The real number $$k$$ for which the equation, $$2x^3 + 3x + k = 0$$ has two distinct real roots in $$[0, 1]$$ belongs to

First we define the polynomial function

$$f(x)=2x^{3}+3x+k.$$

We are asked to find those real numbers $$k$$ for which this equation has two distinct real roots lying in the closed interval $$[0,1]$$.

Because the function is continuous on $$[0,1]$$, the existence of two distinct zeros inside this interval would automatically invoke Rolle’s Theorem. The theorem states that if a function is continuous on $$[a,b]$$, differentiable on $$(a,b)$$, and takes equal values at the endpoints (in particular, both zero), then there must be at least one point $$c\in(a,b)$$ where the derivative vanishes, that is $$f'(c)=0$$.

So, if there are two different points $$x_1,x_2\in[0,1]$$ with $$x_1\neq x_2$$ and

$$f(x_1)=0,\qquad f(x_2)=0,$$

then by Rolle’s Theorem there must exist a number $$c$$ between $$x_1$$ and $$x_2$$ such that

$$f'(c)=0.$$

Therefore we look at the derivative of $$f(x)$$. Using the standard power-rule derivative formula $$\dfrac{d}{dx}(x^n)=nx^{n-1}$$ we obtain

$$f'(x)=\dfrac{d}{dx}(2x^{3})+\dfrac{d}{dx}(3x)+\dfrac{d}{dx}(k) =2\cdot3x^{2}+3+0=6x^{2}+3.$$

We can factor the constant out to see its sign more clearly:

$$f'(x)=3(2x^{2}+1).$$

Now, for every real $$x$$ we always have $$x^{2}\ge 0$$, so $$2x^{2}\ge 0$$, and adding $$1$$ gives $$2x^{2}+1\ge 1.$$ Hence

$$2x^{2}+1>0\quad\text{for all real }x.$$

Multiplying this positive quantity by the positive constant $$3$$ preserves the positivity, so

$$f'(x)=3(2x^{2}+1)>0\quad\text{for all real }x.$$

The derivative is therefore never zero. A strictly positive derivative means that $$f(x)$$ is strictly increasing on the entire real line, and certainly on the interval $$[0,1]$$.

A strictly increasing function cannot come back to the same value after moving forward; consequently it can intersect the horizontal axis at most one point. In particular, it cannot possess two distinct zeros inside any interval, including the interval $$[0,1]$$.

Because $$f'(x)$$ never vanishes, the necessary condition supplied by Rolle’s Theorem can never be satisfied. Therefore, no real value of $$k$$ can force the cubic $$2x^{3}+3x+k=0$$ to have two distinct real roots lying in $$[0,1]$$.

So the required real number $$k$$ does not exist.

Hence, the correct answer is Option B.