Problem source

In this question, you may assume that, if a continuous  function 
takes both positive and negative values in an interval, then
it takes the value $0$ at some point in that interval. 
\begin{questionparts}
\item
The function  $\f$ is continuous and $\f(x)$ is non-zero for some value 
of $x$ in the interval $0\le x \le 1$. Prove by contradiction, or otherwise,
 that if
\[
\int_0^1 \f(x) \d x = 0\,,
\]
then 
 $\f(x)$ takes both positive and negative values 
in the interval $0\le x\le 1$.
\item The function  $\g$ is continuous and
\[
\int_0^1 \g(x) \, \d x = 1\,,
 \quad 
\int_0^1 x\g(x) \, \d x = \alpha\, , 
\quad 
\int_0^1 x^2\g(x) \, \d x = \alpha^2\,.
\tag{$*$}
\] 
 Show, by considering
\[
\int_0^1 (x - \alpha)^2 \g(x) \, \d x
\,,
\]
 that
$\g(x)=0$ for some value of $x$ in the interval $0\le x\le 1$.
Find a function of the  form $\g(x) = a+bx$ that satisfies the conditions
$(*)$ and verify that $\g(x)=0$ for some value of $x$ in the interval 
$0\le x \le 1$.
\item
The function $\h$ has a continuous derivative $\h'$ and
\[
\h(0) = 0\,,
\quad 
\h(1) = 1\,,
\quad 
\int_0^1 \h(x) \, \d x = \beta\,, 
\quad 
\int_0^1 x \h(x) \, \d x = \tfrac{1}{2}\beta (2 - \beta)
\,.
\] 
Use the result in part (ii) to show that $\h^\prime(x)=0$ 
for some value of $x$ in the interval $0\le x\le 1$.
\end{questionparts}

Solution source

\begin{questionparts}
\item Claim: If $f(x)$ non-zero for some $x \in [0,1]$ and $\int_0^1 f(x) \d x =0$ then $f$ takes both positive and negative values in the interval $[0,1]$.

Proof: Suppose not, then WLOG suppose $f(x) > 0$ for some $x \in [0,1]$. Then notice (since $f$ is continuous) that there is some interval where $f(x) > 0$ around the $x$ we have already shown exists. But then $\int_{\text{interval}} f(x) \d x > 0$ and since $f(x) \geq 0$ everywhere $\int_0^1 f(x) \d x > 0$, which is a contradiction.

\item $\,$ \begin{align*}
&& \int_0^1 (x - \alpha)^2 g(x) \d x &= \int_0^1 x^2g(x) \d x - 2 \alpha \int_0^1 x g(x) \d x + \alpha^2 \int_0^1 g(x) \d x \\
&&&= \alpha^2 - 2\alpha \cdot \alpha + \alpha^2 \cdot 1 \\
&&&= 0
\end{align*} Therefore $g(x)(x-\alpha)^2$ is a continuous function which is either exactly $0$ (in which case we've already found our $0$) or it is a continuous function which is both positive somewhere and has $0$ integral, and therefore by part (i) must take both positive and negative values (and therefore takes $0$ in between those points by continuity).

\begin{align*}
&&1 &= \int_0^1 a+bx \d x \\
&&&= a + \frac12 b \\
&& \alpha &= \int_0^1 ax + bx^2 \d x \\
&&&= \frac12 a + \frac13 b \\
&& \alpha^2 &= \int_0^1 ax^2 + bx^3 \d x\\
&&&= \frac13 a + \frac14 b \\ 
\Rightarrow && \frac1{36}(3a+2b)^2 &= \frac1{12}(4a+3b) \\
\Rightarrow && \frac1{36}(3a+2(2-2a))^2 &= \frac1{12}(4a+3(2-2a)) \\
\Rightarrow && (4-a)^2 &= 3(6-2a) \\
\Rightarrow && 16-8a+a^2 &= 18-6a \\
\Rightarrow && a^2-2a-2 &= 0 \\
\Rightarrow && (a-1)^2 - 3 &= 0 \\
\Rightarrow && a &= \pm \sqrt{3}+1 \\
&& b &= \mp 2\sqrt{3}
\end{align*}

So say $a = \sqrt{3}+1, b = -2\sqrt{3}$

This has a root at $-\frac{a}{b} = \frac{1+\sqrt{3}}{2\sqrt{3}} = \frac{\sqrt{3}+3}{6} < 1$ so we have met our condition.

\item Consider $h'$, we must have

\begin{align*}
&& \int_0^1 h'(x)\d x &= h(1)-h(0) =1\\
&& \int_0^1 xh'(x) \d x &= \left [x h(x) \right]_0^1 - \int_0^1 h(x) \d x \\
&&&= 1 - \beta \\
&& \int_0^1 x^2 h'(x) \d x &= \left [ x^2h(x) \right]_0^1 - \int_0^1 2xh(x) \d x \\
&&&= 1 - 2\tfrac12 \beta(2-\beta) \\
&&&= (1-\beta)^2
\end{align*}

Therefore $h'$ satisfies the conditions with $\alpha = 1-\beta$, so $h'$ must have a root in our interval.
\end{questionparts}