what is the range of the function y = 1 + 2sin(x - π)?\n\na. -1 to 1\nb. -2 to 2\nc. 0 to 3\nd. -1 to 3\ne…

Explanation:

Step1: Recall sine - function range

The range of the basic sine function $y = \sin(u)$ is $[- 1,1]$. Here $u=x-\pi$. So, $-1\leqslant\sin(x - \pi)\leqslant1$.

Step2: Multiply by 2

Multiply each part of the inequality by 2. We get $-2\leqslant2\sin(x - \pi)\leqslant2$ (by the property of inequalities: if $a\leqslant b\leqslant c$ and $k>0$, then $ka\leqslant kb\leqslant kc$).

Step3: Add 1

Add 1 to each part of the inequality $-2\leqslant2\sin(x - \pi)\leqslant2$. We have $-2 + 1\leqslant1+2\sin(x - \pi)\leqslant2 + 1$, which simplifies to $-1\leqslant1 + 2\sin(x - \pi)\leqslant3$.

Answer:

D. -1 to 3