Question : $x$ is a negative number such that $k+k^{-1}=-2$, then what is the value of $\frac{k^2+4 k-2}{k^2+k-5}$?
Option 1: 7
Option 2: 1
Option 3: –7
Option 4: –1
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Correct Answer: 1
Solution : Given: $k+k^{-1}=-2$ ⇒ $k+\frac{1}{k}=-2$ ⇒ $k^2+1=-2k$ ⇒ $k^2+2k+1=0$ ⇒ $(k + 1)^2=0$ $\therefore k=-1$ Putting the value of $k$ in the given equation, $\frac{k^2+4 k-2}{k^2+k-5}$ $=\frac{(-1)^2+4 (-1)-2}{(-1)^2+(-1)-5}$ $=\frac{1-4-2}{1-1-5}$ $=\frac{-5}{-5}$ $=1$ Hence, the correct answer is 1.
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Question : If $(x+\frac{1}{x})=–2$, then the value of $(x^7+\frac{1}{x^7})$ is:
Option 1: 1
Option 2: –1
Option 3: 0
Option 4: –2
Question : If $\frac{1}{6}$ of $x$ – $\frac{7}{2}$ of $\frac{3}{7}$ equals to $(-\frac{7}{4})$, then the value of $x$ is:
Option 1: –1.5
Option 2: –3
Option 3: –2.5
Option 4: 6
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: $9 \frac{3}{4}$
Question : If $(-\frac{1}{2}) \times (x-5) + 3 = -\frac{5}{2}$, then what is the value of $x$?
Option 1: 16
Option 2: 4
Option 3: – 6
Option 4: – 4
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: $\frac{3}{5}$
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