How to solve for $x$ in $\sin(\pi x) = x$?

Question

For the fourth option,

Given $x \in(0,1) $ implying that $[x] = 0$. And thus on equating $ f(x_0) = g(x_0) $,

I find that $\sin(\pi x_0 ) = x_0$

I have no idea how to solve for x here. What should I do?

Edit:- On drawing a rough graph, I did find the option to be true, as the graph of $y = \sin(\pi x)$ will intersect the graph of $ y=x $ in the given interval.

So is that enough to prove the option? Or should I find the value at that point as well to verify my answer with reference to the graph?

Answer 1

The value of $x$ that makes

$$\sin(\pi x) = x$$

cannot be solved for explicitly in terms of canonized special functions. However, you do not actually need an explicit solution to prove that a solution exists. It is rather easy to do this with the intermediate value theorem. In particular, modify the equation to

$$\sin(\pi x) - x = 0$$

and then you will want to find two points $x_l$ and $x_h$ so that, setting $f_2(x) := \sin(\pi x) - x$ for convenience of writing, provably $f_2(x_l) < 0$ and $f_2(x_h) > 0$. It will then follow, given that $f_2$ is continuous, from the IVT that a solution exists.

Answer 2

I might

• Show that $f(x) - g(x)$ is continuous on $(0,1)$.
• Study the intervals on which $f(x) - g(x)$ is increasing and decreasing in $(0,1)$. (There is one of each.)
• Show that $f(x) - g(x) > 0$ on the increasing interval, so there is no $x_0$ in that interval. (A short argument uses both that $f(0) = 0$ and the sign of $f'(x) - g'(x)$.)
• Show that $f(x) - g(x)$ changes sign exactly once on the decreasing interval. The intermediate value theorem gives at least one $x_0$ on that interval. An argument similar to the previous step shows there are no more, establishing the uniqueness.

Answer 3

f(x)-g(x) is continuous on (1/2,1) and ranges (-1,1/2). Therefore passes through 0.

Beware that A is false. When x increases to approach 1, f(x) decreases to 0 while g(x) increases to 1. Therefore g(x)>f(x).