JEE 2022 — Mathematics PYQ
JEE | Mathematics | 2022Let c,k∈R. If f(x)=(c+1)x2+(1−c2)x+2k and f(x+y)=f(x)+f(y)−xy, for all x,y∈R, then the value of ∣2(f(1)+f(2)+f(3)+⋯+f(20))∣ is equal to:
Choose the correct answer:
- A.
3395
(Correct Answer) - B.
3495
- C.
3595
- D.
3695
3395
Explanation
Solution
Step 1: Determine the constants
From the functional equation f(x+y)=f(x)+f(y)−xy:
Set x=0,y=0⟹f(0)=f(0)+f(0)−0⟹f(0)=0.
Given f(x)=(c+1)x2+(1−c2)x+2k, since f(0)=0, we have 2k=0⟹k=0.
Now, substitute the form of f(x) into the functional equation:
Expanding the left side:
Comparing the xy terms:
Thus, f(x)=−21x2+(1−c2)x. Since c=−3/2, 1−c2=1−9/4=−5/4.
So, f(x)=−21x2−45x.
Step 2: Calculate the sum
Sum S=∑n=120f(n)=−21∑n2−45∑n.
Using standard summation formulas:
-
∑n=120n2=620(21)(41)=2870
-
∑n=120n=220(21)=210
Final Step:
The expression is ∣2(S)∣=∣2×(−1697.5)∣=3395.
Explanation
Solution
Step 1: Determine the constants
From the functional equation f(x+y)=f(x)+f(y)−xy:
Set x=0,y=0⟹f(0)=f(0)+f(0)−0⟹f(0)=0.
Given f(x)=(c+1)x2+(1−c2)x+2k, since f(0)=0, we have 2k=0⟹k=0.
Now, substitute the form of f(x) into the functional equation:
Expanding the left side:
Comparing the xy terms:
Thus, f(x)=−21x2+(1−c2)x. Since c=−3/2, 1−c2=1−9/4=−5/4.
So, f(x)=−21x2−45x.
Step 2: Calculate the sum
Sum S=∑n=120f(n)=−21∑n2−45∑n.
Using standard summation formulas:
-
∑n=120n2=620(21)(41)=2870
-
∑n=120n=220(21)=210
Final Step:
The expression is ∣2(S)∣=∣2×(−1697.5)∣=3395.