JEE Advanced 2020 Full Test 4 Paper 1

Let there be a value of k for which x3−3x+k=0 has two distinct roots between 0 and 1
Let a,b be two distinct roots of x3−3x+k=0 lying between 0 and 1 such that a<b. Let f(x)=x3−3x+k. Then f(a)=f(b)=0. Since between any two roots of a polynomial f(x) there exist at least one root of its derivative f′(x). Therefore f′(x)=3x2−3 has at least one root between a and b. But f′(x)=0 has two roots equal to ±1 which do not lie between a and b. Hence f(x)=0 has no real roots lying between 0 and 1 for any value of k.