Question

Let ƒ : R ® R be a function such that ƒ $\frac{1}{2}$= $- \frac{1}{2}$for all x and y, and ƒ(0) = 3 and ƒ¢ (0) = 3. Then –

• A. ƒ(x)/x is continuous on R
• B. ƒ(x) is continuous on R
• C. ƒ(x) is bounded on R
• D. None of these

Answer 1

Answer: (B)

ƒ(x) is continuous on R

Answer 2

ƒ(x + h) = ƒ

= $\frac { 1 } { 2 }$[ƒ(2x) + ƒ(2h)]

= $\frac { 1 } { 2 }$ƒ (2x) + $\frac { 1 } { 2 }$ ƒ(2h) = $\frac { 1 } { 2 }$ ƒ(2x) + $\frac { 1 } { 2 }$ƒ(0)

(ƒ is differentiable at 0 so continuous also )

Putting y = 0 in the given equation, we have

ƒ(x) = ƒ $\left( \frac { 2 x } { 2 } \right)$ = $\frac { f ( 2 \mathrm { x } ) + f ( 0 ) } { 2 }$

Hence ƒ(x + h) = ƒ(x)

Since ƒ(x)/x is not defined at x = 0, ƒ(x)/x is not continuous on R. Clearly ƒ(x) need not be bounded on R. e.g. ƒ(x) = x .

ƒsatisfies the given equation but is not bounded on R.