Question : If $x=\sqrt{–\sqrt{3}+\sqrt{3+8 \sqrt{7+4 \sqrt{3}}}}$ where $x > 0$, then the value of $x$ is equal to:
Option 1: 3
Option 2: 4
Option 3: 1
Option 4: 2
New: SSC CGL 2025 Tier-1 Result
Latest: SSC CGL complete guide
Suggested: Month-wise Current Affairs | Upcoming Government Exams
Correct Answer: 2
Solution : Given: $x=\sqrt{–\sqrt{3}+\sqrt{3+8 \sqrt{7+4 \sqrt{3}}}}$ where $x > 0$. Use the algebraic identity, $(a+b)^2=a^2+b^2+2ab$. $⇒x=\sqrt{–\sqrt{3}+\sqrt{3+8 \sqrt{4+3+4 \sqrt{3}}}}$ $⇒x=\sqrt{–\sqrt{3}+\sqrt{3+8 \sqrt{(2+\sqrt{3}}})^2}$ $⇒x=\sqrt{–\sqrt{3}+\sqrt{3+8 (2+\sqrt{3}})}$ $⇒x=\sqrt{–\sqrt{3}+\sqrt{3+16+8\sqrt{3}}}$ $⇒x=\sqrt{–\sqrt{3}+\sqrt{(4+\sqrt{3}})^2}$ $⇒x=\sqrt{–\sqrt{3}+4+\sqrt{3}}$ $\therefore x=\sqrt4=2$ Hence, the correct answer is 2.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : If $x^2+\frac{1}{x^2}=7$, then the value of $x^3+\frac{1}{x^3}$ where x > 0 is equal to:
Option 1: 18
Option 2: 12
Option 3: 15
Option 4: 16
Question : If $x^2-\sqrt{7} x+1=0$, then what is the value of $x^5+\frac{1}{x^5} ?$
Option 1: $19 \sqrt{7}$
Option 2: $21 \sqrt{7}$
Option 3: $25 \sqrt{7}$
Option 4: $27 \sqrt{7}$
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:
Option 1: $a+1$
Option 2: $\frac{1}{2}(a+1)$
Option 3: $\frac{1}{2}(a–1)$
Option 4: $a–1$
Question : If $x=\frac{1}{x-3},(x>0)$, then the value of $x+\frac{1}{x}$ is:
Option 1: $\sqrt{11}$
Option 2: $\sqrt{17}$
Option 3: $\sqrt{15}$
Option 4: $\sqrt{13}$
Question : If $\frac{22 \sqrt{2}}{4 \sqrt{2}-\sqrt{3+\sqrt{5}}}=a+\sqrt{5} b$, with $a, b>0$, then what is the value of $(a b):(a+b)$?
Option 1: 7 : 8
Option 2: 7 : 4
Option 3: 4 : 7
Option 4: 8 : 7
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile