MATHEMATICS 120 PROBLEM SET 7 – SOLUTIONS
1. Find all values of a for which the equation ln x = ax2 has exactly one solution.
Draw the graphs of y = ln x and y = ax2 for several values of a. (see below). If a ≤ 0, the
graphs always intersect at exactly one point. If a > 0, then there is exactly one solution if
and only if the graphs are tangent to each other. Let (x0 , y0 ) be the tangency point. Then
y0 = ln x0 = ax20 . Also, the slopes of the graphs must be equal, hence 1/x0 = 2ax0 , i.e.
2ax20 = 1. We get that x0 = (2a)−1/2 and ln((2a)−1/2 ) = ax20 = 1/2. We solve this for a:
1
1
1
− ln(2a) = , ln(2a) = −1, 2a = e−1 , a = .
2
2
2e
The final answer is: a ≤ 0 and a = (2e)−1 .
2. A radioactive substance decays at a rate proportional to the amount present. If 10% of
the substance decays in 2 years, how long will it take for 80% of it to decay?
Let y(t) be the amount of the radioactive substance present at time t (measured in years),
with y(0) = A. Then y(t) = Aekt for some unknown k. We know that y(2) = 0.9y(0) =
0.9A. Hence Ae2k = 0.9A, 2k = ln 0.9, k = 12 ln 0.9. We need to find t such that y(t) =
0.2A:
1
2 ln 0.2
1
y(t) = Ae 2 t ln 0.9 = 0.2A, t ln 0.9 = ln 0.2, t =
≈ 30.55 years.
2
ln 0.9
x
3. Show that arctan x = arcsin √
for all x.
1 + x2
Let u = arctan x, then tan u = x and −π/2 < u < π/2 (so that cos u > 0). Therefore
p
p
√
1 + x2 = 1 + tan2 u = sec2 u = sec u
and
√
so that u = arcsin √
tan u
x
= tan u · cos u = sin u,
=
2
sec u
1+x
x
.
1 + x2
x2
for all x 6= 0.
2
(−x)2
x2
=1+
, it is enough to prove it for x > 0. We
Since cosh x − cosh(−x) and 1 +
2
2
2
0
x2
have cosh 0 = 1 +
= 0, therefore it suffices to prove that the function cosh x − 1 −
2
2
is increasing for x > 0, i.e.
4. Prove that cosh x > 1 +
d
x2
(cosh x − 1 − ) = sinh x − x > 0.
dx
2
1
We have sinh 0 − 0 = 0, so it again suffices to prove that sinh x − x is increasing for x > 0,
i.e. (sinh x − x)0 = cosh x − 1 > 0. At x = 0 we have cosh 0 − 1 = 0, so it suffices to prove
that (cosh x − x)0 = sinh x > 0 for x > 0. But this is clear, since sinh x = (ex − e−x )/2 and
for x > 0 we have ex > 1 > e−x .
2