यदि (f(x)=\frac{1}{x+1}) और (g(x)=\frac{1}{x-1}) हों, तो ((f+g)(x)) का सरल रूप क्या है?

If (f(x)=\frac{1}{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. \(\frac{2x}{x^2-1},\ x\ne \pm1\)

Step 1

Concept

With common denominator ((x+1)(x-1)), the numerator is ((x-1)+(x+1)=2x). Do not forget excluded denominator values.

Step 2

Why this answer is correct

The correct answer is A. \(\frac{2x}{x^2-1},\ x\ne \pm1\). With common denominator ((x+1)(x-1)), the numerator is ((x-1)+(x+1)=2x). Do not forget excluded denominator values.

Step 3

Exam Tip

समान हर ((x+1)(x-1)) लेने पर अंश ((x-1)+(x+1)=2x) मिलता है। हर के शून्य मान हटाना न भूलें।

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Mathematics Answer, Explanation and Revision Hints

यदि (f(x)=\frac{1}{x+1}) और (g(x)=\frac{1}{x-1}) हों, तो ((f+g)(x)) का सरल रूप क्या है? / If (f(x)=\frac{1}{x+1}) and (g(x)=\frac{1}{x-1}), what is the simplified form of ((f+g)(x))?

Correct Answer: A. \(\frac{2x}{x^2-1},\ x\ne \pm1\). Explanation: समान हर ((x+1)(x-1)) लेने पर अंश ((x-1)+(x+1)=2x) मिलता है। हर के शून्य मान हटाना न भूलें। / With common denominator ((x+1)(x-1)), the numerator is ((x-1)+(x+1)=2x). Do not forget excluded denominator values.

Which concept should I revise for this Mathematics MCQ?

With common denominator ((x+1)(x-1)), the numerator is ((x-1)+(x+1)=2x). Do not forget excluded denominator values.

What exam hint can help solve this Mathematics question?

समान हर ((x+1)(x-1)) लेने पर अंश ((x-1)+(x+1)=2x) मिलता है। हर के शून्य मान हटाना न भूलें।