फलन (f(x)=\frac{x-2}{x+5}) का परिसर ज्ञात कीजिए।
Find the range of (f(x)=\frac{x-2}{x+5}).
Explanation opens after your attempt
A. \(\mathbb{R}-{1}\)
Concept
If \(y=\frac{x-2}{x+5}\), then \(x=\frac{-5y-2}{y-1}\), so \(y\ne 1\). Hence the range is \(\mathbb{R}-{1}\).
Why this answer is correct
The correct answer is A. \(\mathbb{R}-{1}\). If \(y=\frac{x-2}{x+5}\), then \(x=\frac{-5y-2}{y-1}\), so \(y\ne 1\). Hence the range is \(\mathbb{R}-{1}\).
Exam Tip
यदि \(y=\frac{x-2}{x+5}\), तो \(x=\frac{-5y-2}{y-1}\), इसलिए \(y\ne 1\)। अतः परिसर \(\mathbb{R}-{1}\) है।
Login to save your score, XP, coins and progress.
