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The function $y=|2 - 3x|$​

  1. is continuous $∀ x ∈ R$ and differentiable $∀ x ∈ R$
  2. is continuous $∀ x ∈ R$ and differentiable $∀ x ∈ R$ except at $x=\frac{3}{2}$
  3. is continuous $∀ x ∈ R$ and differentiable $∀ x ∈ R$ except at $x=\frac{2}{3}$
  4. is continuous $∀ x ∈ R$ except $x=3$ and differentiable $∀ x ∈ R$
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4 Answers

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46 votes

$y =\begin{cases} 2 - 3x,\quad   \quad \quad     2 - 3x &\geq 0 \\
      3x - 2, \quad       \quad \quad 2 - 3x &< 0 \end{cases}$

$\implies y = \begin{cases}2 - 3x, \quad \quad \quad        x  &\leq \frac{2}{3}\\
     3x - 2, \quad \quad \quad         x &> \frac{2}{3}\end{cases}$

As $y$ is polynomial it is continuous and differentiable at all points but do not know at $x = \frac{2}{3}$.

To check continuity at $x = \frac{2}{3}$
Left limit = $2-3 \times \frac{2}{3} = 0$
Right limit = $ 3 \times \frac{2}{3} - 2 = 0$
$f(a) = f(2/3) = 2-3 \times \frac{2}{3} = 0$

$\because \text{LL = RL} = f(a), y\text{ is continuous } \forall_{x \epsilon R}.$

To check differentiability at $x = \frac{2}{3}$
Left derivative $= 0 - 3 = -3$
Right derivative $= 3 - 0 = 3$

$\because \text{LD} \neq \text{RD}, y\text{ is not differentiable at }x = \frac{2}{3}.$

So, Answer is option C.

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16 votes

We can say from the above graph of function y= |2-3x| that it is continuous for all real x but not differentiable at x=2/3 as its graph is making sharp corner at this point. 

At x= 2/3

LHD= -3 and RHD= 3 which are not equal so not differentiable at x=2/3.

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The function |2-3x| is continuos ∀x∊R except at x=2/3 as if x>2/3 (i.e in RHL case) then function will become f(x)= 2-3x

if x<2/3(i.e in LHL case) then function will become f(x)=3x-2.  Now, if you put any value of x∈R in these equation you will always get some value i.e function is continuos for LHL and RHL both. 

For differentiability, for LHL(Left Hand Limit) i.e. if x<2/3 then value of function at which min/max occur will be 3

and for RHL(Right Hand Limit) for x>2/3 the value of function at which min/max occur will be -3.

Since both are different value hence, function cannot be differentiable at x=2/3

Hence, option C is the answer.

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