Let f:R→R be a function such that f(x+y)/3=(fx+fy)/3, f0=0 and f^(′) 0 =5, then

Correct answer: (C) fx is differentiable in R. We have, f(x+y)/3=(fx+fy)/3,f0=0 and f^(′) 0 =5 f^(′) x = lim h→0 (f x+h −f x)/h =limh→0⁡(f3x+3h)/3-f(3x+0)/3h =limh→0(f3x+f3h)/3-(f3x+f0)/3h = lim h→0 (f 3h −f 0)/3h =f^(′) 0 =5 ∴fx=5x+c∴f0=0⇒c=0 Hence, fx=5x

Q31 / 93 · NTA Abhyas (JEE Main) 2020 · Mathematics · Continuity and Differentiability

Let $f : R \to R$ be a function such that $f (\frac{x + y}{3}) = \frac{f (x) + f (y)}{3}, f (0) = 0$ and $f^{'} (0) = 5,$ then

A
$f (x)$ is a quadratic function
B
$f (x)$ is continuous but not differentiable
Correct answer
C
$f (x)$ is differentiable in $R$
D
$f (x)$ is bounded in $R$

Solution

We have,

f (\frac{x + y}{3}) = \frac{f (x) + f (y)}{3}, f (0) = 0

and

f^{'} (0) = 5

f^{'} (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h}

= \lim_{h \to 0} \frac{f (\frac{3 x + 3 h}{3}) - f (\frac{3 x + 0}{3})}{h}

= \underset{h \to 0}{l i m} \frac{\frac{f (3 x) + f (3 h)}{3} - \frac{f (3 x) + f (0)}{3}}{h}

= lim_{h \to 0} \frac{f (3 h) - f (0)}{3 h} = f^{'} (0) = 5

∴ f (x) = 5 x + c ∴ f (0) = 0 \Rightarrow c = 0

Hence,

f (x) = 5 x

Hence, the correct answer is option: 'C'.

Let f:R→R be a function such that fx+y3=fx+fy3, f0=0 and f ′ 0 =5, then

Solution

More Continuity and Differentiability Questions

Let $f : R \to R$ be a function such that $f (\frac{x + y}{3}) = \frac{f (x) + f (y)}{3}, f (0) = 0$ and $f^{'} (0) = 5,$ then