$\tan y=(x^2+x+a)$ where $\tan y$ is positive in $\left(0, \dfrac{\pi}{2} \right)$ so the quadratic equation is also positive. Now how to get value of $a$?
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Welcome to MSE.Please go through this:https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Hydrous Caperilla Jan 06 '18 at 05:20
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Put the discriminant of this quadratic less than o .You will get the value of a – Hydrous Caperilla Jan 06 '18 at 05:29
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Let $u=x^2+x+a$ then:
$$0<tan^{-1}{u}<{\pi\over 2}\to 0<tan(tan^{-1}{u})<\infty\to 0<x^2+x+a<\infty$$
$x^2+x+a$ is always non negative only if $1-4a\le0$ which concludes:
$$a\ge {1\over 4}$$
Mostafa Ayaz
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The function $g(x)=x^2+x+a$ has a minimum at $-1/2$ and $$ g(-1/2)=\frac{1}{4}-\frac{1}{2}+a=\frac{4a-1}{4} $$ Since both limits at $\infty$ and $-\infty$ are $\infty$, we conclude that $g(x)$ assumes all values greater than or equal to $(4a-1)/4$. Hence the image of $$ f(x)=\arctan(x^2+x+a) $$ is the interval $$ [\arctan((4a-1)/4),\pi/2) $$ so there is no way the image is $(0,\pi/2)$. The image is $[0,\pi/2)$ when $a=1/4$.
egreg
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