Solveeit Logo
Login

Question

Question: The function f is a differentiable function and satisfies the functional equation \( f(x)+f(y)=f(x+y...

The function f is a differentiable function and satisfies the functional equation f(x)+f(y)=f(x+y)−xy−1f(x)+f(y)=f(x+y)-xy-1 for every pair x,y of real numbers. If f(1)=1f(1)=1 , then the number of integers (n≠1)(n\ne 1) for which f(n)=nf(n)=n is
A. 0
B. 1
C. 2
D. 3

Explanation

Solution

Hint : First, substitute the value of y equal to x in the given equation and check what equation is obtained. Then write the function in the general form. After this, make use of the given conditions and obtain the actual equation of the function.

Complete step-by-step answer :
Let us first analyse the given equation f(x)+f(y)=f(x+y)−xy−1f(x)+f(y)=f(x+y)-xy-1 ….. (i)
Let's put the value of y equal to x.
⇒f(x)+f(x)=f(x+x)−x(x)−1\Rightarrow f(x)+f(x)=f(x+x)-x(x)-1
⇒2f(x)=f(2x)−x2−1\Rightarrow 2f(x)=f(2x)-{{x}^{2}}-1 .
Therefore, we get that f(2x)−2f(x)=x2+1f(2x)-2f(x)={{x}^{2}}+1 ….. (ii)
We know that x2+1{{x}^{2}}+1 is a polynomial function with degree 2.
This means that f(2x)−2f(x)f(2x)-2f(x) is also a polynomial function with degree 2.
Subtraction or addition of polynomial functions of linear independent variables result in another polynomial function of the same degree.
Therefore, f(x) is a polynomial function of degree 2 (quadratic polynomial).
A quadratic polynomial can be written in the form f(x)=ax2+bx+cf(x)=a{{x}^{2}}+bx+c ….. (iii),
where a, b and c are real numbers and a≠0a\ne 0 .
Let us find the values of a, b and c.
Substitute the x=0x=0 in (ii).
⇒f(2(0))−2f(0)=02+1\Rightarrow f(2(0))-2f(0)={{0}^{2}}+1
⇒f(0)−2f(0)=1\Rightarrow f(0)-2f(0)=1 .
⇒−f(0)=1\Rightarrow -f(0)=1
⇒f(0)=−1\Rightarrow f(0)=-1 .
Now, put x=0x=0 in (iii).
⇒f(0)=a(0)2+b(0)+c\Rightarrow f(0)=a{{(0)}^{2}}+b(0)+c
⇒−1=0+0+c\Rightarrow -1=0+0+c
⇒c=−1\Rightarrow c=-1
Now, equation (iii) can be written as f(x)=ax2+bx−1f(x)=a{{x}^{2}}+bx-1 …. (iv)
Substitute x=1x=1 and y=−1y=-1 in (i).
⇒f(1)+f(−1)=f(1−1)−(1)(−1)−1\Rightarrow f(1)+f(-1)=f(1-1)-(1)(-1)-1
⇒f(1)+f(−1)=f(0)+1−1\Rightarrow f(1)+f(-1)=f(0)+1-1
⇒f(1)+f(−1)=f(0)\Rightarrow f(1)+f(-1)=f(0)
It is given that f(1)=1f(1)=1 .
⇒1+f(−1)=−1\Rightarrow 1+f(-1)=-1
⇒f(−1)=−2\Rightarrow f(-1)=-2 .
Now, put x=−1x=-1 in (iv).
⇒f(−1)=a(−1)2+b(−1)−1\Rightarrow f(-1)=a{{(-1)}^{2}}+b(-1)-1
⇒−2=a−b−1\Rightarrow -2=a-b-1
⇒a−b=−1\Rightarrow a-b=-1 …. (v).
Substitute x=1x=1 in (iv).
⇒f(1)=a(1)2+b(1)−1\Rightarrow f(1)=a{{(1)}^{2}}+b(1)-1
⇒1=a+b−1\Rightarrow 1=a+b-1
⇒a+b=2\Rightarrow a+b=2 …. (vi).
Now, add (v) and (vi).
⇒a−b+(a+b)=−1+2\Rightarrow a-b+(a+b)=-1+2
⇒2a=1\Rightarrow 2a=1
⇒a=12\Rightarrow a=\dfrac{1}{2}
Substitute the value of ‘a’ in (v)
⇒12−b=−1\Rightarrow \dfrac{1}{2}-b=-1
⇒b=32\Rightarrow b=\dfrac{3}{2} .
Now, substitute the values of a and b in (iv).
⇒f(x)=12x2+32x−1\Rightarrow f(x)=\dfrac{1}{2}{{x}^{2}}+\dfrac{3}{2}x-1 .
Let us define f(x)=xf(x)=x .
⇒12x2+32x−1=x\Rightarrow \dfrac{1}{2}{{x}^{2}}+\dfrac{3}{2}x-1=x
⇒12x2+12x−1=0\Rightarrow \dfrac{1}{2}{{x}^{2}}+\dfrac{1}{2}x-1=0
⇒x2+x−2=0\Rightarrow {{x}^{2}}+x-2=0
Let us now find the values of x that satisfy the above equation.
We can write the above equation as (x+2)(x−1)=0(x+2)(x-1)=0 .
This means that (x+2)=0(x+2)=0 or (x−1)=0(x-1)=0 .
⇒x=−2\Rightarrow x=-2 or x=1x=1
Therefore, the integers for which f(n)=nf(n)=n (n is an integer) are -2 and 1.
But it is given that n≠1n\ne 1 .
Therefore, only one integer (i.e. n=−2n=-2 ) satisfies the above condition.
So, the correct answer is “Option B”.

Note : Wherever you get a question similar to this question, where an equation involving two variables (x and y) is given, first substitute y=xy=x so that the equation becomes dependent on one variable. Then make of the given conditions and derive the function.