Math Class 12

12th Math Chapter 1 Subjective in English

Relations and Functions Subjective

  1. Define universal relation? Give example.

Ans: A Relation R in a set A called universal relation if each element of A is related to every element of A. Ex. Let = {2,3,4}
R = (AA) = {(2,2),(2,3) (2,4) (3,2) (3,3) (3,4) (4,2) (4,3) (4,4) }

  1. What is trivial relation?

Ans: Both the empty relation and the universal relation are some time called trivial relation.

  1. Let S = {1, 2, 3}
    Determine whether the function f: S à S defined as below have inverse.
    f = {(1, 2), (2, 1), (3, 1)}

Ans:  f(2) = 1 f(3) = 1,
f is not one – one, So that that f is not invertible.

  1. Let f, g and h be function from R to R show that (f + g) oh = foh + goh

Ans:  L.H.S = (f + g) oh
= {(f + g) oh} (x)
= (f + g) h (x)
= f [h (x)] + g [h (x)]
= foh + goh

  1. Show that function f: N à N, given by f(x) = 2x, is one – one.

Ans: the function f is one – one, for
f(x1) = f(x2)
2×1 = 2×2
x1 = x2

  1. Let S = {1, 2, 3}
    Determine whether the function f: S à S defined as below have inverse.
    f = {(1, 1), (2, 2), (3, 3)}

Ans: f is one – one and onto, so that f is invertible with inverse f-1 = {(1, 1) (2, 2) (3, 3)}

  1. Find got f(x) = |x|, g(x) = |5x -2|

Ans:  fog (x) = f(g x)
= f{|5x – 2|)
= |5x – 2|

  1. Consider f: {1, 2, 3} à {a, b, c} given by f(1) = a, f(2) = b and f(3) = c find f-1and show
    that (f-1)-1 = f

Ans: f = {(1, a) (2, b) (3, c)}
f-1 = { (a, 1) (b, 2) (c, 3)}
(f -1) -1 = {(1, a) (2, b) (3, c)}
Hence (f-1)-1 = f.

  1. What is a bijective function?

Ans: A function f: X à Y is said to be one – one and onto (bijective), if f is both one – one and onto.

  1. Let f, g and h be function from R + R. Show that (f.g) oh = (foh). (goh)

Ans:  (f. g) oh
(f. g) h (x)
f[h(x)]. g[h(x)]
foh. goh

  1. Let * be a binary operation defined by a * b = 2a + b – 3. find 3 * 4

Ans:  3 * 4 = 2 (3) + 4-3 = 7

  1. show that a one – one function f: {1, 2, 3} à {1, 2, 3} must be onto.

Ans:  Since f is one – one three element of {1, 2, 3} must be taken to 3 different element of the co – domain {1, 2, 3} under f. hence f has to be onto.

Let S = {1, 2, 3}

  1. Determine whether the function f: S à S defined as below have inverse.
    f = { (1, 2) (2, 1) (3, 1) }

Ans: f(2) = 1, f(3) =1
f is not one – one so that f is not invertible
Hence no inverse

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