The number of values of k for which the equation $x^3-3x+k=0$ has two distinct roots lying in the interval $(0,1)$ are

Solution:

Let there be a value of k for which $x^3-3x+k=0$ 

has two distinct roots between 0 and 1.

Let $a,b$ be two distinct roots of $x^3-3x+k=0$ lying between 0 and 1 such that $a<b.$

Let $f\left(x\right)=x^3-3x+k\text{.}$Then $, f\left(a\right)=f\left(b\right)=0$

Between any two roots of a polynomial $f \left(x\right)$ there exists at least one roots of its derivative $f\text{'} \left(x\right).$ Therefore,$f\text{'}\left(x\right)=3 x2-3$has at least one root between a and b. But, $f\text{'}\left(x\right)=0$ has two roots equal to± 1 which do not lie between a and b. Hence, $f\left(x\right)=0$ has no real roots lying between 0 and 1 for any value of $k.$