  Question

If $$x+2|y|=3y$$, where $$y=f(x)$$, then $$f(x)$$ is

A
continuous everywhere  B
differentiable everywhere  C
discontinuous at x=0  D
not differentiable anywhere  Solution

The correct option is A continuous everywhereFor $$f\left( x \right)\ge 0$$, $$\left| f\left( x \right) \right| =f\left( x \right)$$.So, $$\\ x+2\left| f\left( x \right) \right| =3f\left( x \right) \\ \Rightarrow x+2f\left( x \right) =3f\left( x \right) \\ \Rightarrow f\left( x \right)=x$$.But since this is for $$f\left( x \right)\ge 0$$, $$f\left( x \right)=x\quad \forall \quad x\ge 0$$.For $$f\left( x \right) < 0$$, $$\left| f\left( x \right) \right| =-f\left( x \right)$$.So, $$\\ x+2\left| f\left( x \right) \right| =3f\left( x \right) \\ \Rightarrow x-2f\left( x \right) =3f\left( x \right) \\ \Rightarrow f\left( x \right)=\dfrac{x}{5}$$.But since this is for $$f\left( x \right) < 0$$, $$f\left( x \right)=\dfrac{x}{5}\quad \forall \quad x < 0$$.Checking for continuity at $$x=0$$:$$\displaystyle\lim _{ x\rightarrow { 0 }^{ - } }{ f\left( x \right) } =\lim _{ x\rightarrow { 0 }^{ - } }{ \dfrac { x }{ 5 } } =0\\ \displaystyle\lim _{ x\rightarrow { 0 }^{ + } }{ f\left( x \right) } =\lim _{ x\rightarrow { 0 }^{ + } }{ x } =0\\ f(0)=0$$Since it is an algebraic function that changes its definition only at $$x=0$$, the function is continuous everywhere.Checking for differentiability at $$x=0$$: the slope of the function changes at this point, so it is not differentiable here.So, option A is the correct option Mathematics

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