Question : If $\frac {x^2+3x+1}{x^2–3x+1}=\frac{1}{2 }$, then the value of $(x+\frac{1}{x})$ is:
Option 1: 9
Option 2: –9
Option 3: 1
Option 4: 2
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: –9
Solution : Given: $\frac {x^2+3x+1}{x^2–3x+1}=\frac{1}{2}$ Dividing both the numerator and the denominator by $x$ on the left-hand side, we get, $\frac{(x+3+\frac{1}{x})}{(x–3+\frac{1}{x})}=\frac{1}{2}$ ⇒ $\frac{(x+\frac{1}{x})+3}{(x+\frac{1}{x})–3}=\frac{1}{2}$ ⇒ $2(x+\frac{1}{x})+6={(x+\frac{1}{x})-3}$ ⇒ $(x+\frac{1}{x})=-3-6$ ⇒ $(x+\frac{1}{x})=-9$ Hence, the correct answer is –9.
Candidates can download this ebook to know all about SSC CGL.
Admit Card | Eligibility | Application | Selection Process | Preparation Tips | Result | Answer Key
Question : If $(x-\frac{1}{3x})=\frac{1}{3}$, then the value of $3(x-\frac{1}{3x})$ is:
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 $x$?
Question : If $x^{2} -3x +1=0$, then the value of $\frac{\left(x^4+\frac{1}{x^2}\right)}{\left(x^2+5 x+1\right)}$ is:
Question : If $x+\frac{1}{x}=17,$ what is the value of $\frac{x^{4}+\frac{1}{x^{2}}}{x^{2}-3x+1}\; ?$
Question : If $(x+\frac{1}{x})^{2}=3$, then the value of $(x^{3}+\frac{1}{x^{3}})$ is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile