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What is the range of (f(x)=\frac{x^2+1}{x^2-1})?फलन (f(x)=\frac{x^2+1}{x^2-1}) का परिसर क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: ((-\infty,-1]\cup(1,\infty))((-\infty,-1]\cup(1,\infty))
If (t=x^2\ge 0) and (t\ne 1), then (y=\frac{t+1}{t-1}). Using (t=\frac{y+1}{y-1}\ge 0) gives the range ((-\infty,-1]\cup(1,\infty)).यदि (t=x^2\ge 0) और (t\ne 1), तो (y=\frac{t+1}{t-1})। इससे (t=\frac{y+1}{y-1}\ge 0) देकर परिसर ((-\infty,-1]\cup(1,\infty)) मिलता है।
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