Question:

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

Updated On: Jun 14, 2022
  • defined for all real x
  • continuous at x = 0
  • differentiable for all x
  • such that dydx=13\frac{dy}{dx} = \frac{1}{3} for x<0x < 0
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The Correct Option is D

Solution and Explanation

Given that x+y=2yx + | y | = 2y
If y<0y < 0 then xy=2yx - y = 2y
 y=x/3  x<0\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
LL=limh0f(0h)=limh0(h/3)=0L L = \displaystyle \lim_{h \to 0} f\left(0-h\right) =\displaystyle \lim_{h \to 0} \left(-h /3\right) = 0
RL=limh0f(0+h)=limh0h=0 R L = \displaystyle \lim_{h \to 0 } f\left(0+h\right) =\displaystyle \lim_{h \to 0} h = 0
f(0)=0f\left(0\right) = 0
As LL=RL=f(0)LL = RL = f\left(0\right)
f(x)\therefore f\left(x\right) is continuous at x=0x = 0
Differentiability at x = 0
Lf=1/3;Rf=1Lf ' = 1/3 ; Rf' = 1
As LfRff(x)Lf' \neq Rf' \Rightarrow f(x) is not differentiable at x = 0
But for x<0,dydx=13.x < 0 , \frac{dy}{dx} = \frac{1}{3}.
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Concepts Used:

Continuity

A function is said to be continuous at a point x = a,  if

limx→a

f(x) Exists, and

limx→a

f(x) = f(a)

It implies that if the left hand limit (L.H.L), right hand limit (R.H.L) and the value of the function at x=a exists and these parameters are equal to each other, then the function f is said to be continuous at x=a.

If the function is undefined or does not exist, then we say that the function is discontinuous.

Conditions for continuity of a function: For any function to be continuous, it must meet the following conditions:

  • The function f(x) specified at x = a, is continuous only if f(a) belongs to real number.
  • The limit of the function as x approaches a, exists.
  • The limit of the function as x approaches a, must be equal to the function value at x = a.