Given function f : R $$ \to $$ R is

$$f(x) = \left\{ {\matrix{ {{x^5} + 5{x^4} + 10{x^3} + 10{x^2} + 3x + 1,} & {x < 0;} \cr {{x^2} - x + 1,} & {0 \le x < 1;} \cr {{2 \over 3}{x^3} - 4{x^2} + 7x - {8 \over 3},} & {1 \le x < 3;} \cr {(x - 2){{\log }_e}(x - 2) - x + {{10} \over 3},} & {x \ge 3;} \cr } } \right\}$$

So,

$$f'(x) = \left\{ {\matrix{ {{5x^4} + 20{x^3} + 30{x^2} + 20x + 3,} & {x < 0;} \cr {2x - 1,} & {0 \le x < 1;} \cr {2{x^2} + 8x + 7,} & {1 \le x < 3;} \cr {{{\log }_e}(x - 2),} & {x \ge 3;} \cr } } \right\}$$

At x = 1, f"(1^-) = 2 > 0 and f"(1⁺) = 4$$ - $$8 = $$ - $$4 < 0

$$ \therefore $$ f'(x) is not differentiable at x = 1 and f'(x) has a local maximum at x = 1.

For x $$ \in $$ ($$ - $$$$\infty $$, 0)

f'(x) = 5x⁴ + 20x³ + 30x² + 20x + 3

and since

f'($$ - $$1) = 5$$ - $$20 + 20 + 30 $$ - $$ 20 + 3 = $$ - $$2 < 0

So, f(x) is not increasing on x $$ \in $$($$ - $$$$\infty $$, 0).

Now, as the range of function f(x) is R, so f is onto function.

Hence, options (b), (c) and (d) are correct.