Question

if f(x+y)=f(x)+f(y)-xy-1 for all x, y belongs to real numbers and f(1)=1,then the number of solutions of f(x)=n, n belongs to natural numbers​

Answer 1

Given:

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

f( 1 ) = 1 ;

To Find:

number of solutions of f(x)=n, n belongs to natural numbers​.

Solution:

Given that,

• f ( x + y ) =  f ( x ) + f ( y ) - xy - 1
• f(0) = 2f(0) - 1
• f(0) = 1 --(a)

By First principle of derivative,

• f ' ( x ) = lim h->0 ( f ( x + h ) - f ( x ) )/ h
• f ' ( x )  = lim h - > 0 ( f(x) + f(h) -hx - 1 - f(x) ) / h
• f ' ( x ) = lim h->0 (f(h) - hx - 1) /h = 0/0 form when h = 0
• Therefore, by L'Hospital rule,
• derivative of numerator and denominator,
• f ' ( x ) = lim h-> f ' ( h ) - x = f' ( 0 ) - x = k - x .
• Intergrating,
• f ( x ) = kx - x²/2 + C

Now given ,f (1) = 1  and f (0) = 1

Therefore,

• f (0) = C = 1

• f  ( 1 ) = k - 1/2 + 1 = 1
• k = 1/2

Therefore,  f(x) = x/2 - x²/2 + 1 .

For f( x) to be a natural number , x/2 - x²/2 + 1 > 0

• x - x² + 2 > 0
• x² - x - 2 < 0
• x² + x -2x - 2 < 0
• x ( x + 1 ) -2( x + 1 ) < 0
• ( x -2 ) ( x + 1) < 0
• x <  2 and x > -1  -- (a)

For f(x) to be natural number x should be 1 or a  multiple of 2.

Integer values of x based on (a) is 0 , 1 .

Both are possible solutions.

The number of solutions of f(x)=n, n belongs to natural numbers​ , is 2.