$$f(x) = a\cos (|{x^3} - x|) + b|x|\sin (|{x^3} + x|)$$

If $${x^3} - x \ge 0$$

$$ \Rightarrow \cos \left| {{x^3} - x} \right| = \cos ({x^3} - x)$$

and if $${x^3} - x \le 0$$

$$ \Rightarrow \cos |{x^3} - x| = \cos ({x^3} - x)$$

$$\therefore$$ $$\cos (|{x^3} - x|) = \cos ({x^3} - x),\,\forall x \in R$$ ..... (i)

Again, if $${x^3} + x \ge 0$$

$$ \Rightarrow |x|\sin (|{x^3} + x|) = x\sin ({x^3} + x)$$

and if $${x^3} + x \le 0$$

$$ \Rightarrow |x|\sin (|{x^3} + x|) = - x\sin \{ - ({x^3} + x)\} $$

$$\therefore$$ $$|x|\sin (|{x^3} + x|) = x\sin ({x^3} + x),\,\forall x \in R$$ ...... (ii)

$$ \Rightarrow f(x) = a\cos (|{x^3} - x|) + b|x|\sin (|{x^3} + x|)$$

$$\therefore$$ $$f(x) = a\cos ({x^3} - x) + bx\sin ({x^3} + x)$$ ..... (iii)

For a function to be differentiable at x = 0, the function must be continuous.

$$f(0) = a\cos (0 - 0) + b(0)\sin (0) = a$$

Therefore,

$$f({0^ + }) = \mathop {\lim }\limits_{h \to \infty } [a\cos ({h^3} - h) + bh\sin ({h^3} + h)] = 0$$

$$f({0^ - }) = \mathop {\lim }\limits_{h \to \infty } [a(\cos ( - {h^3} + h) + b( - h)\sin ( - {h^3} - h)]$$

$$ = \mathop {\lim }\limits_{h \to \infty } [a\cos ({h^3} - h) + bh\sin ({h^3} + h)] = 0$$

which is continuous at x = 0; hence, f(x) is differentiable for all values of a and b.

Therefore,

$$f(1) = a\cos (1 - 1) + 1\sin (1 + 1) = a + b\sin 2$$

$$f({1^ + }) = \mathop {\lim }\limits_{h \to 0} a\cos {(1 + h)^3} - (1 + h) + b(1 + h)\sin {(1 + h)^3} + (1 + h)$$

$$ = f(1)$$

$$f({1^ - }) = \mathop {\lim }\limits_{h \to 0} a\cos {(1 - h)^3} - (1 - h) + b(1 - h)\sin {(1 - h)^3} + (1 - h)$$

$$ = f(1)$$

Thus, f(x) is continuous and we can also see that f is differentiable at x = 0 and x = 1.