फलन (f(x)=\frac{x-2+1}{x-2-1}) का परिसर क्या है?
What is the range of (f(x)=\frac{x-2+1}{x-2-1})?
Explanation opens after your attempt
Correct Answer
A. ((-\infty,-1]\cup\(1,\infty\))
Concept
If \(t=x^2\ge 0\) and \(t\ne 1\), then \(y=\frac{t+1}{t-1}\). Using \(t=\frac{y+1}{y-1}\ge 0\) gives the range ((-\infty,-1]\cup\(1,\infty\)).
Why this answer is correct
The correct answer is A. ((-\infty,-1]\cup\(1,\infty\)). If \(t=x^2\ge 0\) and \(t\ne 1\), then \(y=\frac{t+1}{t-1}\). Using \(t=\frac{y+1}{y-1}\ge 0\) gives the range ((-\infty,-1]\cup\(1,\infty\)).
Exam Tip
यदि \(t=x^2\ge 0\) और \(t\ne 1\), तो \(y=\frac{t+1}{t-1}\)। इससे \(t=\frac{y+1}{y-1}\ge 0\) देकर परिसर ((-\infty,-1]\cup\(1,\infty\)) मिलता है।
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