# Number of possible solution(s)?

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.

• Equation $f(x)=g(x)$ has nothing to do with equation $f'(x)=g'(x)$... And anyway it should be $\sin(a/x)-\frac ax\cos(a/x)$. – Nicolas FRANCOIS Aug 17 '18 at 12:02
• $\int_0^x(t^2-8t+13){\rm d}t=\frac{x^3}{3}-4x^2+13x=P(x)$ 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. – Nicolas FRANCOIS Aug 17 '18 at 12:05
• @NicolasFRANCOIS extremely ignorant on my part. Thanks for the correction. – 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

$$\frac{x^3}{3}-4x^2+13x=xsin(\frac{a}{x})$$

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

$$\frac{x^2}{3}-4x+13=sin(\frac{a}{x})$$

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:

$$\frac{1}{3}(x^2-12x+36)+1=\frac{1}{3}(x-6)^2+1$$

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.

• It is correct. Thanks a lot. – 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?

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