Let  $a\in [0,4]$. The maximum area bounded by the curves $y=1-|x-1|$ and  $y=|2x-a|$ is A, then 3A value is

Case-I:
If  $a\in [0,1]$ , the curves intersect at  $(\frac{a}{3},\frac{a}{3})$  and (a, a). The bounded region is contained in the triangle with vertices (0, 0), $(\frac{1}{2},0)$  and (1, 1) with area =  $\frac{1}{4}$
Hence, area cannot exceed  $\frac{1}{4}$
Case-II:
If  $a\in [1,3]$ . In this case the bounded region is a quadrilateral with four vertices  $(\frac{a}{3},\frac{a}{3}),(\frac{a}{2},0)$ ,  $(\frac{a+2}{3},\frac{4-a}{3})$  and (1,1). In this case area bounded    

$=1-\frac{1}{2}\cdot \frac{a}{3}\cdot \frac{a}{2}-\frac{1}{2}\left(\frac{4-a}{3}\right)(2-\frac{a}{2})=\frac{1}{3}-\frac{{(a-2)}^2}{6}\le \frac{1}{3}$