# Algebra: Polynomials Test-2

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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 $\displaystyle 5a+3b+6c+c+3e\,\,is$
 A 53.6003 B 53.603 C 153.6 D 213
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
 Question 2
$\displaystyle If\,\,\,x=7-4\sqrt{3\,}\,\,then\,\,\sqrt{x}+\frac{1}{\sqrt{x}}$
 A 1 B 2 C 3 D 4
Question 2 Explanation:
$\displaystyle \sqrt{x}+\frac{1}{\sqrt{x}}$=4
$\begin{array}{l}{{\left( \sqrt{x}+\frac{1}{\sqrt{x}} \right)}^{2}}=x+\frac{1}{x}+2\\=7-4\sqrt{3}+\frac{1}{7-4\sqrt{3}}+2\\=7-4\sqrt{3}+\frac{7+4\sqrt{3}}{\left( 7-4\sqrt{3} \right)\left( 7+4\sqrt{3} \right)}+2\\=7-4\sqrt{3}+\frac{7+4\sqrt{3}}{49-48}+2\\=16\\\sqrt{x}+\frac{1}{\sqrt{x}}=4\end{array}$
 Question 3
$\displaystyle If\,\,\,\frac{x}{y}=\frac{3}{4},\,$ the value of $\displaystyle \frac{6}{7}+\frac{y-x}{y+x}\,\,is:$
 A 1 B $\displaystyle \frac{2}{7}$ C $\displaystyle \frac{3}{7}$ D $\displaystyle 1\frac{3}{7}$
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 $\displaystyle \frac{{{\left( a+b \right)}^{2}}-ab}{{{\left( a+b \right)}^{2}}-3ab}$
 A $\displaystyle \frac{3}{4}$ B $\displaystyle \frac{4}{3}$ C $\displaystyle \frac{3}{5}$ D $\displaystyle \frac{5}{3}$
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
$\displaystyle \begin{array}{l}If\,\,x=\frac{\sqrt{3}}{2},\,\,\,then\,\,\,\,\\\frac{\sqrt{1+x}}{1+\sqrt{x+1}}+\frac{\sqrt{1-x}}{1-\sqrt{1-x}}\,\,\\is\,\,\,equal\,\,\,to\end{array}$
 A 1 B $\displaystyle 2/\sqrt{3}$ C $\displaystyle 2-\sqrt{3}$ D 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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