The function f(x) = (x² – 1) |x² – 3x + 2| + cos (|x|) is not differentiable at :

x = 2

Since cos (–x) = cos x

∴ cos |x| is differentiable for each x ∈ R

Also,  $x^2-3 x+2>0 \Rightarrow(x-2)(x-1)>0$

$\Rightarrow x \in(-\infty, 1) \cup(2, \infty)$

Similarly, $x^2-3 x+2<0 \Rightarrow x \in(1,2)$

If  $g(x)=\left(x^2-1\right)\left|x^2-3 x+2\right|$, then f(x) is not differentiable at points where g(x) is so.

Now,  $g(x)=(x-1)^2(x+1)(x-2), ~\forall ~x \in(-\infty, 1) \cup(2, \infty)$

and  $=-(x-1)^2(x+1)(x-2) ~\forall ~x \in(1,2)$

∴ g(x) is not differentiable at x = 2

⇒ f(x) is not differentiable at x = 2

Hence (4) is correct answer.