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Question-196257




Question Number 196257 by mathlove last updated on 21/Aug/23
Answered by cortano12 last updated on 21/Aug/23
  g(x) ⇒m = 2+b and passes through   the point (1, 1+b+c)   ⇒y−(1+b+c)= (2+b)(x−1)   ⇒y= (2+b)x−2−b+1+b+c   ⇒y= (2+b)x+c−1   (0,0)⇒0 = c−1 ; c = 1   ⇒g(x)= (2+b)x    passes througt the point (1,1)   ⇒1 = (2+b).1 ⇒b=−1     { ((f(x)=x^2 −x+1)),((g(x)= x )) :} ⇒c+b = 0
$$\:\:\mathrm{g}\left(\mathrm{x}\right)\:\Rightarrow\mathrm{m}\:=\:\mathrm{2}+\mathrm{b}\:\mathrm{and}\:\mathrm{passes}\:\mathrm{through} \\ $$$$\:\mathrm{the}\:\mathrm{point}\:\left(\mathrm{1},\:\mathrm{1}+\mathrm{b}+\mathrm{c}\right) \\ $$$$\:\Rightarrow\mathrm{y}−\left(\mathrm{1}+\mathrm{b}+\mathrm{c}\right)=\:\left(\mathrm{2}+\mathrm{b}\right)\left(\mathrm{x}−\mathrm{1}\right) \\ $$$$\:\Rightarrow\mathrm{y}=\:\left(\mathrm{2}+\mathrm{b}\right)\mathrm{x}−\mathrm{2}−\mathrm{b}+\mathrm{1}+\mathrm{b}+\mathrm{c} \\ $$$$\:\Rightarrow\mathrm{y}=\:\left(\mathrm{2}+\mathrm{b}\right)\mathrm{x}+\mathrm{c}−\mathrm{1} \\ $$$$\:\left(\mathrm{0},\mathrm{0}\right)\Rightarrow\mathrm{0}\:=\:\mathrm{c}−\mathrm{1}\:;\:\mathrm{c}\:=\:\mathrm{1} \\ $$$$\:\Rightarrow\mathrm{g}\left(\mathrm{x}\right)=\:\left(\mathrm{2}+\mathrm{b}\right)\mathrm{x}\: \\ $$$$\:\mathrm{passes}\:\mathrm{througt}\:\mathrm{the}\:\mathrm{point}\:\left(\mathrm{1},\mathrm{1}\right) \\ $$$$\:\Rightarrow\mathrm{1}\:=\:\left(\mathrm{2}+\mathrm{b}\right).\mathrm{1}\:\Rightarrow\mathrm{b}=−\mathrm{1} \\ $$$$\:\:\begin{cases}{\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}}\\{\mathrm{g}\left(\mathrm{x}\right)=\:\mathrm{x}\:}\end{cases}\:\Rightarrow\mathrm{c}+\mathrm{b}\:=\:\mathrm{0}\:\: \\ $$$$\:\:\: \\ $$
Commented by mathlove last updated on 21/Aug/23
thanks
$${thanks} \\ $$

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