Matrix Matching: Let  $f:R\to [-1,1]$is defined by  $f\left(x\right)=\sin (2x+1)$. If domain is restricted to 

$f\left(x\right)=\sin (2x+1)$

A)  $\left(-\frac{3\pi }{4}-\frac{1}{2}\right)\le x\le \left(-\frac{\pi }{2}-\frac{1}{2}\right)$

$\left(-\frac{3\pi }{2}-1\right)\le 2x\le (-\pi -1)$

$-\frac{3\pi }{2}\le 2x+1\le -\pi$

$\sin \left(-\frac{3\pi }{2}\right)\le \sin (2x+1)\le \sin (-\pi )$

$+1\le f\left(x\right)\le 0$

$f$ is one-one but not onto in $\left[-\frac{3\pi }{2},-\pi \right]$

B)  $\left[-3\frac{\pi }{4}-\frac{1}{2},\frac{-1}{2}\right]$

$-\frac{3\pi }{4}-\frac{1}{2}\le -\frac{1}{2}$

$-\frac{3\pi }{2}\le 2x+1\le 0$

$\sin \left(-\frac{3\pi }{2}\right)\le \sin (2x+1)\le \sin 0$

$f$ is onto but not one-one in  $\left[-\frac{3\pi }{2},0\right]$

C)  $\left[\frac{\pi }{4}-\frac{1}{2},\frac{3\pi }{4}-\frac{1}{2}\right]$

$\frac{\pi }{4}-\frac{1}{2}\le x\le \frac{3\pi }{4}-\frac{1}{2}c$

$\frac{\pi }{2}\le 2x+1\le \frac{3\pi }{2}$

$\sin \frac{\pi }{2}\le \sin (2x+1)\le \sin \left(\frac{3\pi }{2}\right)$

$\therefore f$ is one-one and onto in

D) $\left[-\frac{3\pi }{4}-\frac{1}{2},\frac{3\pi }{2}-\frac{1}{2}\right]\cup \left[\frac{5\pi }{4}-\frac{1}{2},\frac{3\pi }{2}-\frac{1}{2}\right]$ 

$\frac{5\pi }{4}=\frac{1}{2}\le xk,\frac{3\pi }{2}-\frac{1}{2}$

$\frac{5\pi }{2}\le 2x+1\le 3\pi$

$\sin \left(\frac{5\pi }{2}\right)\le \sin (2x+1)\le \sin \left(3\pi \right)$

One-one but nor onto

$-\frac{3\pi }{4}-\frac{1}{2}\le x\le \frac{3\pi }{2}-\frac{1}{2}$

$-\frac{3\pi }{2}\le 2x+1\le 3\pi$

Onto but not one-one

$\therefore f$ is neither one-one nor onto