JEE MAIN - Mathematics (2013 (Offline) - No. 17)
The real number $$k$$ for which the equation, $$2{x^3} + 3x + k = 0$$ has two distinct real roots in $$\left[ {0,\,1} \right]$$
lies between 1 and 2
lies between 2 and 3
lies between $$ - 1$$ and 0
does not exist.
Explanation
$$f\left( x \right) = 2{x^3} + 3x + k$$
$$f'\left( x \right) = 6{x^2} + 3 > 0$$
$$\forall x \in R$$ $$\,\,\,\,\,\,$$ (as $$\,\,\,\,\,\,$$ $${x^2} > 0$$)
$$ \Rightarrow f\left( x \right)$$ is strictly increasing function
$$ \Rightarrow f\left( x \right) = 0$$ has only one real root, so two roots are not possible.
$$f'\left( x \right) = 6{x^2} + 3 > 0$$
$$\forall x \in R$$ $$\,\,\,\,\,\,$$ (as $$\,\,\,\,\,\,$$ $${x^2} > 0$$)
$$ \Rightarrow f\left( x \right)$$ is strictly increasing function
$$ \Rightarrow f\left( x \right) = 0$$ has only one real root, so two roots are not possible.
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