Question : If $\left(w+\frac{1}{w}\right)=6$, then what will be the value of $\left(w-\frac{1}{w}\right)?$
Option 1: $4 \sqrt{2}$
Option 2: $\sqrt{2}$
Option 3: $3 \sqrt{2}$
Option 4: $2 \sqrt{2}$
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Correct Answer: $4 \sqrt{2}$
Solution : Given: $\left(w+\frac{1}{w}\right)=6$ Squaring both sides, we have, ⇒ $\left(w+\frac{1}{w}\right)^{2}=6^{2}$ ⇒ $w^{2}+\frac{1}{w^{2}}+2=36$ ⇒ $w^{2}+\frac{1}{w^{2}}=34$ ⇒ $w^{2}+\frac{1}{w^{2}}-2=34-2$ ⇒ $(w-\frac{1}{w})^{2}=32$ ⇒ $(w-\frac{1}{w})=\sqrt{32}$ $\therefore(w-\frac{1}{w})=4\sqrt{2}$ Hence, the correct answer is $4\sqrt{2}$.
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Question : If $x=(\sqrt{6}-1)^{\frac{1}{3}}$, then the value of $\left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right)$ is:
Option 1: $\frac{2 \sqrt{6}-6}{5}$
Option 2: $\frac{4 \sqrt{6}-6}{5}$
Option 3: $\frac{4 \sqrt{6}-6}{3}$
Option 4: $\frac{4 \sqrt{3}-6}{5}$
Question : If $\left(y^2+\frac{1}{y^2}\right)=74$ and $y>1$, then find the value of $\left(y-\frac{1}{y}\right)$.
Option 1: $6 \sqrt{2}$
Option 2: $-2 \sqrt{19}$
Option 3: $2 \sqrt{19}$
Option 4: $-6 \sqrt{2}$
Question : If $a=\frac{1}{a-\sqrt{6}}$ and $(a>0)$, then the value of $\left(a+\frac{1}{a}\right)$ is:
Option 1: $\sqrt{6}$
Option 2: $\sqrt{10}$
Option 3: $\sqrt{15}$
Option 4: $\sqrt{7}$
Question : If $\sin \theta \cos \theta=\frac{\sqrt{2}}{3}$,then the value of $\left(\sin ^6 \theta+\cos ^6 \theta\right)$ is:
Option 1: $\frac{1}{3}$
Option 2: $\frac{4}{3}$
Option 3: $\frac{2}{3}$
Option 4: $\frac{5}{3}$
Question : If $\left(x+\frac{1}{x}\right)=5 \sqrt{2}$, and $x>1$, what is the value of $\left(x^6-\frac{1}{x^6}\right) ?$
Option 1: $22970 \sqrt{23}$
Option 2: $23030 \sqrt{23}$
Option 3: $23060 \sqrt{23}$
Option 4: $22960 \sqrt{23}$
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