यदि (f(x)=\frac{x-2}{x+1}) और (g(x)=\frac{1}{x+1}) हैं, तो ((f-g)(x)) का सरल रूप क्या होगा?
If (f(x)=\frac{x-2}{x+1}) and (g(x)=\frac{1}{x+1}), what is the simplified form of ((f-g)(x))?
Explanation opens after your attempt
Correct Answer
A. \(x-1,\ x\ne-1\)
Concept
(\frac{x-2-1}{x+1}=\frac{(x-1)(x+1)}{x+1}=x-1), but (x=-1) is excluded. Use the original denominator to set restrictions.
Why this answer is correct
The correct answer is A. \(x-1,\ x\ne-1\). (\frac{x-2-1}{x+1}=\frac{(x-1)(x+1)}{x+1}=x-1), but (x=-1) is excluded. Use the original denominator to set restrictions.
Exam Tip
(\frac{x-2-1}{x+1}=\frac{(x-1)(x+1)}{x+1}=x-1), पर (x=-1) हटेगा। मूल हर से प्रतिबंध तय करें।
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