Find all points of discontinuity of f, where f is defined by f(x)={(x³− 3, if x ≤ 2) (x²+ 1, if x > 2)

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) = {(x³ − 3, if x ≤ 2) (x² + 1, if x > 2)

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 < 2, then f(c) = c³ − 3

lim_x→c f(x) = lim_x→c (x³ − 3) = c³ − 3

⇒ lim_x→c f(x) = f(c)

Therefore, f is continuous at all points x, such that x < 2.

Case II:

If c = 2, then f(c) = f(2) = 2³ − 3 = 5

lim_x→2⁻ f(x) = lim_x→2⁻ (x³ − 3) = 2³ − 3 = 5

lim_x→2⁺ f(x) = lim_x→2⁺ (x² + 1) = 2² + 1 = 5

⇒ lim_x→2 f(x) = f(2)

Therefore, f is continuous at x = 2 .

Case III:

If c > 2, then f(c) = c² + 1

lim_x→c f(x) = lim_x→c (x² + 1) = c² + 1

⇒ lim_x→c f(x) = f(c)

Therefore, f is continuous at all points x, such that x > 2.

Thus, the given function f is continuous at every point on the real line.

Hence, f has no point of discontinuity

NCERT Solutions Class 12 Maths - Chapter 5 Exercise 5.1 Question 11