1 True or False (1) f(x)= ) 1,xis rational 0,xis irrational is integrable on any bounded closed interval [a,b] (2)The derivative of y=y(x) determined by the formula arctan(y)/(x)=lnsqrt (x^2+y^2) is y'=(x+y)/(x-y)

Let's analyze each statement one by one:<br /><br />(1) The function $f(x)$ is defined as $f(x) = 1$ for rational $x$ and $f(x) = 0$ for irrational $x$. This function is not integrable on any bounded closed interval $[a, b]$ because it is not continuous. The function jumps between values of 0 and 1 at every irrational point in the interval, which means it has an infinite number of discontinuities. Therefore, the statement is False.<br /><br />(2) To find the derivative of $y = y(x)$ determined by the equation $arctan\frac{y}{x} = ln\sqrt{x^2 + y^2}$, we need to differentiate both sides of the equation with respect to $x$. Let's differentiate the left side:<br /><br />$\frac{d}{dx}arctan\frac{y}{x} = \frac{1}{1 + \left(\frac{y}{x}\right)^2} \cdot \frac{d}{dx}\frac{y}{x} = \frac{1}{1 + \frac{y^2}{x^2}} \cdot \left(-\frac{y}{x^2}\right) = -\frac{y}{x^2 + y^2}$<br /><br />Now, let's differentiate the right side:<br /><br />$\frac{d}{dx}ln\sqrt{x^2 + y^2} = \frac{1}{\sqrt{x^2 + y^2}} \cdot \frac{d}{dx}\sqrt{x^2 + y^2} = \frac{1}{\sqrt{x^2 + y^2}} \cdot \frac{x + y}{\sqrt{x^2 + y^2}} = \frac{x + y}{x^2 + y^2}$<br /><br />Equating the derivatives from both sides, we get:<br /><br />$-\frac{y}{x^2 + y^2} = \frac{x + y}{x^2 + y^2}$<br /><br />Multiplying both sides by $x^2 + y^2$, we have:<br /><br />$-y = x + y$<br /><br />Simplifying, we get:<br /><br />$y' = \frac{x + y}{x - y}$<br /><br />Therefore, the statement is True.