QUESTION

# 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 $P\left( 0 \right)=0,P\left( 1 \right)=1,P'\left( x \right)>0\forall x\in \left[ 0,1 \right]$, then(a) $S=\varnothing$ (b) $S=ax+\left( 1-a \right){{x}^{2}}$, $\forall a\in \left( 0,2 \right)$(c) $S=ax+\left( 1-a \right){{x}^{2}}$, $\forall a\in \left( 0,\infty \right)$ (d) $S=ax+\left( 1-a \right){{x}^{2}}$, $\forall a\in \left( 0,1 \right)$

Hint: Assume that the polynomial P(x) is of the form $b{{x}^{2}}+ax+c$. Apply the conditions given in the question on this polynomial. Simplify the equations to find the value of the variables a, b, and c and thus, the set S.

We have a polynomial P(x) of degree less than or equal to 2. We have certain conditions imposed on the polynomial. We have to find the set S of all the polynomials which satisfy the given conditions.
Let’s assume that the polynomial P(x) is of the form $b{{x}^{2}}+ax+c$.
We know that $P\left( 0 \right)=0$.
Substituting $x=0$ in the equation $P\left( x \right)=b{{x}^{2}}+ax+c$, we have $b{{\left( 0 \right)}^{2}}+a\left( 0 \right)+c=0$. Thus, we have $c=0$.
So, we have $P\left( x \right)=b{{x}^{2}}+ax$.
We know that $P\left( 1 \right)=1$.
Substituting $x=1$ in the equation $P\left( x \right)=b{{x}^{2}}+ax$, we have $b{{\left( 1 \right)}^{2}}+a\left( 1 \right)=1$. Thus, we have $a+b=1$.
We can rewrite the above equation as $b=1-a$.
Thus, we can rewrite P(x) as $P\left( x \right)=b{{x}^{2}}+ax=\left( 1-a \right){{x}^{2}}+ax$.
Now, we know that $P'\left( x \right)>0\forall x\in \left[ 0,1 \right]$.
We will differentiate the function $P\left( x \right)=\left( 1-a \right){{x}^{2}}+ax$.
We know that differentiation of function of the form $y=a{{x}^{n}}$ is $\dfrac{dy}{dx}=an{{x}^{n-1}}$.
Thus, we have $\dfrac{dP\left( x \right)}{dx}=\dfrac{d}{dx}\left( \left( 1-a \right){{x}^{2}}+ax \right)=\dfrac{d}{dx}\left( \left( 1-a \right){{x}^{2}} \right)+\dfrac{d}{dx}\left( ax \right)$.
So, we have $\dfrac{dP\left( x \right)}{dx}=\dfrac{d}{dx}\left( \left( 1-a \right){{x}^{2}} \right)+\dfrac{d}{dx}\left( ax \right)=\left( 1-a \right)2x+a$.
Thus, we have $2\left( 1-a \right)x+a>0\forall x\in \left[ 0,1 \right]$.
We will now substitute $x=0$ in the above expression. Thus, we have $2\left( 1-a \right)0+a>0\Rightarrow a>0$.
We will now substitute $x=1$ in the above expression. Thus, we have $2\left( 1-a \right)1+a>0\Rightarrow 2-2a+a>0\Rightarrow 2-a>0\Rightarrow a<2$.
Thus, we have $a>0$ and $a<2$, which means $0Hence, the set S is$S=ax+\left( 1-a \right){{x}^{2}}$,$\forall a\in \left( 0,2 \right)$, which is option (b). Note: One must impose all the conditions given in the question to find all the possible elements of the set S. We can also take P(x) of the form$ax+b$or$a\$ as the polynomial P(x) can have degree less than or equal to 2. However, we will get the same answer as above.