MHT CET202210 Aug 2022Morning ShiftMathematicsFunctionsActual
Let f: R R be a differentiable function with f(0)=1 and satisfying the equation f(x+y)=f(x) f^ (y)+f^ (x) f(y), x, y R , then the value of (f(4)) is
Options
- A1
- B4
- C2
- D1 2
Correct answer
C. 2
Step-by-step solution
f(x+y)=f(x) f^ (y)+f^ (x) f(y) x, y R putting x=y=0 we get aligned & f(0)=2 f(0) f^ (0) & f^ (0)= 1 2 [ given f(0)=1] aligned Now putting x=x and y=0 aligned & f(x)=f(x) f^ (0)+f^ (x) f(0) & f(x)= 1 2 f(x)+f^ (x) [ f(0)=1 and f^ (0)= 1 2 ] & 1 2 f(x)=f^ (x) & f^ (x) f(x) d x= 1 2 ~d x & (f(x))= 1 2 x+c & f(0)=1 & c=0 aligned i.e., (f(x))= 1 2 x putting x=4 (f(4))= 1 2 4=2