यदि (f(x)=x-2+1) और (g(x)=x-2-1) हों, तो (\left\(\frac{f}{g}\right\)(x)) (1) से अधिक कब है?

If (f(x)=x-2+1) and (g(x)=x-2-1), when is (\left\(\frac{f}{g}\right\)(x)) greater than (1)?

Explanation opens after your attempt
Correct Answer

A. (x<-1) या (x>1)(x<-1) or (x>1)

Step 1

Concept

\(\frac{x^2+1}{x^2-1}>1\) gives \(\frac{2}{x^2-1}>0\), so \(x^2-1>0\). Checking the denominator sign is essential.

Step 2

Why this answer is correct

The correct answer is A. (x<-1) या (x>1) / (x<-1) or (x>1). \(\frac{x^2+1}{x^2-1}>1\) gives \(\frac{2}{x^2-1}>0\), so \(x^2-1>0\). Checking the denominator sign is essential.

Step 3

Exam Tip

\(\frac{x^2+1}{x^2-1}>1\) से \(\frac{2}{x^2-1}>0\), अतः \(x^2-1>0\)। हर का संकेत जांचना जरूरी है।

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Mathematics Answer, Explanation and Revision Hints

यदि (f(x)=x-2+1) और (g(x)=x-2-1) हों, तो (\left\(\frac{f}{g}\right\)(x)) (1) से अधिक कब है? / If (f(x)=x-2+1) and (g(x)=x-2-1), when is (\left\(\frac{f}{g}\right\)(x)) greater than (1)?

Correct Answer: A. (x<-1) या (x>1) / (x<-1) or (x>1). Explanation: \(\frac{x^2+1}{x^2-1}>1\) से \(\frac{2}{x^2-1}>0\), अतः \(x^2-1>0\)। हर का संकेत जांचना जरूरी है। / \(\frac{x^2+1}{x^2-1}>1\) gives \(\frac{2}{x^2-1}>0\), so \(x^2-1>0\). Checking the denominator sign is essential.

Which concept should I revise for this Mathematics MCQ?

\(\frac{x^2+1}{x^2-1}>1\) gives \(\frac{2}{x^2-1}>0\), so \(x^2-1>0\). Checking the denominator sign is essential.

What exam hint can help solve this Mathematics question?

\(\frac{x^2+1}{x^2-1}>1\) से \(\frac{2}{x^2-1}>0\), अतः \(x^2-1>0\)। हर का संकेत जांचना जरूरी है।