62. Given examples of two functions f: N → N and g: N → N such that gof is onto but f is not onto.
(Hint: Consider f(x) = x + 1 and
62. Given examples of two functions f: N → N and g: N → N such that gof is onto but f is not onto.
(Hint: Consider f(x) = x + 1 and
Define by
And, by,
We first show that g is not onto.
For this, consider element 1 in co-domain N. it is clear that this element is not an image of any of the elements in domain.
is not onto.
Now, is defined by,
Then, it is clear that for , there exists such that
Hence, gof is onto.
Similar Questions for you
R1 = { (1, 1) (1, 2), (1, 3)., (1, 20), (2, 2), (2, 4). (2, 20), (3, 3), (3, 6), . (3, 18),
(4, 4), (4, 8), . (4, 20), (5, 5), (5, 10), (5, 15), (5, 20), (6, 6), (6, 12), (6, 18), (7. 7),
(7, 14), (8, 8), (8, 16), (9, 9), (9, 18), (10, 10), (10, 20), (11, 11), (12, 12)


⇒ (y, x) ∈ R V (x, y) ∈ R
(x, y) ∈ R ⇒ 2x = 3y and (y, x) ∈ R ⇒ 3x = 2y
Which holds only for (0, 0)
Which does not belongs to R.
∴ Value of n = 0
f is increasing function
x < 5x < 7x

f (x) < f (5x) < f (7x)
->
Given f (k) =
Case I : If x is even then g (x) = x . (i)
Case II : If x is odd then g (x + 1) = x + 1 . (ii)
From (i) & (ii), g (x) = x, when x is even
So total no. of functions = 105 × 1 = 105
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Maths Relations and Functions 2025
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