Question

For a positive constant $t,$ let $\alpha ,\beta$ be the roots of the quadratic equation $x^2+t^{2}x-2t=0$. If the minimum value of $\underset{-1}{\overset{2}{\int }}[(x+\frac{1}{{\alpha }^2})(x+\frac{1}{{\beta }^2})+\frac{1}{\alpha \beta }]dx$ is $\sqrt{\frac{a}{b}}+c$ where $a,b,c\in N,$ then the least possible value of $(a+b+c)$ is

Answer 1

$\alpha$ and $\beta$ are the roots of $x^2+t^{2}x-2t=0$
Then we have $\alpha +\beta =-t^2$ and $\alpha \beta =-2t$
$\Rightarrow \frac{{\alpha }^2+{\beta }^2}{(\alpha \beta {)}^2}=\frac{(\alpha +\beta {)}^2-2\alpha \beta }{(\alpha \beta {)}^2}=\frac{t^2}{4}+\frac{1}{t}$
$\Rightarrow \frac{1}{(\alpha \beta {)}^2}+\frac{1}{\alpha \beta }=\frac{1}{4t^2}-\frac{1}{2t}$

Now,
$I=\underset{-1}{\overset{2}{\int }}[(x+\frac{1}{{\alpha }^2})(x+\frac{1}{{\beta }^2})+\frac{1}{\alpha \beta }]dx$
$=\underset{-1}{\overset{2}{\int }}[x^2+(\frac{t^2}{4}+\frac{1}{t})x+(\frac{1}{4t^2}-\frac{1}{2t})]dx$
$=\frac{3t^2}{8}+\frac{3}{4t^2}+3$
Differentiating with respect to $t,$ we get
$\frac{dI}{dt}=\frac{3t}{4}-\frac{3}{2t^3}=0$
$\Rightarrow t=\sqrt[4]{42}(\because t>0)$
$\therefore I_{min}=\sqrt{\frac{9}{8}}+3$
$\Rightarrow a+b+c=20$