Question : If $a = 64$ and $b = 289$, then the value of $\sqrt{\sqrt{\sqrt{a}+\sqrt{b}}-\sqrt{\sqrt{b}-\sqrt{a}}}$ is:
Option 1: $2^{\frac{1}{2}}$
Option 2: $2$
Option 3: $4$
Option 4: $–1$
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Correct Answer: $2^{\frac{1}{2}}$
Solution : Given: $a = 64$ and $b = 289$ The given expression is $\sqrt{\sqrt{\sqrt{a}+\sqrt{b}}-\sqrt{\sqrt{b}-\sqrt{a}}}$. Substitute the value of $a = 64$ and $b = 289$ in the above equation, we get, $=\sqrt{\sqrt{\sqrt{64}+\sqrt{289}}-\sqrt{\sqrt{289} - \sqrt{64}}}$ $=\sqrt{\sqrt{8+17}-\sqrt{17-8}}$ $=\sqrt{\sqrt{25}-\sqrt{9}}$ $=\sqrt{5-3}$ $=2^{\frac{1}{2}}$ Hence, the correct answer is $2^{\frac{1}{2}}$.
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Question : If $2x=\sqrt{a}+\frac{1}{\sqrt{a}}, a>0$, then the value of $\frac{\sqrt{x^{2}-1}}{x-\sqrt{x^{2}-1}}$ is:
Option 1: $a+1$
Option 2: $\frac{1}{2}(a+1)$
Option 3: $\frac{1}{2}(a–1)$
Option 4: $a–1$
Question : If $(2+\sqrt{3})a=(2-\sqrt{3})b=1$, then the value of $\frac{1}{a}+\frac{1}{b}$ is:
Option 1: $1$
Option 3: $2\sqrt{3}$
Option 4: $4$
Question : If $\text{a}= {\sqrt{2}+1}$ and $\text{b} = {\sqrt{2}–1}$, then the value of $\frac{1}{a+1}+\frac{1}{b+1}$ will be:
Option 1: 0
Option 2: 1
Option 3: 2
Option 4: –1
Question : If $a=\frac{1}{a - 5}(a>0)$, then the value of $a+\frac{1}{a}$ is:
Option 1: $\sqrt{29}$
Option 2: $–\sqrt{27}$
Option 3: $-\sqrt{29}$
Option 4: $\sqrt{27}$
Question : The value of $5–\frac{8+2\sqrt{15}}{4}–\frac{1}{8+2\sqrt{15}}$ is equal to:
Option 1: $\frac{1}{4}$
Option 2: $1$
Option 3: $\frac{2}{3}$
Option 4: $\frac{1}{2}$
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