Question

If f is a real-valued differentiable function satisfying $\left| f(x)-f(y) \right|\le {{(x-y)}^{2}},x,y\in \mathbb{R}$
and $f(0)=0$ , then $f(1)$ is equal to
(a)2
(b)1
(c)-1
(d)0

Answer 1

Hint: Here, we have to take $x=x+h$ and $y=x$, divide the equation by $h$, take limit on both the sides, and then apply the formula of derivative, $f'(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x+h)-f(x)}{h}$. Also apply the property that derivative of a constant is zero.

Complete step-by-step answer:
In the question it is given that,
$\left| f(x)-f(y) \right|\le {{(x-y)}^{2}},x,y\in \mathbb{R}\text{ }....\text{ (1)}$
Now, put $x=x+h$ and $y=x$ in equation (1).
Our equation (1) becomes:
$\begin{align}
  & \left| f(x+h)-f(x) \right|\le {{(x+h-x)}^{2}} \\
 & \left| f(x+h)-f(x) \right|\le {{h}^{2}} \\
\end{align}$
Now, by dividing h on both the sides we get:
$\begin{align}
  & \left| \dfrac{f(x+h)-f(x)}{h} \right|\le \dfrac{{{h}^{2}}}{h} \\
 & \left| \dfrac{f(x+h)-f(x)}{h} \right|\le h\text{ } \\
\end{align}$
Next, by applying limit on both the sides we obtain:
$\underset{h\to 0}{\mathop{\lim }}\,\left| \dfrac{f(x+h)-f(x)}{h} \right|\le \underset{h\to 0}{\mathop{\lim }}\,h\text{ }$
We know that $\underset{h\to 0}{\mathop{\lim }}\,h=0$ .
Therefore, our equation becomes,
$\underset{h\to 0}{\mathop{\lim }}\,\left| \dfrac{f(x+h)-f(x)}{h} \right|\le 0\text{ }....\text{ (2)}$
We know by the definition of derivatives that,
$f'(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x+h)-f(x)}{h}$
Now, by applying modulus on both the sides we get:
$\left| f'(x) \right|=\left| \underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x+h)-f(x)}{h} \right|$
Since, $\underset{h\to 0}{\mathop{\lim }}\,$ is a constant we can take it outside from the modulus, then our equation becomes,
$\left| f'(x) \right|=\underset{h\to 0}{\mathop{\lim }}\,\left| \dfrac{f(x+h)-f(x)}{h} \right|\text{ }....\text{ (3)}$
Now put equation (2) in equation (3) we get:
$\left| f'(x) \right|=\underset{h\to 0}{\mathop{\lim }}\,\left| \dfrac{f(x+h)-f(x)}{h} \right|\le 0$ i.e.
$\left| f'(x) \right|\le 0$
We know that $\left| f'(x) \right|$ is positive therefore it can never be negative. So we can say that,
$\left| f'(x) \right|=0$
If the modulus of a function equals zero then we can say that function also equals zero. Therefore, $f'(x)=0$.
We also know that if the derivative of a function is zero then the function is a constant. i.e. $f(x)$ is a constant.
In the question it is given that $f(0)=0$. Since, $f(x)$ is a constant we can say that $f(1)=0$.
Hence, the correct answer is option (d)

Note: Here we have to apply the definition of derivative and also don’t forget that if the derivative of a function is zero then the function is a constant.