Question : If $x^4+\frac{1}{x^4}=194, x>0$, then find the value of $x^3+\frac{1}{x^3}+x+\frac{1}{x}$
Option 1: 76
Option 2: 66
Option 3: 56
Option 4: 46
New: SSC CHSL Tier 2 answer key released | 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: 56
Solution : Given: $x^4+\frac{1}{x^4}=194$ $⇒x^4+\frac{1}{x^4}+2=194+2$ $⇒(x^2+\frac{1}{x^2})^2=196$ $⇒x^2+\frac{1}{x^2}=\sqrt{196}$ $⇒x^2+\frac{1}{x^2}=14$ $⇒x^2+\frac{1}{x^2}+2=14+2$ $⇒(x+\frac{1}{x})^2=16$ $⇒x+\frac{1}{x}=4$ -----------------------------(1) Now, $x^3+\frac{1}{x^3}=(x+\frac{1}{x})^3-3×x×\frac{1}{x}(x+\frac{1}{x})$ $⇒x^3+\frac{1}{x^3}=4^3-3 ×4$ $⇒x^3+\frac{1}{x^3}=52$--------------------------------(2) Therefore, $x^3+\frac{1}{x^3}+x+\frac{1}{x}$ = $52+4 = 56$ (from equations (1) and (2)) Hence, the correct answer is 56.
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 $\frac{x^2+1}{x}=5$, then find the value of $x^4+\frac{1}{x^4}-36$.
Question : If $x+\frac{1}{x}=2$, then find the value of $x^{1823}+\frac{1}{x^{1929}}$.
Question : If $x=\frac{1}{x-5}(x>0)$, then the value of $x+\frac{1}{x}$ is:
Question : If $x^2-\sqrt{9.76} x+1=0$ and $x>1$, then the value of $\left(x^3-\frac{1}{x^3}\right)$ is:
Question : If $x>1$ and $x+\frac{1}{x}=2\frac{1}{12}$, then the value of $x^{4}-\frac{1}{x^{4}}$ is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile