Relations and Functions Test - 13

p(1) - q(1) = 0 ⇒ a² + b - l - n = 0....(1)

p(2) - q(2) = 1 ⇒ a² + 2b - 4l -2m - n = 1....(2)

p(3) - q(3) = 4 ⇒ a² +3b - 9l - 3m - n = 4....(3)

p(4) - q(4) = a² +4b -16l - 4m - n=.....

(3) - (2) gives b - 5l - m = 3, therefore, b = 5l + m + 3

(2) - (1) gives b - 3l - m = 1, therefore, b = 3l + m +1.

Therefore, 5l + m +3 = 3l + m + 1

⇒ l =-1 

∴ -5 + m + 3 = b,

∴ m = 2+b,

∴ a² + b

= l + m + n gives a² + b

= -1+ 2 + b + n,

∴ a² = n + 1.

p(4) - q(4) = a² + 4b - 16l - 4m - n

= (n+1) + 8 - n = 9.

If the mappings f: A →B and g: B →C are both bijective, then the mapping A→C is defined as composition from A to C. And every composite mapping is bijective , i.e. one-one and onto.

A function f : X → Y is defined to be one – one (or injective),if fx₁ ≠ x₂ in A ⇒ f(x₁) ≠ f(x₂) in B.

If range of f is same as co-domain,the function is said to be onto.i.e. Every element of Y is the image of some element of X under f

If f for each y in Y , there exists some x in X such that y = f(x).

A bijective function may be defined as a function which is both injective(one-one) and surjective(onto).e.g.let  us consider a function f:N→N defined by f(x) = x + 10,we see that 

f(x₁) = f(x₂)⇒ x₁ + 10 = x₂ + 10 ⇒ x₁ = x₂ , f is one-one.  Now for onto, let y = f(x) = x + 10 ⇒ x = y - 10 belongs to N,so f is onto.

Hence f is both one-one and onto ⇒ f is a bijective function.