CLASS 12 ADDITIONAL SUMS ON RELATIONS AND FUNCTIONS

CLASS 12

RELATIONS AND FUNCTIONS

ADDITIONAL SUMS

ONE-ONE AND ONTO FUNCTIONS

1. Prove that the function f:N →  N defined by f(x) = x²+ x + 1 is one - one but not onto.

Let x, y ∊ N such that

f(x) = f(y)

x² +  x + 1  =     y²  + y + 1 

x² - y² + x - y = 0

(x - y)(x + y) + (x - y) = 0

(x - y)(x + y + 1) = 0

Since x+y+1 ≠ 0, we have

x - y = 0

x = y

Therefore f is one-one.

But, since f:N →N, f(x)  cannot take values 1 and 2.  Therefore f is not onto.

2. Show that the function f: R→ R defined by f(x)  = 2x³ - 7 is bijective.

Let x, y ∊ R  

such that f(x) = f(y)

2x³ - 7 = 2y³ - 7

2(x³ - y³) = 0

2(x-y)(x² + x y + y²) = 0

We know x² + x y + y² ≠ 0

we have x- y = 0

which implies x = y

Therefore f is one-one.

Let y = 2x³ - 7

      x³ = (y + 7)/2

      x  = ((y+7)/2) ^ 1/3 

Therefore for all values of y ∊ R,  there exists x ∊ R and so f is onto.

Hence f is bijective.

3. Show that the function f: W→ W defined f(n) = n+1 , if n is even and f(n) = n-1, if n is odd is bijective.

Let x, y be even

such that f(x) = f(y)

Then x +1 = y + 1

    ⇒     x  =  y

Let x, y be odd 

such that f(x) = f(y)

Then x - 1 = y - 1

      ⇒    x  =  y

Let x is even and y is odd, then

x + 1 will be odd and y - 1 will be even and so

if x ≠  y then f(x)   ≠  f(y)

Let x be odd and y be even, then

x - 1 will be even and y + 1 will be odd and so

if x ≠ y then f(x) ≠  f(y)

So f is one-one.

If x is odd, then there exists even number x - 1 such that 

f(x - 1 ) = x - 1 + 1 = x

If x is even, then there exists odd number x + 1 such that

f(x + 1) = x + 1 - 1 = x

So, f is onto.

Therefore f is bijective.

4. Let A = R - {2/3 } and f: A → A be defined by f(x) = (4x+3)/(6x-4).  Show that f is one - one and onto.

Let x, y ∊ A such that f(x) = f(y)

(4x+3)/ (6x -4) = (4y + 3) / (6y - 4)

(4x + 3) (6y - 4)  =  (4y + 3) (6x - 4)

24xy - 16x + 18 y - 12 = 24xy - 16y + 18 x - 12

-16 x + 16 y + 18 y - 18 x = 0

-34 x + 34 y = 0

-34 (x - y ) = 0

which implies x- y = 0

       x = y

So, f is one - one.

Let y = (4x+3)/(6x-4)

y(6x-4) = 4x + 3

6xy - 4y - 4x = 3

x(6y - 4) = 3 + 4y

x =( 4y + 3 )/ (6y -4)

For all y ∊ A = R- {2/3}, there exists x ∊  A = R - {2/3} and so f is onto.

5. Let f: N→  R defined by f(x) = 4x² + 12 x  15 where R is the range of f.  Show that f is bijective.

Let x,  y  ∊ N

such that f(x) = f(y)

4x² + 12 x + 15 = 4 y² + 12 y + 15

4x² + 12 x - 4 y² - 12 y = 0

4(x² -  y² )  + 12 (x - y) = 0

4 (x + y)( x - y) + 12( x - y ) = 0

(x-y)(4x + 4y +12 ) = 0

Clearly 4x + 4y + 12 ≠ 0

So x- y = 0

  ⇒    x = y

Therefore f is one-one.

Let y = 4x² + 12 x + 15

4x² + 12 x + 15 - y = 0

  

x  = (- 12 ± √(144 - 16(15-y)))/8

    = (-12  ±  √(144 - 240 + 16y))/8

    = (-12 ±  √(16y-96))/8

    = (-12 ±  √(16(y-6)))/8

    = (-12 ± 4 √(y-6))/8

    = (-3 ± √(y-6))/2

    = (√(y-6) - 3 )/2

So f is onto.

Hence f is bijective.