# Find all points of discontinuity of f , where f is defined by f(x) = {(|x| + 3, if x ≤ −3) (−2x, if −3 < x < 3) (6x + 2, if x ≥ 3)

**Solution:**

The given function is

f(x) = {(|x| + 3, if x ≤ −3) (−2x, if −3 < x < 3) (6x + 2, if x ≥ 3)

The given function f is defined at all the points of the real line.

Let c be a point on the real line.

Case I:

If c < −3, then f(c) = −c + 3

lim_{x→c} f(x) = lim_{x→c} (−x + 3)

= −c + 3

⇒ lim_{x→c} f(x) = f(c)

Therefore, f is continuous at all points x, such that x < −3.

Case II:

If c = −3, then f(−3) = −(−3) + 3 = 6

lim_{x→−3−} f(x) = lim_{x→−3−} (−x + 3)

= −(−3) + 3 = 6

lim_{x→−3+} f(x) = lim_{x→−3+} (−2x)

= −2(−3) = 6

⇒ limx→−3 f(x) = f(−3)

Therefore, f is continuous at x = −3x.

Case III:

If −3 < c < 3, then f(c)

= −2c

lim_{x→c} f(x) = lim_{x→c} (−2x)

= −2c

⇒ lim_{x→c} f(x) = f(c)

Therefore, f is continuous in (−3, 3).

Case IV:

If c = 3, then the left hand limit of f at x = 3 is,

lim_{x→3−} f(x) = lim_{x→3−} (−2x)

= −2(3) = −6

The right hand limit of ff at x = 3 is,

lim_{x→3+} f(x) = lim_{x→3+ }(6x + 2)

= 6(3) + 2 = 20

It is observed that the left and right hand limit of f at x = 3 do not coincide.

Therefore, f is not continuous at x = 3.

Case V:

If c > 3, then f(c) = 6c + 2

lim_{x→c} f(x) = lim_{x→c} (6x+2)

= 6c+2

⇒ lim_{x→c} f(x) = f(c)

Therefore, f is continuous at all points x, such that x > 3.

Hence, x = 3 is the only point of discontinuity of f

NCERT Solutions Class 12 Maths - Chapter 5 Exercise 5.1 Question 7

## Find all points of discontinuity of f , where f is defined by f(x) = {(|x| + 3, if x ≤ −3) (−2x, if −3 < x < 3) (6x + 2, if x ≥ 3)

**Summary:**

For the function defined by f(x) = {(|x| + 3, if x ≤ −3) (−2x, if −3 < x < 3) (6x + 2, if x ≥ 3) ,x = 3 is the only point of discontinuity of f