यदि (f(x)=\frac{1}{x-1}) और (g(x)=\frac{1}{x+1}) हैं, तो ((f-g)(x)) का सही रूप क्या है?
If (f(x)=\frac{1}{x-1}) and (g(x)=\frac{1}{x+1}), what is the correct form of ((f-g)(x))?
Explanation opens after your attempt
Correct Answer
A. \(\frac{2}{x^2-1},\ x\ne\pm1\)
Concept
(\frac{1}{x-1}-\frac{1}{x+1}=\frac{x+1-(x-1)}{x-2-1}=\frac{2}{x-2-1}). The zeros \(x=\pm1\) of the denominator are excluded.
Why this answer is correct
The correct answer is A. \(\frac{2}{x^2-1},\ x\ne\pm1\). (\frac{1}{x-1}-\frac{1}{x+1}=\frac{x+1-(x-1)}{x-2-1}=\frac{2}{x-2-1}). The zeros \(x=\pm1\) of the denominator are excluded.
Exam Tip
(\frac{1}{x-1}-\frac{1}{x+1}=\frac{x+1-(x-1)}{x-2-1}=\frac{2}{x-2-1})। हर के शून्य \(x=\pm1\) निषिद्ध हैं।
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