The number of continuous functions f:[0,1] ri | vedclass

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(D) Given the condition: $\int_0^1 (f(x))^2 dx = 2 \int_0^1 f(x) dx$. Rearranging the terms,we get: $\int_0^1 (f(x))^2 dx - 2 \int_0^1 f(x) dx = 0$. A

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more than $4$

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(D) Given the condition: $\int_0^1 (f(x))^2 dx = 2 \int_0^1 f(x) dx$. Rearranging the terms,we get: $\int_0^1 (f(x))^2 dx - 2 \int_0^1 f(x) dx = 0$. Adding and subtracting $1$ inside the integral: $\int_0^1 ((f(x))^2 - 2f(x) + 1 - 1) dx = 0$. This simplifies to: $\int_0^1 (f(x) - 1)^2 dx - \int_0^1 1 dx = 0$. Therefore,$\int_0^1 (f(x) - 1)^2 dx = 1$. We are looking for continuous functions $f(x)$ such that the integral of $(f(x) - 1)^2$ over $[0, 1]$ is $1$. Let $g(x) = f(x) - 1$. Then we need $\int_0^1 (g(x))^2 dx = 1$. There are infinitely many continuous functions $g(x)$ that satisfy this condition (e.g.,$g(x) = 1$,$g(x) = -1$,$g(x) = \sqrt{3}x$,$g(x) = \sqrt{5}x^2$,etc.). Since there are infinitely many such functions $g(x)$,there are infinitely many such functions $f(x) = g(x) + 1$. Thus,the number of such functions is more than $4$.

Column $I$	Column $II$
$(A) \int_{-1}^1 \frac{dx}{1+x^2}$	$(p) \frac{1}{2} \log \left(\frac{2}{3}\right)$
$(B) \int_0^1 \frac{dx}{\sqrt{1-x^2}}$	$(q) 2 \log \left(\frac{2}{3}\right)$
$(C) \int_2^3 \frac{dx}{1-x^2}$	$(r) \frac{\pi}{3}$
$(D) \int_1^2 \frac{dx}{x \sqrt{x^2-1}}$	$(s) \frac{\pi}{2}$

The number of continuous functions $f:[0,1] \rightarrow(-\infty, \infty)$ satisfying the condition $\int_0^1 (f(x))^2 dx = 2 \int_0^1 f(x) dx$ is

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