Question

Find the sum of the local maximum and local minimum values of the function $f(x)=\frac{\tan 3x}{{\tan }^{3}x}$ on interval $\left(0,\frac{\pi }{2}\right)$

Answer 1

Answer: 34

$y=\frac{\tan 3x}{{\tan }^{3}x}=\frac{3\tan x-{\tan }^{3}x}{{\tan }^{3}x\left(1-3{\tan }^{2}x\right)}$
$y=\frac{3-{\tan }^{2}x}{{\tan }^{2}x\left(1-3{\tan }^{2}x\right)}=\frac{3-t}{t(1-3t)}\text{ where }{\tan }^{2}x=t>0$
$\left(t-3t^2\right)y=3-t$
$3y^2-(1+y)t+3=0$
$\therefore t>0\Rightarrow D\ge 0;\text{ Sum of roots }>0;\text{ Product of roots }>0$
$\text{ hence }(1+y{)}^2-36y\ge 0;\frac{1+y}{3y}>0\text{ and }\frac{1}{y}>0\text{ hence }y>0$
$y^2-34y-1\ge 0$
$(y-17{)}^2\ge 288$
$(y-17{)}^2-(12\sqrt{2}{)}^2\ge 04<br/>$(y-17-12 \sqrt{2})(y-17+12 \sqrt{2}) \geq 0$<br/>${[y-(17+12 \sqrt{2})][y-(17-12 \sqrt{2})] \geq 0}$<br/><imgclass="img-fluidquestion-image"alt="image"src="https://cdn.tardigrade.in/img/question/mathematics/a7c8e8b075ae89d3609f9a6faf15eae9-.png"/><br/>Hence$y_{\max }=17-12 \sqrt{2} $<br/>$y_{\min }=17+12 \sqrt{2} $<br/>$y_{\max }+y_{\min }=34 \text { which is rational }$