Difficult calculus question that includes composite functions, primes, roots, etc

Question

Let $f(x)$ be a cubic whose coefficient of the leading term is positive and $g(x) = e^{\sin(\pi x)} - 1$. The composite function $h(x) = g(f(x))$ is defined in the set of all numbers, and has a local maximum at $x=0$. In the open interval $(0,3)$ the function $h(x)$ intersects the line $y= 1$ seven times.

Given that $f(3) =1/2$, $f'(3) = 0$, and $f(2) = \frac{q}{p}$, find the value of $p+q$ given that $p$ and $q$ are coprime natural numbers

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This is my working so far:

Since $h(x)$ has a local maximum at $x=0$, this means that at $x=0$ the derivative of $h(x) = 0$, i.e. $$h'(0) = g'f(0) \cdot f'(0) = 0$$

We need to find the derivative of g(x) first to find an expression for the derivative of h(x), so

$$\frac{d}{dx} g(x)$$ $$= \frac{d}{dx} (e^{\sin(\pi x)} - 1)$$ $$g'(x)= \pi \cos(\pi x) \cdot e^{\sin(\pi x)}$$

Hence $$h'(0) = \pi \cdot \cos(\pi f(0)) \cdot e^{\sin(\pi f(0))} \cdot f'(0) = 0$$ $$\cos(\pi f(0)) \cdot e^{\sin(\pi f(0))} \cdot f'(0) = 0$$

Given that any exponential function of the form $e^x$ is only defined for $x>0$, $e^{\sin(\pi f(0))} \neq 0$.

So we are left with

$$\cos(\pi f(0)) \cdot f'(0) = 0$$

Case 1: $$\cos(\pi f(0)) = 0$$

$$\pi f(0) = \frac{\pi}{2} + k\pi$$ where $k \in \mathbb{Z}$

$$f(0) = \frac{1}{2} + k$$

Because $f(x)$ is a cubic, we know it is of the form $f(x) = ax^3 + bx^2 + cx + d$

Hence, $$f(0) = d$$

So $d= \frac{1}{2} + k$

Case 2:

$$f'(0) = 0$$

$f'(x) = 3ax^2 + 2bx + c$, so

$$f'(0) = c = 0$$

So $c=0$

Hence $f(x) = ax^3 + bx^2 + d$, where $a>0$ and $d = \frac{1}{2} + k$

This is a part where I think I might have done something wrong. I assumed both $c=0$ and $d=\frac{1}{2} + k$, but only one of them have to be true right? Because at first I thought both conditions would need to be true, but only one of $\cos(\pi f(0))$ or $f'(0)$ needs to be equal to $0$ for $h'(0) = 0$ to hold true (remember that $h'(0) = \cos(\pi f(0)) \cdot f'(0)$)

Carrying on,

$$f(3) = \frac{1}{2}$$ $$a \cdot 3^{3} + b \cdot 3^{2} + d= \frac{1}{2}$$ $$27a + 8b + d = \frac{1}{2}$$

$$f'(3) = 0$$ $$ 3a \cdot 3^{2} + 2b \cdot 3 = 0$$ $$ 27a + 6b = 0$$

$$f(2) = \frac{q}{p}$$ $$a \cdot 2^{3} + b \cdot 2^{2} + d= \frac{q}{p}$$ $$ 8a + 4b + d = \frac{q}{p}$$

So know we have the system of equations $$27a + 8b + d = \frac{1}{2}$$
$$ 27a + 6b = 0$$ $$ 8a + 4b + d = \frac{q}{p}$$

Let's eliminate $a$ from equations $1$ and $3 $ using equation $2$. First, we solve equation $2$ for $27a$:

$ 27a = -6b $

Now we substitute $ -6b $ for $ 27a $ in equations $1$ and $3$:

1. $ -6b + 8b + d = \frac{1}{2} $
2. $ -\frac{6}{27} \cdot 8b + 4b + d = \frac{q}{p} $

Simplify those equations:

1.$ 2b + d = \frac{1}{2} $ 3. $ -\frac{48}{27}b + 4b + d = \frac{q}{p} $

Let's simplify the coefficients in equation $3$:

$ -\frac{16}{9}b + 4b + d = \frac{q}{p} $

Multiply every term by 9 to get rid of the fraction:

$ -16b + 36b + 9d = \frac{9q}{p} $

Now combine like terms:

$ 20b + 9d = \frac{9q}{p} $

Now we have two equations:

1. $ 2b + d = \frac{1}{2} $
2. $ 20b + 9d = \frac{9q}{p} $

We can multiply the first equation by 10 to align the coefficients of $b$:

$ 20b + 10d = 5 $

Now we have the system:

1. $ 20b + 10d = 5 $
2. $ 20b + 9d = \frac{9q}{p} $

Subtract the second equation from the first:

