Question

Range of the function y = (sin x +1) is​

Answer 1

$\large\underline\blue{\bold{Given \:  Question   }}$

$\tt \:  y \: = \: sinx \: + \: 1$

$\large\underline\blue{\bold{To \:  Find :-  }}$

$\tt \:Range \: of \: y$

$\large\underline\blue{\bold{Solution :- }}$

We know,

$:\implies \tt \: - 1 \leqslant sinx \leqslant 1$

On adding 1 in each term, we get

$:\implies \tt \: 1 - 1 \leqslant sinx + 1 \leqslant 1 + 1$

$:\implies \tt \: 0 \leqslant sinx + 1 \leqslant 2$

$:\implies \tt \: 0 \leqslant y \leqslant 2$

Hence,

$:\implies \boxed{ \pink{\tt \: Range \: of \: y \: \in \: [0, \: 2]}}$

Answer 2

Function range definition : The set of values of the dependent variable for which a function is defined

• $\mathrm{The\:range\:of\:the\:basic}\:\sin \mathrm{function\:is}\:-1\le \sin \left(x\right)\le \:1$

• $\mathrm{Add\:to\:the\:edges\:of\:the\:range:}\:1$

$\to0\le \sin \left(x\right)+1\le \:2$

• $\mathrm{Therefore\:the\:range\:is}$

$\to0\le \:f\left(x\right)\le \:2$

$\boxed{\therefore\:\mathrm{Range\:of\:}\sin \left(x\right)+1:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:0\le \:f\left(x\right)\le \:2\:\\ \:\mathrm{Interval\:Notation:}&\:\left[0,\:2\right]\end{bmatrix}}$