Animated Solution for Mathematics - Differentiation: If $P(x)$ is a polynomial of degree less than or equal to $2$ and $S$ is the set of all such polynomials so that $P(0)=0,P(1)=1$ and $P^{\prime }(x)>0\forall x\in [0,1]$, then

We are tasked with finding a family of polynomials $P(x)$ of degree at most 2 that satisfy three strict conditions: they must pass through the origin $(0,0)$, pass through the point $(1,1)$, and be strictly increasing on the interval $[0,1]$.

We begin with the general form of a polynomial of degree at most 2:

Since $P(0)=0$, substituting $x=0$ immediately reveals that $C=0$. Our polynomial simplifies to $P(x)=A{x}^2+Bx$.

Next, we apply the condition $P(1)=1$. Substituting $x=1$ gives us the relation:

To simplify our analysis, we introduce a single parameter $a$. If we set $B=a$, then $A$ must be $1-a$. Thus, our family of curves is defined by:

To satisfy the condition that the polynomial is strictly increasing, we require $P^{\prime }(x)>0$ for all $x\in [0,1]$. Let us differentiate $P(x)$ with respect to $x$:

Notice that $P^{\prime }(x)$ is a linear function. In the context of JEE mathematics, a linear function is monotonic; therefore, to ensure it is strictly positive on the closed interval $[0,1]$, we only need to ensure that the function is positive at the boundaries of the interval.

We require $P^{\prime }(x)>0$ for all $x\in [0,1]$. This is equivalent to requiring $P^{\prime }(0)>0$ and $P^{\prime }(1)>0$.

Testing the left endpoint: For this to be positive, we must have $a>0$.

Testing the right endpoint: For this to be positive, we must have $2-a>0$, which simplifies to $a<2$.

Combining these two constraints, we find that the parameter $a$ must lie in the open interval $(0,2)$.

The set of all such polynomials is:

We have successfully navigated the constraints using the power of the derivative. Remember, in JEE Advanced, success is found by identifying the algebraic insight that makes the problem collapse into simplicity.