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If $m$ denote the minimum value of $f(x)= \left (\frac{\cos x}{\sin^2x(\cos x-\sin x)}\right)$ where $x\in\left(0,\frac{\pi}{4}\right)$ then find $\lfloor m\rfloor$.

My Attempt:

I am trying to find the minimum value of $f(x)$ though the question wants us to find the two consecutive integers between which the function lies.It is clear $f(x)$ is positive over the given interval.

Let $t=\tan x$. Here $0

$f(x)$ reduces to $\frac{1+t^2}{t^2(1-t)}$.

Let $g(t)=\frac{1+t^2}{t^2(1-t)}$ ; $0

$g'(t)=\frac{t^3+3t-2}{t^3(1-t)}$

Putting $g'(t)=0$. From here I go blank.

Should somehow get some inequality to get the two consecutive integers between which $m$ lies

  • 0
    http://math.stackexchange.com/questions/1846161/if-x-in-left0-frac-pi4-right-then-frac-cos-x-sin2-x-cos-x-si?rq=1. I found the same question but asked in different manner – 2017-01-11

1 Answers 1

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You correctly understood that the problem is equivalent to finding the minimum value of $g(t)=\frac{1+t^2}{t^2(1-t)}$ over $(0,1)$. Since $g'(t)=\frac{t^3+3t-2}{t^3(1-t)^2}$, the problem boils down to finding the only root of $t^3+3t-2$ in $(0,1)$, then evaluating $g(t)$ at such point. By using the cubic formula, we get that the stationary point is given by $$ t = \sqrt[3]{1+\sqrt{2}}-\frac{1}{\sqrt[3]{1+\sqrt{2}}} $$ hence the minimum of $g(t)$ is given by the only positive root of the polynomial $$ q(t) = 4 t^3 - 39 t^2 + 12 t - 4 $$ that has a sign change over the interval $[\color{red}{9},10]$.