Examine the following functions for continuity.
(i) f(x) = x − 5
(ii) f(x) = 1/x − 5, x ≠ 5
(iii) f(x) = x2 − 25/x + 5, x ≠ −5
(iv) f(x) = |x − 5|, x ≠ 5

Solution:

(i)

The given function is f(x) = x − 5

It is evident that f is defined at every real number k and its value at k is k − 5.

It is also observed that

limx→k f(x) = limx→k (x−5)

= k−5 = f(k)

Therefore,

limx→k f(x) = f(k)

Hence,

f is continuous at every real number and therefore, it is a continuous function.

(ii)

The given function is f(x) = 1/x − 5, x ≠ 5

For any real number k ≠ 5, we obtain

limx→k f(x) = limx→k 1/x−5 = 1/k−5

Also,

f(k) = 1/k−5 (As k ≠5)

Therefore,

limx→k f(x) = f(k)

Hence,

f is continuous at every point in the domain of f and therefore, it is a continuous function.

(iii)

The given function is f(x) = x2 − 25/x + 5, x ≠ −5

For any real number c ≠ −5 we obtain

limx →c f(x )= limx →c x2 − 25/x + 5 = limx →c (x+5)(x−5)/x+5= limx →c (x−5) = (c−5)

Also,

f(c) = (c+5)(c−5)/c+5 = (c−5)

Therefore,

limx →c f(x) = f(c) 

Hence,

f is continuous at every point in the domain of f and therefore, it is a continuous function.

(iv)

The given function is f(x) = |x − 5| = {(5 − x, if x < 5), (x − 5, if x ≥ 5)}

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

Let c be a point on a real line.

Then, c < 5, c = 5 or c > 5

Case I: c < 5

Then, f(c) = 5 − c

limx→c f(x) = limx→c (5 − x) = 5 − c

limx→c f(x) = f(c)

Therefore,

f is continuous at all real numbers less than 5.

Case II: c = 5

Then, f(c) = f(5) = (5 − 5) = 0

limx→5− f(x) = limx→5 (5−x) = (5−5) =0

limx→5+ f(x) = limx→5 (x−5) = 0

⇒ limx→c− f(x) = limx→c+ f(x) = f(c)

Therefore,

f is continuous at x = 5

Case III: c > 5

Then, f(c) = f(5) = c − 5

limx→c f(x) = limx→c (x − 5) = c − 5

⇒ limx→c f(x) = f(c)

Therefore,f is continuous at all real numbers greater than 5.

Hence, f is continuous at every real number, and therefore, it is a continuous function

NCERT Solutions Class 12 Maths - Chapter 5 Exercise 5.1 Question 3

Examine the following functions for continuity. (i) f(x) = x − 5 (ii) f(x) = 1/x − 5, x ≠ 5 (iii) f(x) = x2 − 25/x + 5, x ≠ −5 (iv) f(x) = |x − 5|, x ≠ 5

Summary:

(i) f(x) = x − 5 (ii) f(x) = 1/x − 5, x ≠ 5 (iii) f(x) = x2 − 25/x + 5, x ≠ −5 (iv) f(x) = |x − 5|, x ≠ 5, Hence all the functions are continuous in there domain

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