Most Important Selected Qs for JEE AdvancedMathematicsContinuity and Differentiability
Let f (x^2+y )=(f(x))^2+f(y) for all x, y R , then
Options
- Af(x) is odd
- Bf(x) is even f(x)=0 x R
- Cf(x) is continuous at x=0 it is continuous everywhere
- Df(x) is differentiable at x=0 f(x)=x f^ (0), x R
Correct answer
D. f(x) is differentiable at x=0 f(x)=x f^ (0), x R
Step-by-step solution
aligned & (x^2+y )=(f(x))^2+f(y) & f(0)=(f(0))^2+f(0) & for y=0: f (x^2 )=(f(x))^2 (i) & for y=-x^2: 0=f(0)=(f(x))^2+f (-x^2 ) & (f(x))^2=-f (-x^2 ) . . (ii) aligned from (i) and (ii) f (-x^2 )=-f (x^2 ) f(-x)=-f(x) (iii) thus f(x) is an odd function if f(x) is even also, then f(-x)=f(x) (iv) f(x)=0 for all x by (iii) and (iv) since f(x) is continuous at x=0 , aligned & _ h 0 f(h)=0 & _ h^2 0 f (x+h^2 )= _ h^2 0 (f(h))^2+f(x) =f(x) aligned it is continuous everywhere since _ h 0 f(0+h)-f(0) h = _ h 0 f(h) h exists