यदि (f(x)=\frac{x}{x-1}) और (g(x)=\frac{x-1}{x+1}) हों, तो ((fg)(x)) का सरल रूप और प्रतिबंध क्या है?
If (f(x)=\frac{x}{x-1}) and (g(x)=\frac{x-1}{x+1}), what is the simplified form and restriction of ((fg)(x))?
Explanation opens after your attempt
Correct Answer
A. \(\frac{x}{x+1}\), \(x\ne 1,-1\)
Concept
The simplified form is \(\frac{x}{x+1}\), but original denominators keep \(x\ne 1,-1\). Remember restrictions from cancelled factors too.
Why this answer is correct
The correct answer is A. \(\frac{x}{x+1}\), \(x\ne 1,-1\). The simplified form is \(\frac{x}{x+1}\), but original denominators keep \(x\ne 1,-1\). Remember restrictions from cancelled factors too.
Exam Tip
सरल रूप \(\frac{x}{x+1}\) है, पर मूल हरों से \(x\ne 1,-1\) रहेगा। कटे हुए गुणनखंड का प्रतिबंध भी याद रखें।
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