JEE Advanced2011MathematicsContinuity and DifferentiabilityActual
Let f: R R be a function such that f(x+y)=f(x)+f(y), x, y R . If f(x) is differentiable at x=0 , then
Options
- Af(x) is differentiable only in a finite interval containing zero
- Bf(x) is continuous, x R
- Cf^ (x) is constant, x R
- Df(x) is differentiable except at finitely many points
Correct answer
B. f(x) is continuous, x R
Step-by-step solution
f(x+y)=f(x)+f(y) , as f(x) is differentiable at x=0 . aligned & f^ (0)=k Now, f^ (x) & = _ h 0 f(x+h)-f(x) h & = _ h 0 f(x)+f(h)-f(x) h & = _ h 0 f(h) h [ 0 0 from ] aligned Given, f(x+y)=f(x)+f(y), x, y aligned & & f(0) & =f(0)+f(0), & when & x & =y=0 f(0)=0 aligned Using L'Hospital's rule, _ h 0 f^ (h) 1 =f^ (0)=k f^ (x)=k , on integrating both sides, f(x)=k x+C , as f(0)=0 C=0 So, f(x)=k x f(x) is continuous for all x R and f^ (x)=k , i.e. constant for all x R . Hence, both (b) and (c) are correct. Solutions (Q.