Question : If $x=\frac{4\sqrt{15}}{\sqrt{5}+\sqrt{3}}$, the value of $\frac{x+\sqrt{20}}{x–\sqrt{20}}+\frac{x+\sqrt{12}}{x–\sqrt{12}}$ is:
Option 1: $1$
Option 2: $2$
Option 3: $\sqrt{3}$
Option 4: $\sqrt{5}$
New: SSC CHSL tier 1 answer key 2024 out | SSC CHSL 2024 Notification PDF
Recommended: How to crack SSC CHSL | SSC CHSL exam guide
Don't Miss: Month-wise Current Affairs | Upcoming government exams
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: $2$
Solution : Given: $x=\frac{4\sqrt{15}}{\sqrt{5}+\sqrt{3}}=\frac{4\sqrt{5}\sqrt{3}}{\sqrt{5}+\sqrt{3}}$ Now, $\frac{x}{\sqrt{20}}=\frac{4\sqrt{5}\sqrt{3}}{\sqrt{20}(\sqrt{5}+\sqrt{3})} = \frac{2\sqrt{3}}{(\sqrt{5}+\sqrt{3})}$ Using componendo and dividendo $\frac{x+\sqrt{20}}{x-\sqrt{20}}=\frac{\sqrt{5}+3\sqrt{3}}{(\sqrt{3}-\sqrt{5})}$ Similarly, we can find $\frac{x+\sqrt{12}}{x-\sqrt{12}}=\frac{\sqrt{3}+3\sqrt{5}}{(\sqrt{5}-\sqrt{3})}$ Now, $\frac{x+\sqrt{20}}{x–\sqrt{20}}+\frac{x+\sqrt{12}}{x–\sqrt{12}}=\frac{\sqrt{5}+3\sqrt{3}}{(\sqrt{3}-\sqrt{5})}+\frac{\sqrt{3}+3\sqrt{5}}{(\sqrt{5}-\sqrt{3})} =\frac{2\sqrt{3}-2\sqrt{5}}{(\sqrt{3}-\sqrt{5})}$ =2 Hence, the correct answer is $2$.
Candidates can download this e-book to give a boost to thier preparation.
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
Question : If $x=5–\sqrt{21}$, the value of $\frac{\sqrt{x}}{\sqrt{32–2x}–\sqrt{21}}$ is:
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:
Question : If $x=(7+3 \sqrt{5})$, then find the value of $x^2+\frac{1}{x^2}$.
Question : If $x^{4}+\frac{1}{x^{4}}=16$, then what is the value of $x^{2}+\frac{1}{x^{2}}$?
Question : If $x=\sqrt3+\frac{1}{\sqrt3}$, then the value of $(x-\frac{\sqrt{126}}{\sqrt{42}})(x-\frac{1}{x-\frac{2\sqrt3}{3}})$ is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile