If $$x + \frac{1}{x} = 3{\text{,}}$$ where $$x \ne 0{\text{,}}$$ then the value of $$\frac{{{x^4} + 3{x^3} + 5{x^2} + 3x + 1}}{{{x^4} + 1}}$$ = ?
A. 3
B. 5
C. 7
D. 2
Answer: Option A
Solution(By Examveda Team)
$$\eqalign{ & x + \frac{1}{x} = 3 \cr & \Rightarrow {x^2} + 1 = 3x\,.....(i) \cr & \Rightarrow {\left( {{x^2} + 1} \right)^2} = {\left( {3x} \right)^2} \cr & \Rightarrow {x^4} + 1 + 2{x^2} = 9{x^2} \cr & \Rightarrow {x^4} + 1 = 7{x^2}\,.....(ii) \cr & \therefore \frac{{{x^4} + 3{x^3} + 5{x^2} + 3x + 1}}{{{x^4} + 1}} \cr & \Rightarrow \frac{{7{x^2} + 3{x^3} + 5{x^2} + 3x}}{{{x^4} + 1}} \cr & \Rightarrow \frac{{12{x^2} + 3{x^3} + 3x}}{{7{x^2}}} \cr & {\text{From equation (i)}} \cr & \Rightarrow \frac{{12x + 3\left( {{x^2} + 1} \right)}}{{7x}} \cr & \Rightarrow \frac{{12x + 3 \times 3x}}{{7x}} \cr & \Rightarrow \frac{{21x}}{{7x}} \cr & \Rightarrow 3 \cr} $$Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
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