Question : If $2x+\frac{1}{2x}=2,$ what is the value of $\sqrt{2\left (\frac{1}{x}\right)^{4}+\left (\frac{1}{x}\right)^{5}}\; ?$
Option 1: 1
Option 2: 2
Option 3: 4
Option 4: 8
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: 8
Solution : Given: $2x+\frac{1}{2x}=2$ $⇒4x^{2}-4x+1=0$ $⇒\left (2x-1 \right)^{2}=0$ $⇒x=\frac{1}{2}$ So, $\sqrt{2\left (\frac{1}{x}\right)^{4}+\left (\frac{1}{x}\right)^{5}}\;=\sqrt{2\left (2\right)^{4}+\left (2\right )^{5}}\;=\sqrt{32+32}\;=8$ Hence, the correct answer is 8.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : If $x+\left [\frac{1}{(x+7)}\right]=0$, what is the value of $x-\left [\frac{1}{(x+7)}\right]$?
Question : If $\left(x+\frac{1}{x}\right)=\sqrt{6}$ and $x>1$, what is the value of $\left(x^8-\frac{1}{x^8}\right)$?
Question : If $\left(x+\frac{1}{x}\right)=5$, and $x>1$, what is the value of $\left(x^8-\frac{1}{x^8}\right)?$
Question : If $\left(x^2 - \frac{1}{x^2}\right) = 4 \sqrt{6}$ and $x>1$, what is the value of $\left(x^3 - \frac{1}{x^3}\right)?$
Question : If $\left(x-\frac{1}{x}\right)^2=12$, what is the value of $\left(x^2-\frac{1}{x^2}\right)$, given that $x>0$?
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile