Question : If $x^2+\frac{1}{x^2}=66$, the value of $x-\frac{1}{x}$ is:
Option 1: 10
Option 2: 8
Option 3: 9
Option 4: 6
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Correct Answer: 8
Solution : Given, $x^2+\frac{1}{x^2}=66$ ⇒ $x^2+\frac{1}{x^2} -2=66-2$ ⇒ $x^2+\frac{1}{x^2} -2(x)(\frac{1}{x})=64$ We know that $(a-b)^2 = a^2+b^2-2ab$ So, $(x-\frac{1}{x})^2=8^2$ ⇒ $(x-\frac{1}{x})=8$ Hence, the correct answer is 8.
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Question : If $\frac{x}{2}-\frac{\left [4\left (\frac{15}{2}-\frac{x}{3} \right ) \right ]}{3} = –\frac{x}{18}$ then what is the value of $x$?
Option 1: –$10$
Option 2: $\frac{9}{8}$
Option 3: $10$
Option 4: $–\frac{9}{8}$
Question : If $x-\frac{1}{x}=5, x \neq 0$, then what is the value of $\frac{x^6+3 x^3-1}{x^6-8 x^3-1} ?$
Option 1: $\frac{3}{8}$
Option 2: $\frac{13}{12}$
Option 3: $\frac{4}{9}$
Option 4: $\frac{11}{13}$
Question : If $x=8(\sin \theta+\cos \theta)$ and $y=9(\sin \theta-\cos \theta)$, then the value of $\frac{x^2}{8^2}+\frac{y^2}{9^2}$ is:
Option 1: 4
Option 2: 6
Option 3: 8
Option 4: 2
Question : If $x^{2} -3x +1=0$, then the value of $\frac{\left(x^4+\frac{1}{x^2}\right)}{\left(x^2+5 x+1\right)}$ is:
Option 1: $\frac{9}{4}$
Option 2: $\frac{27}{8}$
Option 3: $\frac{5}{2}$
Option 4: $2$
Question : If $2 x^2-7 x+5=0$, then what is the value of $x^2+\frac{25}{4 x^2} ?$
Option 1: $5 \frac{1}{2}$
Option 2: $7 \frac{1}{4}$
Option 3: $9 \frac{1}{2}$
Option 4: $9 \frac{3}{4}$
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