Class problem by Shubhamkar 1 Multi-correct JEE

The solution, too, is given by our teacher.

Multiplying the given equations by $a^2, 2a,1$ respectively, we get

• $\int_{0}^{1}a^2f(x)dx=a^2$
• $\int_{0}^{1}2axf(x)dx=2a^2$
• $\int_{0}^{1}x^2f(x)dx=a^2$

Adding first and third equation and subtrating the third from it yields $\int_{0}^{1}(x-a)^2f(x)dx=0$ .

Now, $(x-a)^2\geq0$ and the previous statement implies $f(x)<0$ , as only then can the definite integral evaluate to zero. The definite integral with distinct limits of a positive function cannot be zero. Therefore, no function exists! (Given in the question that $f(x)$ is non-negative.)

Thus, 1111 is the answer.

Consider a continuous random variable $X$ with pdf given by the function $f(x)$ . Then according to the given condition, we require that $\mathbb{E}(X^2) = (\mathbb{E}(X))^2.$ By Jensen's inequality, we always have, in general, $\mathbb{E}(X^2) \geq (\mathbb{E}(X))^2$ , with the equality holding iff $X$ is constant. In that case $f(x)$ is required to be a point mass ("Dirac Delta"), and thus $f(x)$ can not be continuous. Hence, no such continuous function exists.