Question

The number of continuous functions $f:[0,1]\to [0,1]$ such that $f\left(x\right)<x^2$ for all $x$ and ${\int }_0^{1}f\left(x\right)dx=\frac{1}{3}$ is

• A. $0$
• B. $1$
• C. $2$
• D. Infinite

Answer 1

The correct option is A $0$

since $f\left(x\right)<x^2$ for $(0,1)$.

So, $f\left(x\right)=x^n$ where $n>2$

${\int }_0^{1}f\left(x\right)dx={\int }_0^1{x}^{n}dx={\left[\frac{x^{n+1}}{n+1}\right]}_0^1=\frac{1}{n+1}=\frac{1}{3}3=n+1n=2$

but since $n<2$,so no such function possible.