Question : The value of exponential form of $\sqrt{\sqrt{2}\sqrt{3}}$ is:
Option 1: $6$
Option 2: $6^{\frac{1}{2}}$
Option 3: $6^{-\frac{1}{2}}$
Option 4: $6^{\frac{1}{4}}$
New: SSC CHSL Tier 2 answer key released | 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: $6^{\frac{1}{4}}$
Solution : Given: $\sqrt{\sqrt{2}\sqrt{3}}$ Simplifying this expression, we have, = $\sqrt{\sqrt{6}}$ = $\left (6^{\frac{1}{2}} \right )^\frac{1}{2}$ = $6^{\frac{1}{4}}$ Hence, the correct answer is $6^{\frac{1}{4}}$.
Candidates can download this e-book to give a boost to thier preparation.
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
Question : The simplified value of $\frac{3\sqrt 2 }{\sqrt3 + \sqrt6} - \frac{4 \sqrt 3 }{\sqrt{6}+ \sqrt {2}} + \frac{\sqrt 6}{\sqrt{3}+ \sqrt 2}$ 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 : The value of $\frac{1}{1+\sqrt{2}+\sqrt{3}}+\frac{1}{1-\sqrt{2}+\sqrt{3}}$ is:
Question : If $x=\sqrt{3}-\frac{1}{\sqrt{3}}, y=\sqrt{3}+\frac{1}{\sqrt{3}}$, then the value of $\frac{x^2}{y}+\frac{y^2}{x}$ is:
Question : The value of $x$ in the expression $\tan^{2}\frac{\pi }{4}-\cos^{2}\frac{\pi }{3}=x\sin\frac{\pi }{4}\cos\frac{\pi }{4}\tan\frac{\pi }{3}$ is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile