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11TH · MATH

Question 4

menu_bookSolutionmenu_bookTheorymenu_book•(i)g(x)=5−xg(x)=5-xg(x)=5−xmenu_book•(ii)g(x)=x+2g(x)=\sqrt{x+2}g(x)=x+2​menu_book•(iii)g(x)={6x+7,x≤−24−3x,x>−2g(x)=\begin{cases}6x+7,&x\le -2\\4-3x,&x>-2\end{cases}g(x)={6x+7,4−3x,​x≤−2x>−2​menu_book•(iv)g(x)=∣x−5∣g(x)=|x-5|g(x)=∣x−5∣menu_book•(v)g(x)=x+23−xg(x)=\dfrac{x+2}{3-x}g(x)=3−xx+2​
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11TH · MATH

Question 4

menu_bookSolutionmenu_bookTheorymenu_book•(i)g(x)=5−xg(x)=5-xg(x)=5−xmenu_book•(ii)g(x)=x+2g(x)=\sqrt{x+2}g(x)=x+2​menu_book•(iii)g(x)={6x+7,x≤−24−3x,x>−2g(x)=\begin{cases}6x+7,&x\le -2\\4-3x,&x>-2\end{cases}g(x)={6x+7,4−3x,​x≤−2x>−2​menu_book•(iv)g(x)=∣x−5∣g(x)=|x-5|g(x)=∣x−5∣menu_book•(v)g(x)=x+23−xg(x)=\dfrac{x+2}{3-x}g(x)=3−xx+2​
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Accedmychevron_right11thchevron_rightmathchevron_rightFunctions And Graphschevron_rightExercise 2.1

(v)

Dg=R∖{3},Rg=R∖{−1}\boxed{D_g=\mathbb{R}\setminus\{3\},\quad R_g=\mathbb{R}\setminus\{-1\}}Dg​=R∖{3},Rg​=R∖{−1}​
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