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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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