Question : If $k^4+\frac{1}{k^4}=47$, then what is the value of $k^3+\frac{1}{k^3}$?
Option 1: 4.5
Option 2: 54
Option 3: 18
Option 4: 9
New: SSC CGL 2025 Tier-1 Result
Latest: SSC CGL complete guide
Suggested: Month-wise Current Affairs | Upcoming Government Exams
Correct Answer: 18
Solution : Given, $k^4+\frac{1}{k^4}=47$ Adding 2 on both sides, ⇒ $k^4+\frac{1}{k^4}+2=47+2$ ⇒ $(k^2)^2+\frac{1}{(k^2)^2}+2\times k^2\times \frac{1}{k^2}=49$ ⇒ $(k^2+\frac{1}{k^2})^2=49$ ⇒ $k^2+\frac{1}{k^2}=7$ Again adding 2 on both sides, $k^2+\frac{1}{k^2}+2=7+2$ ⇒ $(k+\frac{1}{k})^2=9$ ⇒ $k+\frac{1}{k}=3$ cubing both sides, we get, ⇒ $k^3+\frac{1}{k^3}+3×k\times \frac{1}{k}(k+\frac{1}{k})=27$ ⇒ $k^3+\frac{1}{k^3}+3×3=27$ $\therefore k^3+\frac{1}{k^3}=27-9=18$ Hence, the correct answer is 18.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : If ${k}+\frac{1}{{k}}=3$, then what is the value of ${k}^2+\frac{1}{{k}^2}$?
Option 1: 9
Option 2: 6
Option 3: 7
Option 4: 5
Question : If $\frac{x}{2}-\frac{\left [4\left (\frac{15}{2}-\frac{x}{3} \right ) \right ]}{3} = –\frac{x}{18}$ then what is the value of $x$?
Option 1: –$10$
Option 2: $\frac{9}{8}$
Option 3: $10$
Option 4: $–\frac{9}{8}$
Question : If $2 x^2-7 x+5=0$, then what is the value of $x^2+\frac{25}{4 x^2} ?$
Option 1: $5 \frac{1}{2}$
Option 2: $7 \frac{1}{4}$
Option 3: $9 \frac{1}{2}$
Option 4: $9 \frac{3}{4}$
Question : If $\sin\theta=\frac{9}{41}$, $0^{\circ}<\theta<90^{\circ}$. Then, what is the value of $\cot \theta $?
Option 1: $\frac{39}{9}$
Option 2: $\frac{47}{8}$
Option 3: $\frac{35}{8}$
Option 4: $\frac{40}{9}$
Question : If $(x-\frac{1}{3})^2+(y-4)^2=0$, then what is the value of $\frac{y+x}{y-x}$?
Option 1: $\frac{11}{13}$
Option 2: $\frac{13}{11}$
Option 3: $\frac{16}{9}$
Option 4: $\frac{9}{16}$
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile