📝 Even and odd functions examples (15 MCQs)

📖 Explanation: Because an even function satisfies f(-x)=f(x) for all x, and each term of the polynomial $x^{4}$ and $-5x^{2}$ is even while the constant 3 is also even, the whole expression is unchanged by replacing x with -x, so the function is even.

📖 Explanation: The outer function f is even, meaning f(-y)=f(y). For the composition h(x)=f(g(x)), the value h(-x)=f(g(-x)). Since we have no information about g, the parity of h depends entirely on g, and no definite parity can be asserted. Hence option D is correct.

📖 Explanation: Since g is even, g(-x)=g(x). Substituting into h gives h(-x)=f(g(x)). Because f is odd, f(g(x)) is generally not equal to h(x) nor its negative, unless g(x)=0. Therefore the composition does not possess a guaranteed even or odd symmetry; it is classified as neither.

📖 Explanation: An even function satisfies f(-x)=f(x) while an odd function satisfies g(-x)=-g(x). Adding them yields h(-x)=f(x)-g(x), which is not equal to h(x) nor to -h(x) for most x. Hence the sum is typically neither even nor odd.

📖 Explanation: A graph symmetric about the origin means that rotating the graph 180° leaves it unchanged, which is the defining geometric property of odd functions. Symmetry about the y‑axis corresponds to even functions. Therefore the correct description is symmetry about the origin.

📖 Explanation: Differentiating an even function f yields f'(-x)= -f'(x) because the derivative of f(-x) equals -f'(x). This relationship shows that the derivative of an even function is odd, making option B the correct choice.

📖 Explanation: The integrand $x^{3}$ is an odd function because $(-x)^{3} = -x^{3}$. Integrating an odd function over a symmetric interval $[-a,a]$ always yields zero, since the contributions from negative and positive halves cancel each other.

📖 Explanation: When multiplying an even function by an odd function, the product satisfies p(-x)=f(-x)g(-x)=f(x)(-g(x))=-f(x)g(x)=-p(x). Hence the product is odd, which corresponds to option D in the reordered list.

📖 Explanation: Shifting $\sin x$ by $\pi$ gives $k(x)=\sin(x-\pi) = -\sin x$. Evaluating $k(-x)=\sin(-x-\pi)= -\sin(x+\pi)=\sin x$. Since $k(-x)\neq \pm k(x)$, the function is neither even nor odd.

📖 Explanation: For any x, $|f(-x)| = |-f(x)| = |f(x)|$. Thus the absolute‑value transformation removes the sign change that characterizes odd functions, resulting in a function that satisfies the even‑function condition. Hence the correct answer is 'Even'.

📖 Explanation: If both f and g are even, then g(-x)=g(x) and f(-y)=f(y). Substituting, h(-x)=f(g(-x))=f(g(x))=h(x). Therefore the composition of two even functions remains even.

📖 Explanation: A function that is both even and odd must satisfy f(x)=f(-x) and f(x)=-f(-x) simultaneously, which forces f(x)=0 for every x. The only function meeting both criteria is the zero function.

📖 Explanation: The Taylor series of $\cos x$ contains only even powers of x ( $1 - x^{2}/2! + x^{4}/4! -\dots$ ), while the series of $\sin x$ contains only odd powers ( $x - x^{3}/3! + x^{5}/5! -\dots$ ). Because replacing x with -x leaves the even‑power terms unchanged and flips the sign of odd‑power terms, cosine is even and sine is odd.

📖 Explanation: Testing the definition, for x=2 we have f(2)=0 and f(-2)=0? Actually f(-2)=(-2)+2=0, while -f(2)=0, but checking other points such as x=0.5 gives f(0.5) = -0.5 and f(-0.5)= -0.5? The values do not satisfy either even or odd conditions consistently, so the function is neither even nor odd.

📖 Explanation: When a function is even on the interval $[-L,L]$, its Fourier expansion contains only cosine terms because cosine functions are even and sine functions are odd. The orthogonality of sine terms forces their coefficients to be zero, leaving a pure cosine series.