Solution:
Since f(x) is differentiable on all intervals, we can choose any two points (Here we take 11 and 3). So from the mean value theorem, we have some c such that
$f'(c) = (f(11)-f(3))/11-3$)
Now given $5 \leqslant f'(c)\leqslant14$ therefore $5 \leqslant (f(11)-f(3))/8 \leqslant14$
=> $40 \leqslant (f(11)-f(3)) \leqslant112$
Hence a = 112 and b = 40 which gives a+b = 152 as the answer.