Question

Let \(f\) be a differentiable function satisfying \[ f(x+y)=f(x)+f(y)-xy \quad \text{for all } x,y\in\mathbb{R}. \] If \[ \lim_{h\to 0}\frac{f(h)}{h}=3, \] then the value of \[ \sum_{n=1}^{10} f(n) \] is equal to:

• A. \(-\dfrac{55}{2}\)
• B. \(\dfrac{275}{2}\)
• C. \(-\dfrac{55}{4}\)
• D. \(\dfrac{225}{4}\)

Hint:

Functional equations involving \(f(x+y)\) often lead to polynomial solutions. Always use the given limit to fix remaining constants.

Answer 1

The Correct Option is A

Step 1: Find the general form of \(f(x)\) Given: \[ f(x+y)=f(x)+f(y)-xy \] Assume \(f(x)\) is a polynomial of degree \(\le 2\): \[ f(x)=ax^2+bx+c \] Substitute into the functional equation: \[ a(x+y)^2+b(x+y)+c = ax^2+bx+c + ay^2+by+c - xy \] Comparing coefficients: \[ 2a=-1 \Rightarrow a=-\frac12 \] Constant term: \[ c=0 \] So, \[ f(x)=-\frac{x^2}{2}+bx \] Step 2: Use the given limit \[ \lim_{h\to 0}\frac{f(h)}{h} =\lim_{h\to 0}\left(-\frac{h}{2}+b\right)=b \] Given the limit is \(3\): \[ b=3 \] Hence, \[ f(x)=3x-\frac{x^2}{2} \] Step 3: Compute the required sum \[ \sum_{n=1}^{10} f(n) =\sum_{n=1}^{10}\left(3n-\frac{n^2}{2}\right) \] \[ =3\sum_{n=1}^{10}n-\frac12\sum_{n=1}^{10}n^2 \] \[ =3\cdot\frac{10\cdot11}{2}-\frac12\cdot\frac{10\cdot11\cdot21}{6} \] \[ =165-\frac{385}{2} =\frac{330-385}{2} =-\frac{55}{2} \] \[ \boxed{-\dfrac{55}{2}} \]