Q. The second degree polynomial $f(x)$, satisfying $f(0) = 0,\ f(1) = 1,\ f^{'}(x) > 0$ for all $x \in (0,\ 1)$
A) $f(x) = \phi$
B) $f(x) = ax + (1 - a)x^{2};\forall a \in (0,\ \infty)$
C) $f(x) = ax + (1 - a)x^{2},\ a \in (0,\ 2)$
D) No such polynomial
Explanation
(c)
Let $f(x) = bx^{2} + ax + c$
Since, $f(0) = 0\ \Rightarrow c = 0$
And $f(1) = 0\ \Rightarrow a + b = 1$
$\therefore\ \ \ f(x) = = ax + (1 - a)x^{2}$
Also, $f^{'}(x) > 0$ for $x \in (0,\ 1)$
$\Rightarrow \ \ \ \ \ a + 2(1 - a)x > 0\ \ \ \ \ \ \Rightarrow \ \ \ \ \ \ a(1 - 2x) + 2x > 0$
$\Rightarrow \ \ \ \ \ \ a > \frac{2x}{2x - 1}\ \Rightarrow \ \ \ \ 0 < a < 2$
Since, $x \in (0,\ 1)$
$\therefore\ \ \ \ \ f(x) = ax + (1 - a)x^{2};0 < a < 2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.