Question

The second degree polynomial $f\left(x\right),$ satisfying $f\left(0\right)=0$, $f\left(1\right)=1$, $f\text{'}\left(x\right)>0,\forall x\in (0,1)$is

• A. $f\left(x\right)=\varnothing$
• B. $f\left(x\right)=ax+(1-a)x^2,\vee a\in (0,10)$
• C. $f\left(x\right)=ax+(1-a)x^2,a\in (0,2)$
• D. No such polynomials

Answer 1

The correct option is C

$f\left(x\right)=ax+(1-a)x^2,a\in (0,2)$

Explanation for correct option:

$f\left(x\right)=a{x}^2+bx+c$ ; [$\because$Second degree polynomial]

Given, $f\left(0\right)=0$

$\begin{array}{rcl}& \Rightarrow & 0=a{\left(0\right)}^2+b\left(0\right)+c\\ & \Rightarrow & c=0\end{array}$

Also given, $f\left(1\right)=1$

$\begin{array}{rcl}& \Rightarrow & 1=a{\left(1\right)}^2+b\left(1\right)+0\left[\because c=0\right]\\ 1& =& a+b...\left(1\right)\end{array}$

$\therefore f\left(x\right)=ax+(1-a)x^2$

Also $f\text{'}\left(x\right)>0,$for $x\in (0,1)$

$$\begin{gathered} \Rightarrow a+2(1-a)x>0 \\ \Rightarrow a(1-2x)+2x>0 \\ \Rightarrow a>\frac{2x}{2x-1}\Rightarrow 0<a<2;x\in (0,1) \\ \therefore f\left(x\right)=ax+(1-a)x^2;0<a<2 \end{gathered}$$

Hence the correct option is option C.