Given fa(x)=tan−12x−3ax+7 is non-decreasing function in (−6π,6π), so
fa′(x)=1+4x22−3a≥0
i.e. a≤(3(1+4x2)2)min
Now, maximum value of a is at x=0
i.e. amax=aˉ=9+π26
Hence, faˉ(8π)=tan−14π−3(9+π26)(8π)+7
=tan−14π−4(π2+9)9π+7=8−4(π2+9)9π