Question

If the least positive value of $p$ for which $|6x+25p|\left(4+x^2\right)\ge 50x\forall x\in R$ is $\left(\frac{a}{b}\right)$, where $a,b\in N$, then find the least value of $(b-5a)$.

Answer 1

Answer: 5

$|6x+25p|\ge \frac{50x}{4+x^2}$
$\left|\frac{3}{25}x+\frac{p}{2}\right|\ge \frac{x}{4+x^2}$
$f(x)=\frac{x}{4+x^2}\Rightarrow f^{\prime }(x)=\frac{\left(4+x^2\right)1-x\cdot 2x}{{\left(4+x^2\right)}^2}=\frac{4-x^2}{{\left(4+x^2\right)}^2}$
$m_{AP}=\frac{k-0}{h+\frac{25p}{6}}=\frac{4-h^2}{{\left(4+h^2\right)}^2}=\frac{3}{25}$

$\frac{4-h^2}{\left(4+h^2\right)}=\frac{3}{25}$
$100-25h^2=3\left(16+8h^2+h^4\right)$
$3h^4+49h^2-52=0$
$\left(3h^2+52\right)\left(h^2-1\right)=0\Rightarrow h=\pm 1$
$\therefore h=1;k=1/5$
$\text{ Now }\frac{k}{h+\frac{25p}{6}}=\frac{3}{25}\Rightarrow \frac{\frac{1}{5}}{1+\frac{25p}{6}}=\frac{3}{25}$
$\Rightarrow 5=3+\frac{25p}{2}$
$b-5a=25-20=5.\Rightarrow p=\frac{4}{25}\equiv \frac{a}{b}$