यदि (f(x)=\frac{x-2+1}{x}) और (g(x)=\frac{x-2-1}{x}) हैं, तो ((f-g)(x)) क्या है?

If (f(x)=\frac{x-2+1}{x}) and (g(x)=\frac{x-2-1}{x}), what is ((f-g)(x))?

Explanation opens after your attempt
Correct Answer

A. \(\frac{2}{x}\), \(x\ne 0\)

Step 1

Concept

With a common denominator, (f-g=\frac{x-2+1-\(x^2-1\)}{x}=\frac{2}{x}), where \(x\ne 0\). In subtraction, both signs of the second numerator change.

Step 2

Why this answer is correct

The correct answer is A. \(\frac{2}{x}\), \(x\ne 0\). With a common denominator, (f-g=\frac{x-2+1-\(x^2-1\)}{x}=\frac{2}{x}), where \(x\ne 0\). In subtraction, both signs of the second numerator change.

Step 3

Exam Tip

समान हर से (f-g=\frac{x-2+1-\(x^2-1\)}{x}=\frac{2}{x}), जहां \(x\ne 0\)। घटाव में दूसरे अंश के दोनों चिन्ह बदलते हैं।

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Mathematics Answer, Explanation and Revision Hints

यदि (f(x)=\frac{x-2+1}{x}) और (g(x)=\frac{x-2-1}{x}) हैं, तो ((f-g)(x)) क्या है? / If (f(x)=\frac{x-2+1}{x}) and (g(x)=\frac{x-2-1}{x}), what is ((f-g)(x))?

Correct Answer: A. \(\frac{2}{x}\), \(x\ne 0\). Explanation: समान हर से (f-g=\frac{x-2+1-\(x^2-1\)}{x}=\frac{2}{x}), जहां \(x\ne 0\)। घटाव में दूसरे अंश के दोनों चिन्ह बदलते हैं। / With a common denominator, (f-g=\frac{x-2+1-\(x^2-1\)}{x}=\frac{2}{x}), where \(x\ne 0\). In subtraction, both signs of the second numerator change.

Which concept should I revise for this Mathematics MCQ?

With a common denominator, (f-g=\frac{x-2+1-\(x^2-1\)}{x}=\frac{2}{x}), where \(x\ne 0\). In subtraction, both signs of the second numerator change.

What exam hint can help solve this Mathematics question?

समान हर से (f-g=\frac{x-2+1-\(x^2-1\)}{x}=\frac{2}{x}), जहां \(x\ne 0\)। घटाव में दूसरे अंश के दोनों चिन्ह बदलते हैं।