From my exam

Let $f(x)-\frac{1}{2} x=g(x)$ .

Note that if $A=(a, \frac{1}{2}a), B=(b, \frac{1}{2}b), C=(c, \frac{1}{2}c)$ then $g(x)$ is of the form $p(x-a)(x-b)(x-c)$ .

Now note that the ratio of the derivative of $g(x)$ at point $(a, 0)$ and point $(c,0)$ can be expressed in the from $(b-a):(c-b) =4-\frac{1}{2}: 5-\frac{1}{2}=7: 9$

Using this, let $b=a+7k, c=a+16k$ . Note that the derivative of $g(x)$ at point $(a,0)$ is $p \times 7k \times 16k =\frac{7}{2} \Leftrightarrow pk^2=\frac{1}{32}$

Using the same method $g(x)$ at point $(b,0)$ can be calculated to be $7k \times 9k \times p= -\frac{63}{32}$ , the derivative of $f(x)$ at point $(b, \frac{1}{2}b )$ is $-\frac{63}{32}+\frac{1}{2}=-\frac{47}{32}$ .

The answer is $32 + 47 =79$ .

Let's propose a very powerful theorem :

$\displaystyle \color{#D61F06} \text{If a polynomial of degree} \ \color{#333333}n \ \color{#D61F06} \text{has real roots} \ \color{#333333} x_1, x_2, x_3,\cdots, x_n \ \color{#D61F06} \text{such that}\ \color{#333333} x_1 < x_2 < x_3 <\cdots <x_n.$

$\displaystyle \color{#D61F06} \text{If slopes of tangents at} \ \color{#333333} x_i \ \color{#D61F06} \text{are}\color{#333333} \ m_i,\ \color{#D61F06} \text{then},$

$\large \frac{1}{m_1} + \frac{1}{m_2} + \frac{1}{m_3} + \cdots + \frac{1}{m_n} = 0$

PROOF

Let $\displaystyle P(x) = a_n x^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + a_{n-3}x^{n-3} + \cdots + a_0$ have roots $x_1, x_2, x_3,\cdots, x_n$ .

Then we can also write it as - $\displaystyle P(x) = a_n (x-x_1) (x-x_2) (x-x_3) (x - x_4) \cdots (x - x_n)$

Let $a = a_n$ ,

$\displaystyle P'(x) = a (x - x_2)(x - x_3) \cdots (x - x_n) + a (x - x_1)(x - x_3) \cdots (x - x_n) + \cdots + a (x - x_1)(x - x_2) \cdots (x - x_{n-1})$

$\displaystyle P'(x_1) = m_1 = a (x_1 - x_2)(x_1 - x_3) \cdots (x_1 - x_n)$

$\displaystyle P'(x_2) = m_2 = a (x_2 - x_1)(x_2 - x_3) \cdots (x_2 - x_n)$

$\vdots$

$\displaystyle P'(x_n) = m_n = a (x_n - x_1)(x_n - x_2) \cdots (x_n - x_{n-1})$

Now, experience tells us that $\frac{1}{m_i}$ 's might be a good way to go.(One can check cases like $n=2, 3$ to feel more comfortable).

$\displaystyle \frac{1}{m_1} = \frac{1}{a (x_1 - x_2)(x_1 - x_3) \cdots (x_1 - x_n)}$

$\displaystyle \frac{1}{m_2} = \frac{1}{a (x_2 - x_1)(x_2 - x_3) \cdots (x_2 - x_n)}$

$\vdots$

$\displaystyle \frac{1}{m_n} = \frac{1}{a (x_n - x_1)(x_n - x_2) \cdots (x_n - x_{n-1})}$

Let's take $\frac{1}{m_1}$ and try to expand it using partial fractions.

$\displaystyle \frac{1}{m_1} = \frac{A_2}{a (x_1 - x_2)} + \frac{A_3}{a (x_1 - x_3)} + \cdots + \frac{A_n}{a (x_1 - x_n)}$

What is the value of $A_2, A_3, \cdots A_n$ ?

We know that

$\displaystyle 1 = A_2 (x_1 - x_3) (x_1 - x_4) \cdots (x_1 - x_n) + A_3 (x_1 - x_2) (x_1 - x_4) \cdots (x_1 - x_n) + \cdots + A_n (x_1 - x_3) (x_1 - x_4) \cdots (x_1 - x_{n-1})$

Thus, if we substitute $x_1 = x_2$ , we get

$A_2 = \frac{1}{(x_2 - x_3)(x_2 - x_4)\cdots (x_2 - x_n)}$ and so on.

Our expression for $\frac{1}{m_1}$ therefore becomes -

$\displaystyle \frac{1}{m_1} = \frac{\frac{1}{(x_2 - x_3)(x_2 - x_4)\ldots (x_2 - x_n)}}{a (x_1 - x_2)} + \frac{\frac{1}{(x_3 - x_2)(x_3 - x_4)\ldots (x_3 - x_n)}}{a (x_1 - x_3)} + \ldots + \frac{\frac{1}{(x_n - x_2)(x_n - x_3)\ldots (x_n - x_{n-1})}}{a (x_1 - x_n)}$

Observe carefully and one will get

$\displaystyle \frac{1}{m_1} = \frac{-1}{m_2} + \frac{-1}{m_3} + \frac{-1}{m_4} + \cdots + \frac{-1}{m_n}$

$\large \text{QED}$

So, this problem just becomes $g(x) = f(x) - \frac{x}{2}$ has roots equal to $x_1, x_2, x_3$ .

$m_1 = 4 - \frac{1}{2} = \frac{7}{2}$

$m_3 = 5 - \frac{1}{2} = \frac{9}{2}$

Using above result, $m_2 = \frac{-63}{32}$

Our required slope is slope on curve $f(x)$ , which becomes - $\frac{-63}{32} + \frac{1}{2} = \frac{-47}{32}$ .

Let the points of intersections are $A=(p,p/2) ; B=(q,q/2); C=(r,r/2)$ where $p<q<r$ .As there are three intersections then

$2f'(x)-x=a(x-p)(x-q)(x-r)$ .Now differentiating w.r.t $x$

$2f'(x)-1=a[(x-p)(x-q)+(x-q)(x-r)+(x-p)(x-r)]$

Put $x=p$ then $2f'(p)-1=a(p-q)(p-r) => a(p-q)(p-r)=7$

Put $x=r$ then $2f'(r)-1=a(r-p)(r-q) => a(r-p)(r-q)=9 \implies a(p-r)(q-r)=9$

Adding these two equations $a(p-r)(p-q+q-r)=16 \implies a(p-r)^2=16 \implies (p-r)=|\dfrac{4}{\sqrt{a}}|$

Putting these values in the previous equations $p-q=\dfrac{7}{a} \times |\dfrac{4}{\sqrt{a}}|$ and $r-q=\dfrac{9}{a} \times |\dfrac{4}{\sqrt{a}}|$ )

Now put $x=q$ in the first equation $2f'(q)-1=a(q-p)(q-r)$ .Substituting these values

$2f'(q)-1=a \times \dfrac{7}{a} \times |\dfrac{4}{\sqrt{a}}| \times \dfrac{9}{a} \times |\dfrac{4}{\sqrt{a}}|$

Now the important part is to decide the modulus part when it will be negative or positive

As $p-r<0$ so , $p-r=- \frac{4}{\sqrt{a}}$

So, $2f'(q)-1=-\dfrac{63}{16}=1-\dfrac{63}{16}=-\dfrac{47}{16} \implies f'(q)=-\dfrac{47}{32}$