The value of $f(4)-f(3)$ is

If $f(x) = 4x^3 - x^2 - 2x + 1$ and $g(x) = \begin{cases} \min \{f(t) : 0 \le t \le x\} & ; 0 \le x \le 1 \\ 3 - x & ; 1 < x \le 2 \end{cases}$,then the value of $g\left( \frac{1}{4} \right) + g\left( \frac{3}{4} \right) + g\left( \frac{5}{4} \right)$ is:

$(i)$ $f(x)$ is continuous and defined for all real numbers.
$(ii)$ $f'(-5) = 0$; $f'(2)$ is not defined and $f'(4) = 0$.
$(iii)$ $(-5, 12)$ is a point which lies on the graph of $f(x)$.
$(iv)$ $f''(2)$ is undefined,but $f''(x)$ is negative everywhere else.
$(v)$ The signs of $f'(x)$ are given below:
$f'(x)$ sign chart:
- For $x < -5$,$f'(x) > 0$
- For $-5 < x < 2$,$f'(x) < 0$
- For $2 < x < 4$,$f'(x) > 0$
- For $x > 4$,$f'(x) < 0$
From the possible graph of $y = f(x)$,we can say that:

Let $a_1, a_2, \ldots, a_{100}$ be non-zero real numbers such that $a_1+a_2+\ldots+a_{100}=0$. Then,

Let $f(x) = x|x|$,$g(x) = \sin x$ and $h(x) = (g \circ f)(x)$. Then

$\mathop {Lim}\limits_{\lambda \to 0} \,{\left( {\int\limits_0^1 {{{(1 + x)}^\lambda }dx} } \right)^{\frac{1}{\lambda }}}$ is equal to