Question : If $K+\frac{1}{K}=-3$, then what is the value of $\left(\frac{K^6+1}{K^3}\right)+\left(\frac{K^4+1}{K^2}\right)$?
Option 1: 27
Option 2: – 29
Option 3: 29
Option 4: – 27
New: SSC CHSL tier 1 answer key 2024 out | SSC CHSL 2024 Notification PDF
Recommended: How to crack SSC CHSL | SSC CHSL exam guide
Don't Miss: Month-wise Current Affairs | Upcoming government exams
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: – 29
Solution : Given equation, $K+\frac{1}{K}=-3$ ..........(1) Squaring the equation, we get, $(K+\frac{1}{K})^2=(-3)^2$ $K^2 + \frac{1}{K^2} + 2 = 9$ ⇒ $K^2 + \frac{1}{K^2} = 7$ ...........(2) Cubing equation (1), we get $(K+\frac{1}{K})^3 = (-3)^3$ ⇒ $K^3 + \frac{1}{K^3} + 3\cdot K\cdot\frac{1}{K}(K+\frac{1}{K}) = -27$ ⇒ $K^3+\frac{1}{K^3}+3(-3) = -27$ [Putting value from equation (1)] ⇒ $K^3+\frac{1}{K^3} = -36$ ..........(3) Now consider, $\left(\frac{K^6+1}{K^3}\right)+\left(\frac{K^4+1}{K^2}\right)$ $= \frac{K^6}{K^3}+\frac{1}{K^3} + \frac{K^4}{K^2}+\frac{1}{K^2}$ $= K^3+\frac{1}{K^3} + K^2+\frac{1}{K^2}$ $=- 36 + 7 = -29$ Hence, the correct answer is – 29.
Candidates can download this e-book to give a boost to thier preparation.
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
Question : If $\left(\mathrm{k}+\frac{1}{\mathrm{k}}\right)^2=9$, then what is the value of $\mathrm{k}^3+\frac{1}{\mathrm{k}^3} ?$
Question : What is the value of $\left(k-\frac{1}{k}\right)\left(k^2+\frac{1}{k^2}\right)\left(k^4+\frac{1}{k^4}\right)\left(k^8+\frac{1}{k^8}\right)\left(k^{16}+\frac{1}{k^{16}}\right) ?$
Question : If $x=(\sqrt{6}-1)^{\frac{1}{3}}$, then the value of $\left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right)$ is:
Question : If $\left(a+\frac{1}{a}\right)=6$, then what is the value of $\frac{3}{4}\left(a^2+\frac{1}{a^2}\right)$?
Question : What is the value of $\left(\mathrm{k}+\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}-\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}^2+\frac{1}{\mathrm{k}^2}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)$?
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile