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What is the range of (f(x)=\frac{x^2+4x+8}{x^2+4})?फलन (f(x)=\frac{x^2+4x+8}{x^2+4}) का परिसर क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: ( [0,2] )( [0,2] )
From (y=\frac{x^2+4x+8}{x^2+4}), real (x) is required in ((y-1)x^2-4x+4y-8=0). The discriminant condition (16-4(y-1)(4y-8)\ge 0) gives (0\le y\le 2).(y=\frac{x^2+4x+8}{x^2+4}) से ((y-1)x^2-4x+4y-8=0) में वास्तविक (x) चाहिए। विविक्तकर (16-4(y-1)(4y-8)\ge 0) से (0\le y\le 2) मिलता है।
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