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: $12\sqrt{10}$
Latest: SSC CGL Tier 1 Result 2024 Out | SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL Tier 1 Scorecard 2024 Released | SSC CGL complete guide
Suggested: Month-wise Current Affairs | Upcoming Government Exams
Correct Answer: $-24\sqrt{2}$
Solution : Use (a – b) 2 = (a + b) 2 – 4ab So, $(x^2-\frac{1}{x^2})^2 = (x^2+\frac{1}{x^2})^2 -4$ ⇒ $(x^2-\frac{1}{x^2})^2 = 6^2 -4$ ⇒ $(x^2-\frac{1}{x^2}) = - \sqrt {32}\ \ \ [\because 0<x<1]$ Also, $x^4 - \frac1{x^4} = (x^2+\frac{1}{x^2})(x^2-\frac{1}{x^2})$ ⇒ $x^4 - \frac1{x^4} = - 6 \times \sqrt{32}$ $\therefore x^4 - \frac1{x^4} = - 24 \sqrt{2}$ Hence, the correct answer is $-24\sqrt2$.
Candidates can download this ebook to know all about SSC CGL.
Result | Eligibility | Application | Selection Process | Preparation Tips | Admit Card | Answer Key
Question : If $\left(x^2+\frac{1}{x^2}\right)=7$, and $0<x<1$, find the value of $x^2-\frac{1}{x^2}$.
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 $x>0$?
Question : If $\left(x+\frac{1}{x}\right)=2 \sqrt{2}$ and $x>1$, what is the value of $\left(x^6-\frac{1}{x^6}\right)$?
Question : If $(x+\frac{1}{x})=6$ and $x>1$, find the value of $(x^2–\frac{1}{x^2})$.
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile