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If (f(x)=\frac{1}{x-1}) and (g(x)=\frac{1}{x+1}), what is the correct form of ((f-g)(x))?यदि (f(x)=\frac{1}{x-1}) और (g(x)=\frac{1}{x+1}) हैं, तो ((f-g)(x)) का सही रूप क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: (\frac{2}{x^2-1},\ x\ne\pm1)(\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.(\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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