यदि (f(x)=\frac{1}{x-2}) और (g(x)=\frac{1}{x+2}) हैं तो ((f-g)(x)) क्या है?
If (f(x)=\frac{1}{x-2}) and (g(x)=\frac{1}{x+2}) then what is ((f-g)(x))?
Explanation opens after your attempt
Correct Answer
A. \(\frac{4}{x^2-4}), \(x\ne \pm 2\)
Concept
((f-g)(x)=\frac{1}{x-2}-\frac{1}{x+2}=\frac{4}{x-2-4}), where \(x\ne \pm 2\). Keep restrictions while multiplying denominators.
Why this answer is correct
The correct answer is A. \(\frac{4}{x^2-4}), \(x\ne \pm 2\). ((f-g)(x)=\frac{1}{x-2}-\frac{1}{x+2}=\frac{4}{x-2-4}), where \(x\ne \pm 2\). Keep restrictions while multiplying denominators.
Exam Tip
((f-g)(x)=\frac{1}{x-2}-\frac{1}{x+2}=\frac{4}{x-2-4}), जहाँ \(x\ne \pm 2\)। हरों का गुणन करते समय प्रतिबंध रखें।
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