Find all points of discontinuity of f, where f is defined by f(x) = {(x¹⁰ − 1, if x ≤ 1) (x², if x > 1)

Solution:

The given function is f(x) = {x¹⁰−1, if x ≤ 1 ; x², 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 < 1,

then f(c) = c¹⁰ − 1

lim_x→c f(x) = lim_x→c (x¹⁰ − 1)

= c¹⁰ − 1

⇒ lim_x→c f(x) = f(c)

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

Case II:

If c = 1, then the left hand limit of f at x = 1 is,

lim_x→1⁻ f(x) = lim_x→1⁻ (x¹⁰ − 1)

= 1¹⁰ − 1

= 1 − 1 = 0

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

⇒ lim_x→1⁺ f(x) = lim_x→1⁺ (x²)

= 1²

= 1

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

Therefore, f is not continuous at x = 1.

Case III:

If c > 1,

then f(c) = c²

lim_x→c f(x) = lim_x→c (x²)

= c²

⇒ lim_x→c f(x) = f(c)

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

Thus from the above observation,

it can be concluded that x = 1 is the only point of discontinuity of f.

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

Find all points of discontinuity of f, where f is defined by f(x) = {(x¹⁰ − 1, if x ≤ 1) (x², if x > 1)

Summary:

It can be concluded that x = 1 is the only point of discontinuity of f defined by f(x) = {(x¹⁰ − 1, if x ≤ 1) (x², if x > 1)