GATE CSE
First time here? Checkout the FAQ!
x
+2 votes
260 views

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

asked in Calculus by (41 points)  
edited by | 260 views

3 Answers

+6 votes
Best answer

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 (9.5k points)  
selected by
+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 (24.9k 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 (901 points)  
edited by
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.

Related questions



Top Users Jun 2017
  1. Bikram

    3704 Points

  2. Arnab Bhadra

    1502 Points

  3. Hemant Parihar

    1502 Points

  4. Niraj Singh 2

    1481 Points

  5. junaid ahmad

    1432 Points

  6. Debashish Deka

    1384 Points

  7. Rupendra Choudhary

    1220 Points

  8. rahul sharma 5

    1220 Points

  9. Arjun

    1168 Points

  10. srestha

    1010 Points

Monthly Topper: Rs. 500 gift card
Top Users 2017 Jun 26 - Jul 02
  1. Arjun

    208 Points

  2. akankshadewangan24

    152 Points

  3. Debashish Deka

    138 Points

  4. Hira Thakur

    130 Points

  5. Soumya29

    106 Points


23,399 questions
30,111 answers
67,490 comments
28,426 users