What is the number of possible solution(s)of the equation $\displaystyle\int_{0}^{x} t^2-8t+13\,\mathrm dt=x\sin\left(\dfrac ax\right)$?

I tried applying Leibniz's rule and differentiated both sides. I got the equation $x^2-8x+13=\sin\left(\dfrac ax\right)-\dfrac ax\cos\left(\dfrac ax\right)$. How should I proceed now? Thanks in advance.

  • $\begingroup$ Equation $f(x)=g(x)$ has nothing to do with equation $f'(x)=g'(x)$... And anyway it should be $\sin(a/x)-\frac ax\cos(a/x)$. $\endgroup$ – Nicolas FRANCOIS Aug 17 '18 at 12:02
  • $\begingroup$ $\int_0^x(t^2-8t+13){\rm d}t=\frac{x^3}{3}-4x^2+13x=P(x)$ is a polynomial function of third degree. So there is $M>0$ such that $\left|P(x)\right|>\left|x\right|$ for all $x\notin[-M,M]$. You may study the variations of functions $P$ and the other member, to look for ideas. $\endgroup$ – Nicolas FRANCOIS Aug 17 '18 at 12:05
  • $\begingroup$ @NicolasFRANCOIS extremely ignorant on my part. Thanks for the correction. $\endgroup$ – Arka Seth Aug 17 '18 at 12:06

Here is my attempt. First integrate the LHS and conclude:

$$\int_0^x t^2-8t+13 dt= \frac{t^3}{3}-4t^2+13t\lvert_{t=0}^{t=x}=\frac{x^3}{3}-4x^2+13x$$

Then the equation becomes


Since the x cannot be zero, divide both sides by x and we have


Thus for any $x$ that is a solution to the original solution, it also has to solve the above equation. Now observe the LHS can be rewritten as:


Conclude that the LHS achieves its minimum 1 at $x=6$.

Now since $sin(\frac{a}{x})$ has range $[-1,1]$. The LHS and the RHS can only intersect at most once. Thus we can consider two cases:

  1. For all $a$ such that $sin(a/6)=1$. The solution to the equation is x=6.

  2. For all other values of a, no solution exists.

Please let me know if you find any error.

  • $\begingroup$ It is correct. Thanks a lot. $\endgroup$ – Arka Seth Aug 17 '18 at 12:16

Hint: You will get the equation

$$\frac{1}{3}x^2+4x+13=\sin \left(\frac{a}{x}\right)$$ Now it must be $$\left|\sin \left(\frac{a}{x}\right)\right|\le 1$$

Can you proceed?

  • 1
    $\begingroup$ Should be $-4x$. $\endgroup$ – Nicolas FRANCOIS Aug 17 '18 at 12:06
  • $\begingroup$ Yes I can proceed now... Thanks a lot $\endgroup$ – Arka Seth Aug 17 '18 at 12:06

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