TIFR-2020-Maths-A: 2

Question

Consider the set of continuous functions $f:\left [ 0,1 \right ]\rightarrow \mathbb{R}$ that satisfy:

$$\int_{0}^{1}f\left ( x \right )\left ( 1-f\left ( x \right ) \right )dx=\frac{1}{4}.$$

Then the cardinality of this set is:

• A. $0$.
• B. $1$.
• C. $2$.
• D. more than $2$.

Answer 1

$\int_{0}^{1}f(x)(1-f(x))\;dx = \frac{1}{4}$

$\int_{0}^{1}(f(x)-(f(x))^2)\;dx = \frac{1}{4}$

$\int_{0}^{1}-((f(x))^2 -f(x))\;dx = \frac{1}{4}$

$\int_{0}^{1}-\left ( \left ( f(x)-\frac{1}{2} \right )^2 -\frac{1}{4} \right )\;dx = \frac{1}{4}$

$\int_{0}^{1}\left ( - \left ( f(x)-\frac{1}{2} \right )^2 + \frac{1}{4} \right )\;dx = \frac{1}{4}$

$ \int_{0}^{1} - \left ( f(x)-\frac{1}{2} \right )^2 \; dx + \int_{0}^{1} \frac{1}{4} \; dx = \frac{1}{4}$

$ \int_{0}^{1} - \left ( f(x)-\frac{1}{2} \right )^2 \; dx + \frac{1}{4} = \frac{1}{4}$

$ \int_{0}^{1} - \left ( f(x)-\frac{1}{2} \right )^2 \; dx = 0$

$\text{It means $- \left ( f(x)-\frac{1}{2} \right )^2 =0 \Rightarrow f(x)= \frac{1}{2}$}$

$\text{So, cardinality of the set is $1$}$