f (x) = |x - 3| = {

3−x	x<3
x−3	x≥3

Let c be a real number.
Case I: c < 3. Then f(c) = 3 – c.

lim

x→c

f (x) =

lim

x→c

(3 - x) = 3 - x
Since,

lim

x→c

f (x) = f (c), f is continuous at all negative real numbers.
Case II: c = 3. Then f(c) = 3 – 3 = 0

lim

x→c

f (x) =

lim

x→c

f (x) (x - 3) = 3 - 3 = 0
Since,

lim

x→c

f (x) f (x) = f (3), f is continuous at x = 3.
Case III: c > 3. Then f(c) = c – 3.

lim

x→c

f (x) =

lim

x→c

(x - 3) = x - 3
Since,

lim

x→c

f (x) = f (c), f is continuous at all positive real numbers
Therefore, f is continuous function.
Now, we need to show that f(x) = |x - 3|, x ∊ R is not differentiable at x = 3.
Consider the left hand limit of f at x = 3

lim

h→0−

f(3+h)−f(3)

h

=

lim

h→0−

|3+h−3|−|3−3|

h

=

lim

h→0−

|h|−0

h

=

lim

h→0−

−h

h

= - 1
h < 0 ⇒ |h| = - h
Consider the right hand limit of f at x = 3

lim

h→0+

f(3+h)−f(3)

h

lim

h→0+

|3+h−3|−|3−3|

h

=

lim

h→0+

|h|−0

h

=

lim

h→0+

h

h

= 1
h > 0 ⇒ |h| = h
Since the left and right hand limits are not equal, f is not differentiable at x = 3.
OR
y = a (cost+logtan

t

2

) find

d2y

dx2

⇒

dy

dt

= a |

d

dt

+cost+

d

dt

(logtan

t

2

)|
= a |−sint+cot

t

2

×sec3

t

2

×

1

2

|
= a |−sint+

1

2sin t 2 cos t 2

|
= a (−sint+

1

sint

) = a (

−sin2t+1

sint

) = a

cos2t

sint

x = a sin t

dx

dt

= a

d

dt

sin t = a cos t
∴

dy

dx

=

( dy dx )

( dx dt )

=

(a cos2t sint )

acost

=

cost

sint

= cot t

d2y

dx2

= - cosec2t

dt

dx

= - cosec2t×

1

acost

= -

1

asin2tcost