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If (f(x)=\sqrt{x+1}) and (g(x)=\frac{1}{x-2}), what is the real domain of ((g\circ f)(x))?यदि (f(x)=\sqrt{x+1}) और (g(x)=\frac{1}{x-2}), तो ((g\circ f)(x)) का वास्तविक प्रान्त कौन-सा होगा?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: ([-1,\infty)-{3})([-1,\infty)-{3})
Step 1: First, for (f(x)=\sqrt{x+1}), we need (x+1\ge0), so (x\ge-1). Step 2: ((g\circ f)(x)=\frac{1}{\sqrt{x+1}-2}), so the denominator must not be zero. Step 3: (\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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