Question

Y=x^2+(1+cosx/2) четная или нечетная

Answer 1

$y=x^2+\dfrac{1+\cos x}{2}\medskip\\y(-x)=(-x)^2+\dfrac{1+\cos(-x)}{2}=x^2+\dfrac{1+\cos x}{2}=y(x)$

Т.к. $y(-x)=y(x)$, то функция $y$ - чётная.

Answer 2

To determine whether the function $y = x^2 + \left(1 + \frac{\cos x}{2}\right)$ is even or odd, we need to check if it satisfies the conditions for even or odd functions:

1. An even function satisfies $f(-x) = f(x)$ for all $x$ in its domain.
2. An odd function satisfies $f(-x) = -f(x)$ for all $x$ in its domain.

Let's check each condition for the given function:

1. Even condition: $f(-x) = f(x)$

Substitute $-x$ for $x$ in the function:

$f(-x) = (-x)^2 + \left(1 + \frac{\cos(-x)}{2}\right)$

Simplify the expression:

$f(-x) = x^2 + \left(1 + \frac{\cos(x)}{2}\right)$

This is not equal to the original function $y = x^2 + \left(1 + \frac{\cos x}{2}\right)$. Therefore, the function is not even.

2. Odd condition: $f(-x) = -f(x)$

Substitute $-x$ for $x$ in the function:

$f(-x) = (-x)^2 + \left(1 + \frac{\cos(-x)}{2}\right)$

Simplify the expression:

$f(-x) = x^2 + \left(1 + \frac{\cos(x)}{2}\right)$

This is the same as the original function $y = x^2 + \left(1 + \frac{\cos x}{2}\right)$ but with the same sign. Therefore, the function is not odd either.

In conclusion, the given function $y = x^2 + \left(1 + \frac{\cos x}{2}\right)$ is neither even nor odd.