Let the function f(x) = (x² – 1)|x² – ax + 2| + cos|x| be not differentiable at  two points x = $\mathrm{\alpha }$ = 2 and x = $\mathrm{\beta }$. Then the distance of the point ($\mathrm{\alpha },\mathrm{\beta }$) from the line 12x + 5y + 10 = 0 is equal to :

cos|x| is always differentiable
$\therefore$ we have to check only for |x² – ax + 2|
$\therefore$ It is not differentiable at x² – ax + 2 = 0
One root is given, $\mathrm{\alpha }$ = 2
$\therefore$ 4 – 2a + 2 = 0
$\Rightarrow$a = 3
$\therefore$ other root $\mathrm{\beta }$ = 1
but for x = 1 f(x) is differentiable
(Drop)