Question

let $f$ and $g$ have continuous first and second derivatives everywhere. if $f(x) \leq g(x)$ for all real $x$, which of the following must be true?
i. $f(x) \leq g(x)$ for all real $x$
ii. $f(x) \leq g(x)$ for all real $x$.
iii. $\int_{0}^{1} f(x)dx \leq \int_{0}^{1} g(x)dx$
a none
b i only
c iii only
d i and ii only
e i, ii, and iii

Explanation:

Step1: Test Statement I with examples

Let $f(x) = 0$ and $g(x) = 1$. Then $f(x) \leq g(x)$ for all $x$, but $f'(x)=0$ and $g'(x)=0$, so this case holds. Now take $f(x) = x$ and $g(x) = x^2$ for $x \geq 2$; adjust to make $f(x) \leq g(x)$ everywhere: let $f(x)=x$ and $g(x)=x^2 + 1$. Then $f(x) \leq g(x)$ for all $x$, but $f'(x)=1$, $g'(x)=2x$. For $x=0$, $g'(0)=0 < 1 = f'(0)$, so $f'(x) \leq g'(x)$ is not always true.

Step2: Test Statement II with examples

Let $f(x) = 0$ and $g(x) = x^2$. Then $f(x) \leq g(x)$ for all $x$, $f''(x)=0$, $g''(x)=2$, so this case holds. Now take $f(x) = -x^2$ and $g(x) = 0$. Then $f(x) \leq g(x)$ for all $x$, but $f''(x) = -2$, $g''(x)=0$, so $f''(x) = -2 < 0 = g''(x)$ here. Now take $f(x) = x^2$ and $g(x) = x^2 + 1$: $f''(x)=2$, $g''(x)=2$. Now take $f(x) = 2x^2$ and $g(x) = x^2 + 10$: $f(x) \leq g(x)$ when $x^2 \leq 10$, adjust to $f(x)=2x^2$ and $g(x)=3x^2 + 1$: $f(x) \leq g(x)$ for all $x$, $f''(x)=4$, $g''(x)=6$, which holds. Now a counterexample: let $f(x) = 0$ and $g(x) = -x^4$. Wait no, $g(x) = -x^4 \leq 0 = f(x)$, which violates $f(x) \leq g(x)$. Correct counterexample: $f(x) = -x^4$ and $g(x) = 0$. Then $f(x) \leq g(x)$ for all $x$, $f''(x) = -12x^2$, $g''(x)=0$. For any $x
eq 0$, $f''(x) = -12x^2 < 0 = g''(x)$, but wait we need $f''(x) > g''(x)$ to disprove. Let $f(x) = x^4$ and $g(x) = x^2$. For $|x| \geq 1$, $x^4 \geq x^2$, so adjust: $f(x) = x^2$ and $g(x) = x^4 + 1$. Now $f(x) \leq g(x)$ for all $x$. $f''(x)=2$, $g''(x)=12x^2$. For $x=0$, $g''(0)=0 < 2 = f''(x)$, so $f''(x) \leq g''(x)$ is not always true.

Step3: Verify Statement III via integral property

By the monotonicity property of definite integrals: if $f(x) \leq g(x)$ for all $x \in [a,b]$, then $\int_{a}^{b} f(x)dx \leq \int_{a}^{b} g(x)dx$. Here $a=0$, $b=1$, and $f(x) \leq g(x)$ for all real $x$, so it holds on $[0,1]$. Thus $\int_{0}^{1} f(x)dx \leq \int_{0}^{1} g(x)dx$ must be true.

Answer:

C. III only