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If $$x + \frac{1}{x} = 2,$$ x ≠ 0, then the value of $${x^2}{\text{ + }}\frac{1}{{{x^3}}}$$ is equal to?
Answer & Solution
Correct Answer:
Option
B
$$\eqalign{
& {\text{ }}x + \frac{1}{x} = 2{\text{, }}\,\,\,x \ne 0 \cr
& {\text{Put }}x = 1 \cr
& 1 + 1 = 2 \cr
& \therefore {x^2}{\text{ + }}\frac{1}{{{x^2}}} \cr
& = 1 + 1 \cr
& = 2 \cr} $$
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