The number of points in $$\left( {1,\,3} \right)$$  where $$f\left( x \right) = {a^{\left[ {{x^2}} \right]}},\,a > 1,$$    is not differentiable, where $$\left[ x \right]$$ denotes the integral part of $$x.$$

Solution :
Here $$1 < x < 3$$   and in this interval $${x^2}$$ is an increasing functions, thus $$1 < {x^2} < 9$$
$$\eqalign{ & \left[ {{x^2}} \right] = 1,\,1 \leqslant x < \sqrt 2 \cr & = 2,\,\sqrt 2 \leqslant x < \sqrt 3 \cr & = 3,\,\sqrt 3 \leqslant x < 2 \cr & = 4,\,2 \leqslant x < \sqrt 5 \cr & = 5,\,\sqrt 5 \leqslant x < \sqrt 6 \cr & = 6,\,\sqrt 6 \leqslant x < \sqrt 7 \cr & = 7,\,\sqrt 7 \leqslant x < \sqrt 8 \cr & = 8,\,\sqrt 8 \leqslant x < 3 \cr} $$
Clearly, $$\left[ {{x^2}} \right]$$ and $${a^{\left[ {{x^2}} \right]}}$$ is discontinuous and not differentiable at only $$7$$ points, $$x = \sqrt 2 ,\,\sqrt 3 ,\,2,\,\sqrt 5 ,\,\sqrt 6 ,\,\sqrt 7 ,\,\sqrt 8 $$