Let f : R → R be the Signum Function defined as f (x) = {1, x > 0; 0, x = 0; - 1, x <0} and g : R → R be the greatest integer function given by g (x) = [x], where [x] is greatest integer less than or equal to x. Then does fog and gof coincide in (0, 1] ?

Solution:

It is given that

f : R → R be the Signum Function defined as

f (x) = {1, x > 0; 0, x = 0; - 1, x < 0}

Also,

g : R → R is defined as

g (x) = [x] is greatest integer less than or equal to x.

Now let x ∈ (0, 1],

[x] = 1if x = 1 and

[x] = 0 if 0 < x < 1.

⇒ fog (x) = f (g (x)) = f (|x|) {f (1), if x = 1; f (0),

if x ∈ (0, 1)} = {1, if x = 1; 0, if x ∈ (0, 1)}

⇒ gof (x) = g ( f (x))

= g (1) [x > 0]

= [1] = 1

Thus, when x ∈ (0, 1), we have fog ( x) = 0 and gof ( x) = 1.

Hence, fog and gof do not coincide in (0, 1]

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NCERT Solutions for Class 12 Maths - Chapter 1 Exercise ME Question 18

Let f : R → R be the Signum Function defined as f (x) = {1, x > 0; 0, x = 0; - 1, x <0} and g : R → R be the greatest integer function given by g (x) = [x], where [x] is greatest integer less than or equal to x. Then does fog and gof coincide in (0, 1] ?

Summary:

Given that f : R → R be the Signum Function defined as f (x) = {1, x > 0; 0, x = 0; - 1, x <0} and g : R → R be the greatest integer function given by g (x) = [x], where [x] is greatest integer less than or equal to x. When x ∈ (0, 1), we have fog ( x) = 0 and gof ( x) = 1. Hence, fog and gof do not coincide in (0, 1]