Question : If $x+ \frac{1}{9x}=4$, then the value of $9x^2+ \frac{1}{9x^2}$ is:
Option 1: 140
Option 2: 142
Option 3: 144
Option 4: 146
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Correct Answer: 142
Solution : Given: $x+ \frac{1}{9x}=4$ Multiply by 3 on both sides of the above equation, we get, ⇒ $3x+ \frac{1}{3x}=12$ Squaring both sides of the above equation, we get, ⇒ $(3x+ \frac{1}{3x})^2=12^2$ ⇒ $9x^2+ \frac{1}{9x^2}+2=144$ ⇒ $9x^2+ \frac{1}{9x^2}=144–2$ ⇒ $9x^2+ \frac{1}{9x^2}=142$ Hence, the correct answer is 142.
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Question : If $x=2+\sqrt3$, then the value of $\frac{x^{2}-x+1}{x^{2}+x+1}$ is:
Option 1: $\frac{2}{3}$
Option 2: $\frac{3}{4}$
Option 3: $\frac{4}{5}$
Option 4: $\frac{3}{5}$
Question : If $x+\frac{1}{x}=1$, then the value of $\frac{x^2+7x+1}{x^2+11x+1}$:
Option 1: $\frac{3}{4}$
Option 2: $\frac{2}{3}$
Option 3: $\frac{1}{3}$
Option 4: $\frac{1}{4}$
Question : If $(x+\frac{1}{x})$ = 5, then the value of $\frac{5x}{x^{2}+5x+1}$ is:
Option 1: $\frac{1}{3}$
Option 2: $\frac{1}{4}$
Option 3: $\frac{1}{2}$
Option 4: $\frac{1}{5}$
Question : If $x+\frac{1}{x}=-14$, and $x<-1$, what will be the value of $x^2-\frac{1}{x^2}$?
Option 1: $-112 \sqrt{3}$
Option 2: $112 \sqrt{3}$
Option 3: $-140 \sqrt{2}$
Option 4: $140 \sqrt{2}$
Question : If $x+\frac{1}{x}=6$, then find the value of $\frac{3 x}{2 x^2-5 x+2}$.
Option 1: $1$
Option 2: $\frac{3}{7}$
Option 3: $\frac{2}{3}$
Option 4: $0$
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