(This is a solution using the properties of distributions. A
shorter solution exists with the use of logarithms)
To begin solving this problem I sketched out some possible
curve-intersection scenarios based on the assumption that a normal
distribution curve looks 'bell-shaped'. I remember that any normal
distribution is exactly characterised by its mean $\mu$ and
standard deviation $\sigma$ and the probability density function in
this case is
$$
f(x) =
\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}
$$
I thought of these plausible cases, where I have indicated the
asymptotic behaviour of the two curves, from which I could start a
more careful analysis:
This is a problem which will benefit from a 'case-by-case'
analysis.
For zero intersections (case A) one of the density functions must
be below the other at all points. Since the integrals (areas
beneath the curves) of the density functions must equal
$1$ this is impossible. Thus, the two density functions must
intersect at least once: zero
intersections is impossible.
For case B there is one intersection. To prove that this is the
case imagine starting with two $N(0,1)$ distributions and
increasing the mean of one of them slightly: this translates the
curve to the right a little. Since any normal density function
decreases with distance from the mean we can deduce that the form
of our picture is correct, so 1
intersection is possible.
The sliding argument used previously might work for case C. Let's
see. We'd need to change the standard deviation of one of the
distributions. As I reduce the standard deviation of the red curve
it would get more spiked, but both of the tails are flattened (you
can see this by looking at the distribution function). For
example, for large enough $x$ we have:
$$
N(0,1)\sim \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}>
\frac{1}{\sqrt{2\pi\times 10}}e^{-\frac{(x-1)^2}{2\times 10}}\sim
N(1, 10)
$$
whereas for $x=1$ we have
$$
\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}<
\frac{1}{\sqrt{2\pi\times 10}}e^{-\frac{(x-1)^2}{2\times 10}}
$$
Thus, C is possible. Case E is very similar. Thus, two intersections are
possible.
For the red curve of case $D$ to peak above the blue curve we would
need the red curve to have a smaller standard deviation; however,
this means that the tails would have to be below those of the blue
curve; thus case D is impossible. However, this does not rule out
rather improbable cases such as
So, what is it about this configuration which makes it impossible
(which I feel that it is)? Well, the gradient of one of the blue
curve would need to increase and decrease. This would mean
that we would need more than two points of inflection; a quick
differentiation shows that this is impossible.