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What is the domain of \(f(x)=\frac{1}{(x-1)(x+2)}\)?\(f(x)=\frac{1}{(x-1)(x+2)}\) का domain क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \(\mathbb{R}\setminus\{1,-2\}\)\(\mathbb{R}\setminus\{1,-2\}\)
For \(f(x)=\frac{1}{(x-1)(x+2)}\), the denominator \((x-1)(x+2)\) cannot be zero.
So \(x-1\neq 0\Rightarrow x\neq 1\) and \(x+2\neq 0\Rightarrow x\neq -2\).
Therefore the domain is \(\mathbb{R}\setminus\{1,-2\}\).\(f(x)=\frac{1}{(x-1)(x+2)}\) में हर (denominator) \((x-1)(x+2)\) शून्य नहीं होना चाहिए।
अतः \(x-1\neq 0\Rightarrow x\neq 1\) और \(x+2\neq 0\Rightarrow x\neq -2\)।
इसलिए domain \(\mathbb{R}\setminus\{1,-2\}\) है।
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