JEE Main 2017 Mathematics question on Sets Relations and Functions
Let a,b,c∈R. If f(x)=ax2+bx+c be such that a+b+c=3 and f(x+y)=f(x)+f(y)+xy, ∀x,y∈R, and ∑n=110f(n) is equal to
A.
330
B.
165
C.
190
D.
255
Show answer and solution
Correct option: A
Solution
From f(x+y) = f(x) + f(y) + xy, set x = y = 0: f(0) = 2f(0), so f(0) = 0, meaning c = 0.
Set y = x: f(2x) = 2f(x) + x². Also f(2x) = 4ax² + 2bx, and 2f(x) + x² = 2ax² + 2bx + x². Comparing: 4a = 2a + 1, so a = 1/2. Since a + b + c = 3 and c = 0: 1/2 + b = 3, so b = 5/2.
Thus f(n) = n²/2 + 5n/2. The sum: Σf(n) from 1 to 10 = (1/2)Σn² + (5/2)Σn = (1/2)(385) + (5/2)(55) = 192.5 + 137.5 = 330.
[JEE Main 2017]
Practise Sets Relations and Functions the way it is asked
A timed set of real JEE Main Sets Relations and Functions questions, graded as the exam grades. No sign-up.
