Question

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

• A. defined for all real x
• B. continuous at x = 0
• C. differentiable for all x
• D. such that $\frac{dy}{dx} = \frac{1}{3}$ for $x < 0$

Answer 1

Answer: (D)

such that $\frac{dy}{dx} = \frac{1}{3}$ for $x < 0$

Answer 2

Given that $x + | y | = 2y$
If $y < 0$ then $x - y = 2y$
$\Rightarrow \ y = x/3 \ \Rightarrow \ x < 0$
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 = $\begin{cases} x/3 & x < 0 \\\ x & x \ge 0 \end{cases}$
Continuity at x = 0
$L L = \displaystyle \lim_{h \to 0} f\left(0-h\right) =\displaystyle \lim_{h \to 0} \left(-h /3\right) = 0$
$R L = \displaystyle \lim_{h \to 0 } f\left(0+h\right) =\displaystyle \lim_{h \to 0} h = 0$
$f\left(0\right) = 0$
As $LL = RL = f\left(0\right)$
$\therefore f\left(x\right)$ is continuous at $x = 0$
Differentiability at x = 0
$Lf ' = 1/3 ; Rf' = 1$
As $Lf' \neq Rf' \Rightarrow f(x)$ is not differentiable at x = 0
But for $x < 0 , \frac{dy}{dx} = \frac{1}{3}.$