Question : If $x+\frac{1}{x}=3$, then the value of $\frac{3x^{2}-4x+3}{x^{2}-x+1}$ is:
Option 1: $\frac{4}{3}$
Option 2: $\frac{3}{2}$
Option 3: $\frac{5}{2}$
Option 4: $\frac{5}{3}$
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: $\frac{5}{2}$
Solution : Given: $x+\frac{1}{x}=3$ To find: $\frac{3x^{2}-4x+3}{x^{2}-x+1}$ Dividing with $x$ on both numerator and denominator, = $\frac{3x-4+\frac{3}{x}}{x-1+\frac{1}{x}}$ = $\frac{3(x+\frac{1}{x})-4}{(x+\frac{1}{x})-1}$ = $\frac{(3×3)-4}{3-1}$ = $\frac{5}{2}$ Hence, the correct answer is $\frac{5}{2}$.
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 $x+\frac{1}{x}=2$, then find the value of $x^5+\frac{1}{x^5}$.
Question : If $\frac{x}{y}=\frac{4}{5}$, then the value of $(\frac{4}{7}+\frac{2y–x}{2y+x})$ is:
Question : If $x+\frac{2}{x}=1$, then the value of $\frac{x^2+7x+2}{x^2+13x+2}$ is:
Question : If $4(x+5)-3>6-4x\geq x-5$, then the value of $x$ is:
Question : If $2x+\frac{1}{4x}=1$, then the value of $x^{2}+\frac{1}{64x^{2}}$ is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile