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

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(B) Given the equation $x + |y| = 2y$.Case $1$: If $y \ge 0,$ then $|y| = y.$$x + y = 2y \Rightarrow y = x.$Case $2$: If $y < 0,$ then $|y| = -y.$$x -

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continuous but not differentiable

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(B) Given the equation $x + |y| = 2y$.Case $1$: If $y \ge 0,$ then $|y| = y.$$x + y = 2y \Rightarrow y = x.$Case $2$: If $y $x - y = 2y \Rightarrow x = 3y \Rightarrow y = x/3.$Thus,the function is defined as $f(x) = \begin{cases} x, & x \ge 0 \ x/3, & x Continuity at $x = 0$:$\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^+} x = 0$.$\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^-} x/3 = 0$.Since $f(0) = 0,$ the function is continuous at $x = 0$.Differentiability at $x = 0$:$LHD = \lim_{h 	o 0^+} \frac{f(0-h) - f(0)}{-h} = \lim_{h 	o 0^+} \frac{(0-h)/3 - 0}{-h} = 1/3$.$RHD = \lim_{h 	o 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0^+} \frac{h - 0}{h} = 1$.Since $LHD 
eq RHD,$ the function is not differentiable at $x = 0$.

If $x + |y| = 2y,$ then $y$ as a function of $x,$ at $x = 0$ is

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