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← JEE Main 2022 · 29 Jun · Shift 1
MathematicsInteger+4 / -1Question 27 of 90

JEE Main 2022 · 29 Jun · Shift 1, Mathematics question 27

Let c,k∈Rc,k\in\mathbf R. If f(x)=(c+1)x2+(1−c2)x+2kf(x)=(c+1)x^2+\left(1-c^2\right)x+2k and f(x+y)=f(x)+f(y)−xyf(x+y)=f(x)+f(y)-xy, for all x,y∈Rx,y\in\mathbf R, then the value of ∣2(f(1)+f(2)+f(3)+⋯+f(20))∣\left|2\left(f(1)+f(2)+f(3)+\dots+f(20)\right)\right| is equal to ______.

Answer and worked solution

Official answer: 3395

Identity. 2(c+1)xy=−xy2(c+1)xy=-xy gives c=−32c=-\frac32; constants give k=0k=0. So f(x)=−x22−54xf(x)=-\frac{x^2}2-\frac54x.

Sum. ∑120f=−28702−54⋅210=−1697.5\sum_1^{20}f=-\frac{2870}2-\frac54\cdot210=-1697.5; ∣2×sum∣=3395|2\times\text{sum}|=3395.

Answer: 3395

Topics:Types of functionsSpecial series sums
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