यदि (f(x)=\frac{1}{x}) और (g(x)=x-2) हैं, तो ((f+g)(x)=2) के वास्तविक हल कितने हैं?
If (f(x)=\frac{1}{x}) and (g(x)=x-2), how many real solutions does ((f+g)(x)=2) have?
Explanation opens after your attempt
Correct Answer
A. (2)
Concept
The equation \(x^2+\frac{1}{x}=2\) gives (x-3-2x+1=0=(x-1)\(x^2+x-1\)). Since \(x\neq0\), count all valid real roots carefully.
Why this answer is correct
The correct answer is A. (2). The equation \(x^2+\frac{1}{x}=2\) gives (x-3-2x+1=0=(x-1)\(x^2+x-1\)). Since \(x\neq0\), count all valid real roots carefully.
Exam Tip
समीकरण \(x^2+\frac{1}{x}=2\) से (x-3-2x+1=0=(x-1)\(x^2+x-1\)) मिलता है। \(x\neq0\) होने पर तीन में से दो वास्तविक हल अलग-अलग हैं?
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