JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023Let f:R→R satisfy f(x+y)=f(x)+f(y)−1. If f′(0)=2, find ∣f(−2)∣=.
Choose the correct answer:
- A.
3
(Correct Answer) - B.
2
- C.
1
- D.
0
3
Explanation
Solution:
-
Differentiate with respect to y: f′(x+y)=f′(y).
-
Set y=0: f′(x)=f′(0)=2.
-
Integrate: f(x)=2x+c.
-
Find c: From the original relation, f(0+0)=f(0)+f(0)−1⟹f(0)=1.
1=2(0)+c⟹c=1So, f(x)=2x+1.
-
Calculate ∣f(−2)∣:
∣f(−2)∣=∣2(−2)+1∣=∣−3∣=3
Final Answer:
∣f(−2)∣=3
Explanation
Solution:
-
Differentiate with respect to y: f′(x+y)=f′(y).
-
Set y=0: f′(x)=f′(0)=2.
-
Integrate: f(x)=2x+c.
-
Find c: From the original relation, f(0+0)=f(0)+f(0)−1⟹f(0)=1.
1=2(0)+c⟹c=1So, f(x)=2x+1.
-
Calculate ∣f(−2)∣:
∣f(−2)∣=∣2(−2)+1∣=∣−3∣=3
Final Answer:
∣f(−2)∣=3

