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## Algebra: Polynomials Test-2

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Question 1 |

$ \displaystyle If\,\,\,47.2506=4a+\frac{7}{b}+2c+\frac{5}{c}+6e,\,$
then the value of $ \displaystyle 5a+3b+6c+c+3e\,\,is$

53.6003 | |

53.603 | |

153.6003 | |

213.0003 |

Question 1 Explanation:

Matching the right and left hand sides

We get,40=4a

a = 10

7=7b

b = 1

2c = 2/10 = c = 0.1

5/d = 5/100 = d = 100

6e = 6/10000 = e = 0.0001

$ \displaystyle 5a+3b+6c+d+3e$

=50 + 3 + 0.6 + 100 +0.0003=153.6003

We get,40=4a

a = 10

7=7b

b = 1

2c = 2/10 = c = 0.1

5/d = 5/100 = d = 100

6e = 6/10000 = e = 0.0001

$ \displaystyle 5a+3b+6c+d+3e$

=50 + 3 + 0.6 + 100 +0.0003=153.6003

Question 2 |

$ \displaystyle If\,\,\,x=7-4\sqrt{3\,}\,\,then\,\,\sqrt{x}+\frac{1}{\sqrt{x}}$

1 | |

2 | |

3 | |

4 |

Question 2 Explanation:

$ \displaystyle \sqrt{x}+\frac{1}{\sqrt{x}}$=4

$ \begin{array}{l}{{\left( \sqrt{x}+\frac{1}{\sqrt{x}} \right)}^{2}}=x+\frac{1}{x}+2\\=7-4\sqrt{3}+\frac{1}{7-4\sqrt{3}}+2\\=7-4\sqrt{3}+\frac{7+4\sqrt{3}}{\left( 7-4\sqrt{3} \right)\left( 7+4\sqrt{3} \right)}+2\\=7-4\sqrt{3}+\frac{7+4\sqrt{3}}{49-48}+2\\=16\\\sqrt{x}+\frac{1}{\sqrt{x}}=4\end{array}$

$ \begin{array}{l}{{\left( \sqrt{x}+\frac{1}{\sqrt{x}} \right)}^{2}}=x+\frac{1}{x}+2\\=7-4\sqrt{3}+\frac{1}{7-4\sqrt{3}}+2\\=7-4\sqrt{3}+\frac{7+4\sqrt{3}}{\left( 7-4\sqrt{3} \right)\left( 7+4\sqrt{3} \right)}+2\\=7-4\sqrt{3}+\frac{7+4\sqrt{3}}{49-48}+2\\=16\\\sqrt{x}+\frac{1}{\sqrt{x}}=4\end{array}$

Question 3 |

$ \displaystyle If\,\,\,\frac{x}{y}=\frac{3}{4},\,$
the value of $ \displaystyle \frac{6}{7}+\frac{y-x}{y+x}\,\,is:$

1 | |

$ \displaystyle \frac{2}{7}$ | |

$ \displaystyle \frac{3}{7}$ | |

$ \displaystyle 1\frac{3}{7}$ |

Question 3 Explanation:

$ \begin{array}{l}let,x=3k\\then\,y=4k\\\frac{y-x}{y+x}=\frac{4k-3k}{4k+3k}=\frac{k}{7k}=\frac{1}{7}\\=>\frac{6}{7}+\frac{y-x}{y+x}=1\end{array}$

Question 4 |

$latex \displaystyle If\,\,\,a=\frac{\sqrt{5}+1}{\sqrt{5}-1}\,\,\,and\,\,\,b\,=\frac{\sqrt{5}-1}{\sqrt{5}+1},$
then the value of $ \displaystyle \frac{{{\left( a+b \right)}^{2}}-ab}{{{\left( a+b \right)}^{2}}-3ab}$

$ \displaystyle \frac{3}{4}$ | |

$ \displaystyle \frac{4}{3}$ | |

$ \displaystyle \frac{3}{5}$ | |

$ \displaystyle \frac{5}{3}$ |

Question 4 Explanation:

$ \displaystyle \begin{array}{l}a=\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{\sqrt{5}+1}{\sqrt{5}-1}\times \frac{\sqrt{5}+1}{\sqrt{5}+1}\\=\frac{{{\left( \sqrt{5}+1 \right)}^{2}}}{5-1}=\frac{5+1+2\sqrt{5}}{4}\\=\frac{3+\sqrt{5}}{2}\\b=\frac{\sqrt{5}-1}{2}=\frac{3-\sqrt{5}}{2}\\Therefore\,\,\,a+b\\=\frac{3+\sqrt{5}}{2}+\frac{3-\sqrt{5}}{2}=3\\and\,\,\,ab=\frac{\sqrt{5}+1}{\sqrt{5}-1}\times \frac{\sqrt{5}-1}{\sqrt{5}+1}\\Therefore\,\,\,\exp ression\\=\frac{{{a}^{2}}+ab+{{b}^{2}}}{{{a}^{2}}-ab+{{b}^{2}}}=\frac{{{\left( a+b \right)}^{2}}-ab}{{{\left( a+b \right)}^{2}}-3ab}\\=\frac{9-1}{9-3}=\frac{8}{3}=\frac{4}{3}\end{array}$

Question 5 |

$ \displaystyle \begin{array}{l}If\,\,x=\frac{\sqrt{3}}{2},\,\,\,then\,\,\,\,\\\frac{\sqrt{1+x}}{1+\sqrt{x+1}}+\frac{\sqrt{1-x}}{1-\sqrt{1-x}}\,\,\\is\,\,\,equal\,\,\,to\end{array}$

1 | |

$ \displaystyle 2/\sqrt{3}$ | |

$ \displaystyle 2-\sqrt{3}$ | |

2 |

Question 5 Explanation:

$ \displaystyle \begin{array}{l}=\,\frac{\sqrt{1+x}}{1+\sqrt{x+1}}+\frac{\sqrt{1-x}}{1-\sqrt{1-x}}\,\,\\=\,\frac{\sqrt{1+x}}{1+\sqrt{x+1}}\times \frac{1-\sqrt{1+x}}{1-\sqrt{1+x}}\,\\+\,\frac{\sqrt{1-x}}{1-\sqrt{x-1}}\times \frac{1+\sqrt{1-x}}{1+\sqrt{1-x}}\,\,\,\\=\frac{\sqrt{1+x}-1-x}{1-1-x}\\+\frac{\sqrt{1-x}+1-x}{1-1+x}\\=\frac{\sqrt{1-x}+1-x}{x}\\-\frac{\sqrt{1+x}-1-x}{x}\\=\frac{\sqrt{1-x}+1-x-\sqrt{1+x}+1+x}{x}\\=\frac{2+\sqrt{1-x}-\sqrt{1+x}}{x}\\=\frac{2+\sqrt{1-\frac{\sqrt{3}}{2}}-\sqrt{1+\frac{\sqrt{3}}{2}}}{\frac{\sqrt{3}}{2}}\\=\frac{2+\sqrt{\frac{2-\sqrt{3}}{2}}-\sqrt{\frac{2+\sqrt{3}}{2}}}{\frac{\sqrt{3}}{2}}\\=\frac{2+\sqrt{\frac{4-2\sqrt{3}}{2}}-\sqrt{\frac{4+2\sqrt{3}}{2}}}{\frac{\sqrt{3}}{2}}\\=\frac{4+\sqrt{3}-1-\sqrt{3}-1}{\sqrt{3}}\\=\frac{2}{\sqrt{3}}\end{array}$

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