यदि (f(x)=\sqrt{x+1}) और (g(x)=\frac{1}{x-2}), तो (\(g\circ f\)(x)) का वास्तविक प्रान्त कौन-सा होगा?
If (f(x)=\sqrt{x+1}) and (g(x)=\frac{1}{x-2}), what is the real domain of (\(g\circ f\)(x))?
Explanation opens after your attempt
Correct Answer
A. \([-1,\infty\)-{3})
Concept
First, for (f(x)=\sqrt{x+1}), we need \(x+1\ge0\), so \(x\ge-1\).
Why this answer is correct
(\(g\circ f\)(x)=\frac{1}{\sqrt{x+1}-2}), so the denominator must not be zero.
Exam Tip
\(\sqrt{x+1}=2\) gives (x=3), hence the domain is \([-1,\infty\)-{3}). चरण 1: पहले (f(x)=\sqrt{x+1}) के लिए \(x+1\ge0\), इसलिए \(x\ge-1\)। चरण 2: (\(g\circ f\)(x)=\frac{1}{\sqrt{x+1}-2}) होगा, इसलिए हर शून्य नहीं होना चाहिए। चरण 3: \(\sqrt{x+1}=2\) से (x=3), अतः प्रान्त \([-1,\infty\)-{3}) है।
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