$ (20b + 10d) - (20b + 9d) = 5 - \frac{9q}{p} $

This simplifies to:

$ d = 5 - \frac{9q}{p} $

Now we can substitute this value for $d$ back into one of the previous equations to find $b$. Let's use the modified version of equation 1:

$ 2b + (5 - \frac{9q}{p}) = \frac{1}{2} $

Solve for $b$:

$ 2b = \frac{1}{2} - (5 - \frac{9q}{p}) $

$ 2b = \frac{1}{2} - 5 + \frac{9q}{p} $

$ 2b = -\frac{9}{2} + \frac{9q}{p} $

$ b = -\frac{9}{4} + \frac{9q}{2p} $

Now we can substitute the value of $b$ back into equation 2 to solve for $a$:

$ 27a + 6b = 0 $

$ 27a + 6(-\frac{9}{4} + \frac{9q}{2p}) = 0 $

$ 27a - \frac{27}{2} + \frac{27q}{p} = 0 $

$ 27a = \frac{27}{2} - \frac{27q}{p} $

$ a = \frac{1}{2} - \frac{q}{p} $

Now we have expressions for $a$, $b$, and $d$:

$$ a = \frac{1}{2} - \frac{q}{p} $$ $$ b = -\frac{9}{4} + \frac{9q}{2p} $$ $$ d = 5 - \frac{9q}{p} $$

But remember that $d=\frac{1}{2} + k$, where $k$ is an integer, so $$ \frac{1}{2} + k = 5 - \frac{9q}{p} $$

$$ k = \frac{9}{5} - \frac{9q}{p} $$

To make $k$ an integer, $\frac{9}{5} - \frac{9q}{p}$ must be an integer. This implies that $\frac{9q}{p}$ must be a fraction that can be expressed with a denominator of 5 since the first fraction has a denominator of 5. This is necessary for their difference to be an integer.

So we can write:

$$ \frac{9q}{p} = \frac{m}{5} $$

where $m$ is an integer such that when subtracted from $9$, the result is a multiple of $5$ (ex. $m=4$). This is because when we subtract $\frac{m}{5}$ from $\frac{9}{5}$, we get an integer.

$$ 5 \cdot 9q = m \cdot p $$

Since $q$ and $p$ are coprime, $p$ must be a divisor of $5$ to satisfy this equation. The only natural number divisors of $5$ are $1$ and $5$ itself, but $1$ isn't prime, so $p=5$

If $p = 5$, then $m = 9q$

Since we're looking for coprime $p$ and $q$, since $p = 5$, $q$ must not include the factor $5$, and it can be any other natural number.

$$f(x) = ax^3 + bx^2 + d$$ $$f(x)= (\frac{1}{2} - \frac{q}{p})x^3 + (-\frac{9}{4} + \frac{9q}{2p})x^2 + 5 - \frac{9q}{p}$$ $$ f(x)= (\frac{1}{2} - \frac{q}{5})x^3 + (-\frac{9}{4} + \frac{9q}{10})x^2 + 5 - \frac{9q}{5}$$

Now I'm stuck

The number of different real roots of the equation $h(x) = 1$ in the open interval $(0,3)$ is $7$.

I don't even know what to do with this information

I tried setting $h(x) = 1$ and solving for $x$

$$e^{\sin(\pi f(x))} - 1 = 1$$ $$e^{\sin(\pi f(x))} = 2$$ $$\sin(\pi f(x)) = \ln(2)$$

but this led nowhere again

Please can someone help me solve this

Answer 1

This answer proves that there are infinitely many $(p,q)$ satisfying the given conditions.

Proof :

Let $f(x):=ax^3+bx^2+cx+d$ with $a\gt 0$.

Then, we have $$\begin{align}f'(x)&=3ax^2+2bx+c \\\\f''(x)&=6ax+2b \\\\g'(x)&=\pi e^{\sin(\pi x)}\cos(\pi x) \\\\g''(x)&=\pi^2 e^{\sin(\pi x)}\bigg(\cos^2(\pi x)-\sin(\pi x)\bigg) \\\\h'(x)&=g'(f(x))f'(x) \\\\h''(x)&=g''(f(x))(f'(x))^2+g'(f(x))f''(x)\end{align}$$

Since $h(x)$ has a local maximum at $x=0$, we have to have $h'(0)=0$ and $h''(0)\le 0$. Since $$h'(0)=\pi e^{\sin(\pi d)}\cos(\pi d)c$$ $$h''(0)=\pi^2c^2 e^{\sin(\pi d)}\bigg(\cos^2(\pi d)-\sin(\pi d)\bigg)+2b\pi e^{\sin(\pi d)}\cos(\pi d)$$ we have $$\cos(\pi d)c=0\tag1$$ $$-\pi c^2\sin(\pi d)+2b\cos(\pi d)\le 0\tag2$$

In the following, let us consider the case $\cos(\pi d)\not=0$, i.e. $d\not=\frac{1}{2}+k$ where $k$ is an integer.

