# Discuss the continuity of the function f , where f is defined by f(x)= {(2x, if x < 0) (0, if 0 ≤ x ≤ 1) (4x, if x > 1)

**Solution:**

A function is said to be continuous when the graph of the function is a single unbroken curve.

The given function is

f(x)= {(2x, if x < 0) (0, if 0 ≤ x ≤ 1) (4x, if x > 1)

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 < 0, 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 at all points x, such that x < 0.

Case II:

If c = 0, then f(0) = 0

The left-hand limit of f at x = 0 is,

lim_{x→0−} f(x) = lim_{x→0−} (2x)

= 2(0) = 0

The right hand limit of f at x = 0 is,

⇒ lim_{x→0+} f(x) = lim_{x→0+} (0) = 0

Therefore, f is not continuous at x = 0*.*

Case III:

If 0 < c < 1, then f(x) = 0

lim_{x→c} f(x) = lim_{x→c }(0) = 0

⇒ lim_{x→c} f(x) = f(c)

Therefore, f is continuous at in the interval (0,1).

Case IV:

If c = 1, then f(1) = 0

The left-hand limit of f at x = 1 is,

lim_{x→1−} f(x) = lim_{x→1−} (0) = 0

The right-hand limit of f at x = 1 is,

lim_{x→1+} f(x) = lim_{x→1+} (4x) = 4(1) = 4

It is observed that the left and right-hand limits of f at x = 1 do not coincide.

Therefore, f is not continuous at x = 1*.*

Case V:

If c < 1, then f(c) = 4c

lim_{x→c} f(x) = lim_{x→c}(4x)

= 4c

⇒ lim_{x→c} f(x) = f(c)

Therefore, f is continuous at all points x such that x > 1.

Hence, f is not continuous only at x = 1

NCERT Solutions Class 12 Maths - Chapter 5 Exercise 5.1 Question 15

## Discuss the continuity of the function f , where f is defined by f(x)= {(2x, if x < 0) (0, if 0 ≤ x ≤ 1) (4x, if x > 1)

**Summary:**

For the function defined by f(x)= {(2x, if x < 0) (0, if 0 ≤ x ≤ 1) (4x, if x > 1), f is continuous at all points x such that x > 1. Hence, f is not continuous only at x = 1

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