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**Hint:**Start by taking the given equation as function of x or f(x) and differentiate with respect to x . Check whether f’(x) obtained is increasing or decreasing , if it is increasing then it will have at most 1 real root.

__Complete step-by-step answer__:Given, $2{x^3} + 3x + k = 0$

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

Differentiating with respect to x , we get

Here we will use the formula $\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}}$

$

f'(x) = 6{x^2} + 3 \\

f'(x) > 0 \\

$

As for any value of x, f’(x) can never be negative because of the square term involved.

f’(x) is a strictly increasing function and has at most 1 real root .

And we know , if a polynomial of odd degree, in this case it is 3, has exactly 1 real root.

So, f(x) = has exactly one real root.

We see that the results found do not satisfy the conditions.

Therefore, k does not exist.

So , option C is the correct answer.

**Note**: Students must know the principle of differentiation , nature of function , graph plotting etc in order to solve such similar problems. Questions can also be asked in such a manner which would demand the application of Lagrange’s mean value theorem(LMVT) , Intermediate value theorem(IVT) , Rolle’s theorem, and are recommended to be practised very well as they make the approach to the solution very easy meanwhile giving valuable information about the function too.

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