If \( \Large x+\frac{1}{x}=2,\ then\ x- \frac{1}{x} \)is ?

A) 0

B) 1

C) 2

D) -2

Correct answer:
A) 0

Description for Correct answer:
Given that, \( \Large x+\frac{1}{x}=2 \)

On squaring both sides, we get

\( \Large \left(x+\frac{1}{x}\right)^{2}=4 \)

=>\( \Large x^{2}+\frac{1}{x^{2}}+2=4 \)

=>\( \Large x^{2}+\frac{1}{x^{2}}=2 \)

Now, we have

\( \Large \left(x-\frac{1}{x}\right)^{2}= \left(x^{2}+\frac{1}{x^{2}}\right)-2 \)

= 2 - 2 = 0 [from Eq. (ii) ]

Therefore, \( \Large x-\frac{1}{x}=0 \)


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