If the function $f(x)=\mathrm{sign} (\sin^2x-\sin x-1)$ has exactly 4 points of discontinuity for all $x\in (0,n\pi)$, find $n$

Question

For discontinuity $$\sin^2x-\sin x-1=0$$ $$\sin x = \frac{1\pm \sqrt 5}{2}$$

In $(0,2\pi)$, there are 4 such values of $x$ which satisfy the given condition. But the given answer is $(0,4\pi)$. Where am I going wrong?

There are no points where $\sin x=(1+\sqrt5)/2$, and two where $\sin x=(1-\sqrt5)/2$. — Gerry Myerson, Sep 13 '20 at 09:17
But it's hard to tell where you went wrong, since you haven't shown us what you did. — Gerry Myerson, Sep 13 '20 at 09:18

score 1 · Accepted Answer · answered Sep 13 '20 at 09:20

1

In $(0,2\pi)$ there are exactly two values of $x$ such that $\sin x=\frac{1+\sqrt 5}2\lor \sin x=\frac{1-\sqrt5}2$: they are $x=2\pi+\arcsin\frac{1-\sqrt5}2$ and $x=\pi-\arcsin\frac{1-\sqrt5}2$.

answered Sep 13 '20 at 09:20

If the function $f(x)=\mathrm{sign} (\sin^2x-\sin x-1)$ has exactly 4 points of discontinuity for all $x\in (0,n\pi)$, find $n$

1 Answers1