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

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(B) Given the equation: $\int_0^1 x f(x) dx = \frac{1}{3} + \frac{1}{4} \int_0^1 (f(x))^2 dx$ Rearranging the terms,we get: $\frac{1}{4} \int_0^1 (f(x

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$1$

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(B) Given the equation: $\int_0^1 x f(x) dx = \frac{1}{3} + \frac{1}{4} \int_0^1 (f(x))^2 dx$ Rearranging the terms,we get: $\frac{1}{4} \int_0^1 (f(x))^2 dx - \int_0^1 x f(x) dx + \frac{1}{3} = 0$ Multiplying by $4$,we have: $\int_0^1 (f(x))^2 dx - 4 \int_0^1 x f(x) dx + \frac{4}{3} = 0$ Adding and subtracting $\int_0^1 (2x)^2 dx = \int_0^1 4x^2 dx = [\frac{4x^3}{3}]_0^1 = \frac{4}{3}$,we get: $\int_0^1 (f(x))^2 dx - 4 \int_0^1 x f(x) dx + \int_0^1 4x^2 dx - \frac{4}{3} + \frac{4}{3} = 0$ $\int_0^1 (f(x)^2 - 4x f(x) + 4x^2) dx = 0$ $\int_0^1 (f(x) - 2x)^2 dx = 0$ Since $f(x)$ is a continuous function,$(f(x) - 2x)^2$ is non-negative and continuous. For the integral of a non-negative continuous function to be $0$,the integrand must be identically zero. Therefore,$f(x) - 2x = 0 \Rightarrow f(x) = 2x$. Thus,there is exactly $1$ such continuous function.

The number of continuous functions $f:[0,1] \rightarrow \mathbb{R}$ that satisfy $\int_0^1 x f(x) dx = \frac{1}{3} + \frac{1}{4} \int_0^1 (f(x))^2 dx$ is

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