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Question

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


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

The correct option is A continuous everywhere
For $$f\left( x \right)\ge 0$$, $$\left| f\left( x \right)  \right| =f\left( x \right)$$.
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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