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: $\frac{1}{2\sqrt{ab}}$
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: $\frac{2}{\sqrt{ab}}$
Solution : Given: $\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}=0$ ⇒ $(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}})^2=0^2$ ⇒ $(\frac{1}{{a}}+\frac{1}{{b}}-\frac{2}{\sqrt{ab}})=0$ $\therefore (\frac{1}{{a}}+\frac{1}{{b}})=\frac{2}{\sqrt{ab}}$ Hence, the correct answer is $\frac{2}{\sqrt{ab}}$.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : If $x=\frac{4\sqrt{ab}}{\sqrt a+ \sqrt b}$, then what is the value of $\frac{x+2\sqrt{a}}{x-2\sqrt a}+\frac{x+2\sqrt{b}}{x-2\sqrt b}$(when $a\neq b$)?
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:
Question : If $c+ \frac{1}{c} =\sqrt{3}$, then the value of $c^{3}+ \frac{1}{c^{3}}$ is equal to:
Question : If $(2+\sqrt{3})a=(2-\sqrt{3})b=1$, then the value of $\frac{1}{a}+\frac{1}{b}$ is:
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:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile