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 2: $2$
Option 3: $2\sqrt{3}$
Option 4: $4$
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Correct Answer: $4$
Solution : $(2+\sqrt{3})a=(2-\sqrt{3})b=1$ $(2+\sqrt{3})a=1$ $⇒ a = \frac{1}{2+\sqrt{3}}$ $⇒ \frac{1}{a} = 2-\sqrt{3}$ Again, $(2-\sqrt{3})b=1$ $⇒ b = \frac{1}{2-\sqrt{3}}$ $⇒ \frac{1}{b} = 2+\sqrt{3}$ $\frac{1}{a} + \frac{1}{b}=2-\sqrt{3}+2+\sqrt{3}$ $\frac{1}{a} + \frac{1}{b}=4$ Hence, the correct answer is $4$.
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Question : If $\sqrt{1+\frac{\sqrt{3}}{2}}-\sqrt{1-\frac{\sqrt{3}}{2}}=c$, then the value of $\mathrm{c}$ is:
Option 1: 1
Option 2: 4
Option 3: 3
Option 4: 2
Question : $\frac{1}{\sqrt a}-\frac{1}{\sqrt b}=0$, then the value of $\frac{1}{a}+\frac{1}{ b}$ is:
Option 1: $\frac{1}{\sqrt{ab}}$
Option 2: $\sqrt{ab}$
Option 3: $\frac{2}{\sqrt{ab}}$
Option 4: $\frac{1}{2\sqrt{ab}}$
Question : If $n= 7+4\sqrt{3}$, the value of $(\sqrt{n}+\frac{1}{\sqrt{n}})$ is:
Option 1: $2\sqrt{3}$
Option 2: $4$
Option 3: $-4$
Option 4: $-2\sqrt{3}$
Question : If $\sin(\theta+30^{\circ})=\frac{3}{\sqrt{12}}$, then the value of $\cos^{2}\theta$ is:
Option 1: $\frac{1}{4}$
Option 2: $\frac{\sqrt{3}}{2}$
Option 3: $\frac{3}{4}$
Option 4: $\frac{1}{2}$
Question : The value of $\frac{1}{4-\sqrt{15}}-\frac{1}{\sqrt{15}-\sqrt{14}}+\frac{1}{\sqrt{14}-\sqrt{13}}-\frac{1}{\sqrt{13}-\sqrt{12}}+\frac{1}{\sqrt{12}-\sqrt{11}}-\frac{1}{\sqrt{11}-\sqrt{10}}+\frac{1}{\sqrt{10}-3}-\frac{1}{3-\sqrt{8}}$ is:
Option 1: $2-2 \sqrt{2}$
Option 2: $4+2 \sqrt{2}$
Option 3: $4-2 \sqrt{2}$
Option 4: $2+2 \sqrt{2}$
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