Question
Mathematics Question on Functions
Let the function f:R→R be defined by
f(x)=eπxsinx(x2−x+3)(x)2023+2024x+2025+eπx2(x2−x+3)(x2023+2024x+2025)
Then the number of solutions of f (x) = 0 in R is __.
Answer
The correct answer is: 1.
Let the function f:R→R be defined by
f(x)=eπxsinx(x2−x+3)(x)2023+2024x+2025+eπx2(x2−x+3)(x2023+2024x+2025)
Then the number of solutions of f (x) = 0 in R is __.
The correct answer is: 1.