Then, we have $$c=0\tag3$$ $$b\cos(\pi d)\le 0\tag4$$

We have $f(3)=\frac 12,f'(3)=0$ and $f(2)=\frac qp$, so $$27a+9b+d=\frac 12$$ $$27a+6b=0$$ $$8a+4b+d=\frac qp$$ So, we can represent $a,b,d$ by $\frac qp$. $$a=\frac 27\bigg(\frac qp-\frac 12\bigg)\tag5$$ $$b=-\frac{9}{7}\bigg(\frac qp-\frac{1}{2}\bigg)\tag6$$ $$d=\frac{27q}{7p}-\frac{10}{7}\tag7$$

We have $a\gt 0$, so $$\frac qp\gt \frac 12\tag8$$ $$b\lt 0\tag9$$ $$\cos(\pi d)\gt 0\tag{10}$$ $$\frac{q}{p}\not=\frac{7}{27}k+\frac 12\tag{11}$$

Now, $h(x)=1$ is equivalent to $$\begin{align}&g(f(x))=1 \\\\&\iff e^{\sin(\pi f(x))} =2 \\\\&\iff \sin(\pi f(x))=\ln 2 \\\\&\iff \pi f(x)=\arcsin(\ln 2)+2m_1\pi\ \text{or}\ \pi-\arcsin(\ln 2)+2m_2\pi \\\\&\iff f(x)=\frac{\arcsin(\ln 2)}{\pi}+2m_1\ \text{or}\ -\frac{\arcsin(\ln 2)}{\pi}+2m_2+1\end{align}$$ where $m_1,m_2$ are integers.

Here, $$0\lt\frac{\arcsin(\ln 2)}{\pi}\lt \frac 12\tag{12}$$ holds since $1\lt 2\lt e$.

Since $f'(0)=f'(3)=0$, we see that $y=f(x)$ is strictly decreasing in $(0,3)$.

Since $h(x)$ intersects the line $y= 1$ seven times in $(0,3)$ with $f(0)=\frac{27q}{7p}-\frac{10}{7}$ and $f(3)=\frac 12$, we have $$7-\frac{\arcsin(\ln 2)}{\pi}\lt \frac{27q}{7p}-\frac{10}{7}\le \frac{\arcsin(\ln 2)}{\pi}+8$$ i.e. $$\frac{59}{27}-\frac{7}{27}\cdot\frac{\arcsin(\ln 2)}{\pi}\lt\frac qp\le\frac{22}{9}+\frac{7}{27}\cdot\frac{\arcsin(\ln 2)}{\pi}\tag{13}$$

From $(10)$, we have $$-\frac{\pi}{2}+2s\pi\lt\pi d\lt \frac{\pi}{2}+2s\pi$$ i.e. $$\frac{13}{54}+\frac{14}{27}s\lt \frac{q}{p}\lt \frac{1}{2}+\frac{14}{27}s\tag{14}$$ where $s$ is an integer.

From $(13)$ and $(14)$, we finally get $$(2.3\dot 14\dot 8=)\ \color{red}{\frac{125}{54}\lt\frac qp\le \frac{22}{9}+\frac{7}{27}\cdot\frac{\arcsin(\ln 2)}{\pi}}\ (\approx 2.5076)$$

Therefore, we see that there are infinitely many $(p,q)$ satisfying the given conditions.$\ \blacksquare$

(For example, $(p,q)=(10^{n+1},\frac{7\cdot 10^{n+1}-1}{3})$ work where $n$ is a positive integer. For $n=1,2,3$, we have $(p,q)=(100,233),(1000,2333),(10000,23333)$.)

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Example 1 : For $(p,q)=(5,12)$ where $(a,b,c,d)=(\frac{19}{35},-\frac{171}{70},0,\frac{274}{35})$, $h(x)$ intersects $y= 1$ seven times in $(0,3)$. (see here)

Example 2 : For $(p,q)=(2,5)$ where $(a,b,c,d)=(\frac 47,-\frac{18}{7},0,\frac{115}{14})$, $h(x)$ intersects $y=1$ seven times in $(0,3)$. (see here)

Example 3 : For $(p,q)=(25,58)$ where $(a,b,c,d)=(\frac{13}{25},-\frac{117}{50},0,\frac{188}{25})$, $h(x)$ intersects $y= 1$ seven times in $(0,3)$. (see here)

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Comment :

I think we need at least one more information to determine $p,q$. (For example, if the question says "..., and has a local maximum $\color{red}0$ at $x=0$", then $c=0$ follows and $(p,q)=(9,22)$ is the only solution.)