Question : If $x^{2}+1=2x$, then the value of $\frac{x^{4}+\frac{1}{x^{2}}}{x^{2}-3x+1}$ is:
Option 1: 0
Option 2: 1
Option 3: 2
Option 4: –2
Correct Answer: –2
Solution : $x^{2}+1=2x$ $⇒x+\frac{1}{x}= 2$----------(i) Now, $\frac{x^{4}+\frac{1}{x^{2}}}{x^{2}-3x+1}$ = $\frac{x(x^{3}+\frac{1}{x^{3}})}{x(x+\frac{1}{x}-3)}$ = $\frac{(x+\frac{1}{x})^3-3(x+\frac{1}{x})}{x+\frac{1}{x}-3}$ Substituting the value of equation (i), we get, = $\frac{2^3-3×2}{2-3}$ = $-2$ Hence, the correct answer is –2.
Result | Eligibility | Application | Selection Process | Cutoff | Admit Card | Preparation Tips
Question : If $\frac{3x-1}{x}+\frac{5y-1}{y}+\frac{7z-1}{z}=0$, what is the value of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}?$
Question : If for a non-zero $x$, $3x^{2}+5x+3=0,$ then the value of $x^{3}+\frac{1}{x^{3}}$ is:
Question : if $x+\frac{1}{x}=2$, then the value of $x^4+\frac{1}{x^4}$=__________.
Question : If $x=\frac{8ab}{a+b}(a\neq b),$ then the value of $\frac{x+4a}{x–4a}+\frac{x+4b}{x–4b}$ is:
Question : If $x^2-3x+1=0$, what is the value of $(x^4+\frac{1}{x^4})$?
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile