Correct Answer - Option (D)
Here, the given function is $f(x) = a^{[x^2]}$
Let $g(x) = a^x$
and $h(x) = [x]$
hence, h(x) represents Greatest Integer Function.
Now, we know that $g(x)$ is continuous and differentiable over domain of real numbers,
So differentiability of $f(x)$ is solely dependent on the differentiability of $h(x)$.
Now, by definition, Greatest Integer Functions are not continuous over integer values.
Hence, they are not differentiable over integer values. Take a look at graph of this function, you can see that left hand limit and right hand limit are not equal for integers.
So, $h(x)$ will not be differentiable when x is an integer.
hence,
$h(x^2) = [x^2]$ will not be differentiable when $x^2$ is an integer.
hence, the given function $f(x)$ will not be differentiable when $x^2$ is an integer.
So, we have to find values of $ \sqrt x$ over set $S = \{x| x \in R, 1<x<3, x^2 \in Z\}$ where $R$ and $S$ represent set of Real and Integer numbers respectively.
So, $f(x)$ will not be differentiable when,
$x^2 = 2 $ i.e. $x = 1.41 \in S$
$x^2 = 3 $ i.e. $x = 1.72 \in S$
$x^2 = 4 $ i.e. $x = 2 \in S$
$x^2 = 5 $ i.e. $x = 2.236 \in S$
$x^2 = 6 $ i.e. $x = 2.5 \in S$
$x^2 = 7 $ i.e. $x = 2.64 \in S$
$x^2 = 8 $ i.e. $x = 2.82 \in S$
$x^3 = 9$ i.e. $x =3 \not \in S$ since it's open interval $(1,3)$, similar reason for $x=1$.
Hence, $f(x)$ will not be differentiable for $7$ values in the interval $(1,3)$.