Question : The value of $\frac{7+3 \sqrt{5}}{3+\sqrt{5}}-\frac{7-3 \sqrt{5}}{3-\sqrt{5}}$ lies between:
Option 1: 3 and 3.5
Option 2: 2 and 2.5
Option 3: 1.5 and 2
Option 4: 2.5 and 3
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Correct Answer: 3 and 3.5
Solution : Given: $\frac{7+3 \sqrt{5}}{3+\sqrt{5}}-\frac{7-3 \sqrt{5}}{3-\sqrt{5}}$ = $\frac{(7+3 \sqrt{5})(3-\sqrt{5})-(7-3\sqrt5)(3+\sqrt{5})}{(3)^2-(\sqrt{5})^2}$ = $\frac{21+9\sqrt{5}-7\sqrt5-15+21+7\sqrt5-9\sqrt5-15}{9-5}$ = $\frac{6+6}{4}$ = $\frac{12}{4}$ = 3 Thus, 3 lies between 3 and 3.5 Hence, the correct answer is 3 and 3.5
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Question : If $x=\frac{\sqrt{5}+1}{\sqrt{5}-1}$ and $y=\frac{\sqrt{5}-1}{\sqrt{5}+1}$, then the value of $\frac{x^{2}+xy+y^{2}}{x^{2}-xy+y^{2}}$ is:
Option 1: $\frac{3}{4}$
Option 2: $\frac{4}{3}$
Option 3: $\frac{3}{5}$
Option 4: $\frac{5}{3}$
Question : What is the value of $\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}–\sqrt{5}} \div \frac{\sqrt{14}+\sqrt{10}}{\sqrt{14}–\sqrt{10}}+\frac{\sqrt{10}}{\sqrt{5}}$?
Option 1: $\sqrt{2}+2$
Option 2: $2 \sqrt{2}+2$
Option 3: $\sqrt{2}+1$
Option 4: $2 \sqrt{2}+1$
Question : If $\left(x^2+\frac{1}{x^2}\right)=7$, and $0<x<1$, find the value of $x^2-\frac{1}{x^2}$.
Option 1: $3 \sqrt{5}$
Option 2: $4 \sqrt{5}$
Option 3: $-4\sqrt{3}$
Option 4: $-3\sqrt{5}$
Question : If $x+\left [\frac{1}{(x+7)}\right]=0$, what is the value of $x-\left [\frac{1}{(x+7)}\right]$?
Option 1: $3\sqrt{5}$
Option 2: $3\sqrt{5}-7$
Option 3: $3\sqrt{5}+7$
Option 4: $8$
Question : If $3\sqrt{\frac{1-a}{a}}+9=19-3\sqrt{\frac{a}{1-a}};$ then, what is the value of $a?$
Option 1: $\frac{3}{10}$ and $\frac{7}{10}$
Option 2: $\frac{1}{10}$ and $\frac{9}{10}$
Option 3: $\frac{2}{5}$ and $\frac{3}{5}$
Option 4: $\frac{1}{5}$ and $\frac{4}{5}$
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