यदि (f(x)=\frac{1}{x-2+1}) और (g(x)=x-2+1) हैं, तो (fg) किस डोमेन पर स्थिर फलन (1) है?
If (f(x)=\frac{1}{x-2+1}) and (g(x)=x-2+1), on which domain is (fg) the constant function (1)?
Explanation opens after your attempt
Correct Answer
A. \( \mathbb{R} \)
Concept
Since \(x^2+1>0\) for every real (x), both (f) and (g) are defined on all \(\mathbb{R}\). Hence (fg=1) on all \(\mathbb{R}\).
Why this answer is correct
The correct answer is A. \( \mathbb{R} \). Since \(x^2+1>0\) for every real (x), both (f) and (g) are defined on all \(\mathbb{R}\). Hence (fg=1) on all \(\mathbb{R}\).
Exam Tip
क्योंकि \(x^2+1>0\) हर वास्तविक (x) के लिए है, (f) और (g) दोनों पूरे \(\mathbb{R}\) पर परिभाषित हैं। अतः (fg=1) पूरे \(\mathbb{R}\) पर है।
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