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

Answer:

Explanation:

Let f(x)= $2x^{3}+3x+k$

 On differentiating  w.r,t x , we get

$f'(x)= 6x^{2}+3> 0,\forall x\epsilon R$

   $\Rightarrow$    f(x) is strictly increasing function.

$\Rightarrow$    f(x)= 0 has only one real root, so two roots are not possible