CBSE Guess > Papers > Important Questions > Class XII > 2010 > Maths > Mathematics By Mr. M.P.Keshari

CBSE CLASS XII

1.2 Types of Function.

Q.1. State whether the function f : N→ N given by f(x) = 5x is injective, surjective or both.

Solution :

Given function f(x) = 5x is injective (one-one).
Explanation : The function f is one-one, for
f(x₁) = f(x₂)
=> 5x₁ = 5x₂
=> x₁ = x₂.
Further f is not onto, since 1 ε N, there does not exist any x ε N such that f(x) = 2x = 1.
Therefore, f is one-one (injective) but not one-one onto (bijective). [Proved.]

Q.2. Let f : N→ N be defined by
∫(n + 1)/2, if n is odd
f(n) = | for all n ε N
∫ n/2, if n is even
Find whether the function f is bijective.

Solution :

We have, f(n) = (n + 1)/2 if n is odd and f(n) = n/2 if n is even.
Since, f(1) = (1 + 1)/2 = 1, f(2) = 2/2 = 1 , f(3) = (3 + 1)/2 = 2, f(4) = 4/2 = 2.
In general, f(2k – 1) = (2k – 1 + 1)/2 = k and f(2k) = 2k/2 = k.
Therefore, f(2k – 1) = f(2k), where k is any positive integer.
Therefore, f is not one-one.
Also Rf = N, therefore, f is onto.
[As, for any n ε N, there is 2n ε N, such that f(2n) = 2n/2 = n]
Thus f is onto but not one-one.
Hence, f is not bijective. [Proved.]

1.3 Composition of Function and Invertible Function.

Q.1. If f(x) = x + 7 and g(x) = x – 7, x ε R, find (fog)(7).

Solution :

We have, f(x) = x + 7 and g(x) = x – 7.
Then (fog)(x) = fo(g(x)) = f(x – 7) = (x – 7) + 7 = x.
Therefore, (fog)(7) = 7. [Ans.]

Q.2. If f : R→ R and g : R→ R are defined respectively as f(x) = x2 + 3x + 1 and g(x) = 2x – 3, find (a) fog , (b) gof.

Solution :

We have, f(x) = x² + 3x + 1 and g(x) = 2x – 3.

1. (fog)(x) = f(g(x))
   = f(2x – 3)
   = (2x – 3)² + 3(2x – 3) + 1
   = 4x² – 12x + 9 + 6x – 9 + 1
   = 4x² – 6x + 1. [Ans.]
   
   
2. (gof)(x) = g(f(x))
   = g(x² + 3x + 1)
   = 2(x² + 3x + 1) – 3
   = 2x² + 6x + 2 – 3
   = 2x² + 6x – 1. [Ans.]

Paper By Mr. M.P.Keshari
Email Id : [email protected]
Ph No. : 09434150289