# Find all points of discontinuity of f, where f is defined by f(x)={(x^{3 }− 3, if x ≤ 2) (x^{2 }+ 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) = {(x^{3} − 3, if x ≤ 2) (x^{2} + 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) = c^{3} − 3

lim_{x→c} f(x) = lim_{x→c} (x^{3} − 3) = c^{3} − 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) = 2^{3} − 3 = 5

lim_{x→2−} f(x) = lim_{x→2−} (x^{3} − 3) = 2^{3} − 3 = 5

lim_{x→2+} f(x) = lim_{x→2+} (x^{2} + 1) = 2^{2} + 1 = 5

⇒ lim_{x→2} f(x) = f(2)

Therefore, f is continuous at x = 2 .

Case III:

If c > 2, then f(c) = c^{2} + 1

lim_{x→c} f(x) = lim_{x→c} (x^{2} + 1) = c^{2} + 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

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

**Summary:**

The given function f defined by f(x)={(x^{3 }− 3, if x ≤ 2) (x^{2 }+ 1, if x > 2) is continuous at every point on the real line. Hence, f has no point of discontinuity

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