If A be the area bounded by y = f(x), $y=f^{-1}\left(x\right)$  and line 4x + 4y – 5 = 0 where f(x) is a polynomial of 2nd degree passing through the origin and having maximum value of $\frac{1}{4}$  at $x=1$, then the value of  $96A$ is________

Let  $f\left(x\right)=a{x}^2+bx$
 $\frac{1}{4}=a+b$ ……… (1)
 $f^{/}\left(x\right)=2ax+b$
$\Rightarrow 2a+b=0$  ……… (2)
From (1) and (2),
 $a=-\frac{1}{4},b=\frac{1}{2}$
 $f\left(x\right)=\frac{2x-x^2}{4}$
Since 4x + 4y – 5 = 0 passes through $A(1,\frac{1}{4})$  and $B(\frac{1}{4},1)$ . So, area bounded is  $OAB=2\times OAC$
= 2[area (OCP) + area (CPQA) – OAQ]
=  $2[\frac{1}{2}\times \frac{5}{8}\times \frac{5}{8}+(\frac{5}{8}\times \frac{1}{4})\times \frac{1}{2}\times (1-\frac{5}{8})-\underset{0}{\overset{1}{\int }}\frac{2x-x^2}{4}dx]$
= $2[\frac{25}{128}+\frac{7}{8}\times \frac{3}{16}-\frac{1}{6}]=\frac{37}{96}=A$, then 96A = 37