finding value of $p,q$ such that function $f(x)$ is continuous at $x=-1$

Question

If $$f(x)=\left\{\begin{matrix} \sin\bigg(\pi(x+p)\bigg)\;\; , &x<-1 \\ q\bigg(\lfloor x \rfloor^2+\lfloor x \rfloor\bigg)+1\;\;,\;\;& x\geq -1 \end{matrix}\right.$$ where $\lfloor x \rfloor $ represent the integer part of $x,$ . Then what values of $p,q$ function $f(x)$ is continuous at $x=-1$

what i try: If function $f(x)$ is continuous at $x=-1.$ Then

$$f(-1^{-})=f(-1^+)=f(1)$$

Here $f(-1^{-})=\lim_{h\rightarrow 0}f(-1-h)=\lim_{h\rightarrow 0}\sin(\pi(-1-h)+\pi p)=\sin(-\pi +\pi p)=-\sin(\pi p)$

and $f(-1^+)=\lim_{h\rightarrow 0}f(-1+h)=\lim_{h\rightarrow 0}q\bigg(\lfloor h-1\rfloor^2+\lfloor h-1\rfloor\bigg)+1=-q+1$

and $f(-1)=q\bigg(1-1\bigg)+1=1$

so for continuity at $x=-1,$

$-\sin(\pi p)=1-q=1$, we get $q=0$ and $\displaystyle \sin(\pi p)=-1=-\sin\bigg(2n\pi+\frac{3\pi}{2}\bigg)\Longrightarrow p=2n+\frac{3}{2}$

so we have $p=2n+1.5$ and $q=0,$ But answer given as $q\in \mathbb{R}$

please help me How $q\in \mathbb{R}$ and whats wrong in my answer. Thanks

Answer 1

In your step for calculating $f(-1+h)$, you have miscalculated

$$ [-1+h] = -1 $$

$$\implies f(-1+h) = q( (-1)^2 + (-1)) +1$$

$$\implies f(-1+h) = q( 1 -1) + 1 = 1$$

Answer 2

When $\;x\ge-1\;$ but very close to $\;-1\;$ , say $\;x\in[-1,-1+\epsilon)\,,\,\,0<\epsilon<<1\;$, we have $\;\lfloor x\rfloor=-1\;$ ,so

$$q\left(\lfloor x\rfloor^2+\lfloor x\rfloor\right)=q(1-1)=0\implies\lim_{x\to-1^+} f(x)=1$$

whereas

$$\lim_{x\to-1^+}f(x)=\lim_{x\to-1^+}\sin\left(\pi(x+p)\right)=\sin\left(-\pi+p\pi\right)=-\sin p\pi$$

Thus, it must be

$$-\sin p\pi=1\iff\sin p\pi=-1\iff p=\frac n2\,,\,\,n=3\pmod 4\,,\,\,q\in\Bbb R$$

Of course, instead of $\;p=\frac n2,\,n=3\pmod 4\,,\,$ we can simply write $\;p=-\cfrac\pi2+2k\pi\,,\,\,k\in\Bbb Z$