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If $x+|y|=2y$, then $y$ as a function of x is

$\begin{array}{1 1}(a)\;\text{defined for all real x}\\(b)\;\text{continuous at x=1}\\(c)\;\text{differentiable for all x}\\(d)\;\text{Such that }\large\frac{dy}{dx}=\frac{1}{2} \normalsize\text{for x < 0}\end{array}$

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Given that $x+|y|=2y$
If $y < 0$ then $x-y=2y$
$\Rightarrow y=\large\frac{x}{3}$
$\Rightarrow x < 0$
If $y=0$ then $x=0$ if $ y > 0$ then $x+y=2y$
$\Rightarrow y=x\Rightarrow x > 0$
Thus we can define $f(x)=y=\left\{\begin{array}{1 1}x/3 &x < 0\\x &x \geq 0\end{array}\right.$
Continuity at $x=0$
LL=$\lim\limits_{h\to 0}f(0-h)=\lim\limits_{h\to 0}(-h/3)=0$
RL=$\lim\limits_{h\to 0}f(0+h)=\lim\limits_{h\to 0}h=0$
$f(0)=0$
As LL=RL=f(0)
$\therefore f(x)$ is continuous at $x=0$
Hence (a) is the correct answer.
answered Jan 3, 2014 by sreemathi.v
 

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