Question : If $(x+\frac{1}{x})=6$ and $x>1$, find the value of $(x^2–\frac{1}{x^2})$.
Option 1: $18 \sqrt{2}$
Option 2: $30 \sqrt{2}$
Option 3: $24 \sqrt{2}$
Option 4: $12 \sqrt{10}$
New: SSC CGL 2025 Tier-1 Result
Latest: SSC CGL complete guide
Suggested: Month-wise Current Affairs | Upcoming Government Exams
Correct Answer: $24 \sqrt{2}$
Solution : Given: $(x+\frac{1}{x})=6$, and $x>1$. We know, $(x–\frac{1}{x})^2=(x+\frac{1}{x})^2–4×x×\frac{1}{x}$ ⇒ $(x–\frac{1}{x})^2=(6)^2–4$ ⇒ $(x–\frac{1}{x})=\sqrt{32}=4\sqrt2$ ⇒ $x^2–\frac{1}{x^2}=(x+\frac{1}{x})(x–\frac{1}{x})$ ⇒ $x^2–\frac{1}{x^2}=(6)(4\sqrt2)=24\sqrt2$ Hence, the correct answer is $24\sqrt2$.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : If $\left(x^2 - \frac{1}{x^2}\right) = 4 \sqrt{6}$ and $x>1$, what is the value of $\left(x^3 - \frac{1}{x^3}\right)?$
Option 1: $20 \sqrt{2}$
Option 2: $24 \sqrt{2}$
Option 3: $18 \sqrt{2}$
Option 4: $22 \sqrt{2}$
Question : If $\left(x^2+\frac{1}{x^2}\right)=6$ and $0<x<1$, what is the value of $x^4-\frac{1}{x^4}$?
Option 1: $24\sqrt{2}$
Option 2: $-24\sqrt{2}$
Option 3: $-12\sqrt{10}$
Option 4: $12\sqrt{10}$
Question : If $\left(x+\frac{1}{x}\right)=2 \sqrt{2}$ and $x>1$, what is the value of $\left(x^6-\frac{1}{x^6}\right)$?
Option 1: $140\sqrt{2}$
Option 2: $116\sqrt{2}$
Option 3: $144\sqrt{2}$
Option 4: $128\sqrt{2}$
Question : If $x^{2} + \frac{1}{x^{2}} = 18$ and $x > 0$, what is the value of $x^{3} + \frac{1}{x^{3}}?$
Option 1: $36 \sqrt{5}$
Option 2: $40 \sqrt{5}$
Option 3: $46 \sqrt{5}$
Option 4: $34 \sqrt{5}$
Question : If $\cos27^{\circ}$ = $x$, then the value of $\tan 63°$ is:
Option 1: $\frac{x}{\sqrt{1–x^2}}$
Option 2: $\frac{x}{\sqrt{1+x^2}}$
Option 3: $\frac{\sqrt{1–x^2}}{x}$
Option 4: $\frac{\sqrt{1+x^2}}{x}$
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile