JEE Main 2017 · 2 Apr · Shift 1 · Mathematics Q69

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MathematicsSingle correct+4 / -1Question 69 of 90

Let a, b, c ∈ ℝ. If $f(x)=ax^2+bx+c$ is such that $a+b+c=3$ and $f(x+y)=f(x)+f(y)+xy$ , ∀ x, y ∈ ℝ, then $\displaystyle\sum_{n=1}^{10}f(n)$ is equal to :

165

190

255

330

Answer and worked solution

Official answer: D

Comparing coefficients in $f(x+y)=f(x)+f(y)+xy$ gives $2a=1$ and $c=0$ , so $a=\frac12$ and $b=3-\frac12=\frac52$ .
$\sum_{n=1}^{10}\left(\frac{n^2}2+\frac{5n}2\right)=\frac{385}2+\frac{275}2=330.$

Topics:Types of functions Special series sums

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JEE Main 2017 · 2 Apr · Shift 1, Mathematics question 69