrt{3}+1}{\sqrt{3}-1}\\=\frac{\left( \sqrt{3}+1 \right)}{\sqrt{3}-1}.\frac{\left( \sqrt{3}+1 \right)}{\left( \sqrt{3}+1 \right)}\\=\frac{3+1+2\sqrt{3}}{2}\\=\frac{4+2\sqrt{3}}{2}=2+\sqrt{3}\\similarly,\\y=\frac{\sqrt{3}-1}{\sqrt{3}+1}\\=\frac{\left( \sqrt{3}-1 \right)}{\sqrt{3}+1}.\frac{\left( \sqrt{3}-1 \right)}{\left( \sqrt{3}-1 \right)}=\frac{3+1-2\sqrt{3}}{2}=2-\sqrt{3}\\{{x}^{2}}+{{y}^{2}}={{\left( 2+\sqrt{3} \right)}^{2}}{{\left( 2-\sqrt{3} \right)}^{2}}\\=4+3+4\sqrt{3}+4+3-4\sqrt{3}=14\end{array}$

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