Question

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

Answer 1

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

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