Question Number 184739 by SLVR last updated on 11/Jan/23
![Number of linear functions be defined f:[−1, 1]→[0,2] is a)1 b)2 c)3 d)4](Q184739.png)
$${Number}\:{of}\:{linear}\:{functions}\: \\ $$$${be}\:{defined}\:{f}:\left[−\mathrm{1},\:\mathrm{1}\right]\rightarrow\left[\mathrm{0},\mathrm{2}\right]\:{is} \\ $$$$\left.{a}\left.\right)\left.\mathrm{1}\left.\:\:\:\:{b}\right)\mathrm{2}\:\:\:\:{c}\right)\mathrm{3}\:\:\:{d}\right)\mathrm{4} \\ $$
Commented by SLVR last updated on 11/Jan/23
$${kindly}\:{help}\:{me}..{answer}\:{says} \\ $$$$\:\mathrm{2}\:{only}..{how} \\ $$
Answered by mr W last updated on 11/Jan/23
![in domain [−1,1] the linear function f(x)=ax+b has the range [−a+b,a+b] if a>0 (case 1) or [a+b,−a+b] if a<0 (case 2) since the range should be [0,2], case 1: −a+b=0 a+b=2 ⇒a=1, b=1 ⇒f(x)=x+1 case 2: a+b=0 −a+b=2 ⇒a=−1, b=1 ⇒ f(x)=−x+1 i.e. there are 2 functions satisfying the given conditions: f(x)=x+1 or f(x)=−x+1](Q184742.png)
$${in}\:{domain}\:\left[−\mathrm{1},\mathrm{1}\right]\:{the}\:{linear}\:{function} \\ $$$${f}\left({x}\right)={ax}+{b}\:{has}\:{the}\:{range} \\ $$$$\left[−{a}+{b},{a}+{b}\right]\:{if}\:{a}>\mathrm{0}\:\left({case}\:\mathrm{1}\right) \\ $$$${or} \\ $$$$\left[{a}+{b},−{a}+{b}\right]\:{if}\:{a}<\mathrm{0}\:\left({case}\:\mathrm{2}\right) \\ $$$${since}\:{the}\:{range}\:{should}\:{be}\:\left[\mathrm{0},\mathrm{2}\right], \\ $$$${case}\:\mathrm{1}: \\ $$$$−{a}+{b}=\mathrm{0} \\ $$$${a}+{b}=\mathrm{2} \\ $$$$\Rightarrow{a}=\mathrm{1},\:{b}=\mathrm{1}\:\Rightarrow{f}\left({x}\right)={x}+\mathrm{1} \\ $$$${case}\:\mathrm{2}: \\ $$$${a}+{b}=\mathrm{0} \\ $$$$−{a}+{b}=\mathrm{2} \\ $$$$\Rightarrow{a}=−\mathrm{1},\:{b}=\mathrm{1}\:\Rightarrow\:{f}\left({x}\right)=−{x}+\mathrm{1} \\ $$$${i}.{e}.\:{there}\:{are}\:\mathrm{2}\:{functions}\:{satisfying} \\ $$$${the}\:{given}\:{conditions}: \\ $$$${f}\left({x}\right)={x}+\mathrm{1}\:{or}\:{f}\left({x}\right)=−{x}+\mathrm{1} \\ $$
Commented by SLVR last updated on 11/Jan/23
$${So}\:{great}\:{of}\:{you}\:{sir}... \\ $$$${so}\:{kind}\:{of}\:{you}\:{sir}..{Thanks}\:{again} \\ $$