If x€R and alpha=x²/1+x⁴,then alpha lies in the interval
Answers
Answered by
1
Start with
![<br />0\le (x^2-1)^2<\infty\\<br />\Rightarrow 0\le x^4+1-2x^2<\infty\\<br />\Rightarrow 2x^2\le x^4+1<\infty\\<br />\Rightarrow 2\le\dfrac{1+x^4}{x^2}<\infty\\<br />\Rightarrow \dfrac{1}{2}\ge\dfrac{x^2}{1+x^4}>0\\<br />\Rightarrow\alpha\in\left (0,\dfrac{1}{2}\right]\\<br />](https://tex.z-dn.net/?f=%3Cbr+%2F%3E0%5Cle+%28x%5E2-1%29%5E2%26lt%3B%5Cinfty%5C%5C%3Cbr+%2F%3E%5CRightarrow+0%5Cle+x%5E4%2B1-2x%5E2%26lt%3B%5Cinfty%5C%5C%3Cbr+%2F%3E%5CRightarrow+2x%5E2%5Cle+x%5E4%2B1%26lt%3B%5Cinfty%5C%5C%3Cbr+%2F%3E%5CRightarrow+2%5Cle%5Cdfrac%7B1%2Bx%5E4%7D%7Bx%5E2%7D%26lt%3B%5Cinfty%5C%5C%3Cbr+%2F%3E%5CRightarrow+%5Cdfrac%7B1%7D%7B2%7D%5Cge%5Cdfrac%7Bx%5E2%7D%7B1%2Bx%5E4%7D%26gt%3B0%5C%5C%3Cbr+%2F%3E%5CRightarrow%5Calpha%5Cin%5Cleft+%280%2C%5Cdfrac%7B1%7D%7B2%7D%5Cright%5D%5C%5C%3Cbr+%2F%3E "<br />0\le (x^2-1)^2<\infty\\<br />\Rightarrow 0\le x^4+1-2x^2<\infty\\<br />\Rightarrow 2x^2\le x^4+1<\infty\\<br />\Rightarrow 2\le\dfrac{1+x^4}{x^2}<\infty\\<br />\Rightarrow \dfrac{1}{2}\ge\dfrac{x^2}{1+x^4}>0\\<br />\Rightarrow\alpha\in\left (0,\dfrac{1}{2}\right]\\<br />")
Similar questions