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The second question:

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The third question:

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The fourth question:

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This part of the extended materials mainly examines the knowledge points of even-numbered functions:

If any x in the domain of function f(x) has f(x) = f (-x), then function f(x) is an even function.

If the function expression is known, f(x) = F (-x), such as Y = X * X, is satisfied for any x in the domain of function f(x); If the image is known, the image of an even function is symmetrical about y (straight line x=0), and the symmetry of the domain d about the origin is a necessary and sufficient condition for the function to become an even function.

Algebraic judgment method, mainly according to the definition of parity function, first judges whether the domain is symmetric about the origin, if it is asymmetric, it is odd or even, if it is symmetric, f(-x)=-f(x) is odd function; F(-x)=f(x) is an even function.

Odd function must satisfy f(0)=0 (because the expression f(0) indicates that 0 is within the scope of the definition domain, and f(0) must be 0), so it is not necessary for odd function to have F(0), but if there is F(0), F(0) must be equal to 0, and it is not necessary to have f(0)=0, so the odd function is deduced. In this case, the function is not necessarily an odd function, for example.

Odd function f(x) defined on r must satisfy f (0) = 0; Because the domain is on r, there is f(0) at x=0. If you want to be symmetrical about the origin, you can only get a y value at the origin, which can only be f(0)=0. This is a conclusion that can be directly used: when x can take 0 and f(x) is odd function, f(0)=0).

If and only if f(x)=0 (the domain is symmetric about the origin), f(x) is both a odd function and an even function. On the symmetric interval, the definite integral whose integrand is odd function is zero.