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The function y=|2 - 3x|​

a) is continuous  x  R and differentiable  x  R

b) is continuous  x  R and differentiable  x  R except at x= 3/2

c) is continuous  x  R and differentiable  x  R except at x= 2/3

d) is continuous  x  R except x=3 and differentiable  x  R

edited | 292 views

y = 2 - 3x        2 - 3x $\geq$ 0
3x - 2         2 - 3x $<$ 0

y = 2 - 3x        x $\leq \frac{2}{3}$
3x - 2         x $> \frac{2}{3}$

as y is polynomial it is continuous and differentiable at all points but don't know at x = $\frac{2}{3}$

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) = $2-3 \times \frac{2}{3} = 0$

$\therefore$ LL = RL = f(a) so y is continuous $\forall x \epsilon R$

Differentiability at x = $\frac{2}{3}$
left derivative = 0 - 3 = -3
right derivative = 3 - 0 = 3

$\therefore$ LD $\neq$ RD, so y is not differentiable at  $x = \frac{2}{3}$

So Answer is option c

answered by Boss (9.6k points)
selected
+1 vote

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.

answered by Veteran (25k points)
–1 vote

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.

answered by Junior (921 points)
edited
You say if we put any value such that x belongs to R, LHL and RHL will be the same. Lets say x=4

Then 2-3x = 2-3*4 = -10

and 3x-2 = 3*4-2 = 10

which are NOT the same.

I am not saying that at any value of x∊R, LHL and RHL will be same. I am saying that at any value of x function will be continuous i.e it will have some non-zero value except at x=2/3.

+1 vote