Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

If $f$ is a real-valued differentiable function satisfying

$\left | f\left ( x \right ) -f\left ( y \right )\right |\leq \left ( x-y \right )^{2},x,y \in R$ and $f\left ( 0 \right )= 0$ then $f\left ( 1 \right )$ equals

Option 1)
$1$

Option 2)
$2$

Option 3)
$0$

Option 4)
$-1$

Answers (1)

best_answer

As we learnt in

Differentiability -

Let f(x) be a real valued function defined on an open interval (a, b) and $x\epsilon$ (a, b).Then the function f(x) is said to be differentiable at $x_{\circ }$ if

$\lim_{h\rightarrow 0}\:\frac{f(x_{0}+h)-f(x_{0})}{(x_{0}+h)-x_{0}}$

$or\:\:\:\lim_{h\rightarrow 0}\:\frac{f(x)-f(x_{0})}{x-x_{0}}$

-

$\lim_{x\rightarrow y}\frac{\left | f\left ( x \right )-f\left ( y \right ) \right |}{\left | x-y \right |}\leq \lim_{x\rightarrow y}\left | x-y \right |$

$\Rightarrow \left | f{}'\left ( y \right ) \right |\leq 0 \Rightarrow f{}'\left ( y \right )=0$

$\Rightarrow f\left ( y \right )= constant$

Since $f\left ( 0 \right )=0$

$\therefore f\left ( y \right )=0$

Hence $f\left ( 1 \right )=0$

Option 1)

$1$

Incorrect option

Option 2)

$2$

Incorrect option

Option 3)

$0$

Correct option

Option 4)

$-1$

Incorrect option

Posted by

divya.saini

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2024 Topper: Farmer’s son Gajare Nilkrishna says ‘keep questioning until you understand’

anam-khan

NIT Puducherry Cut-off 2024: Computer Science and Engineering had highest closing rank last year

anam-khan

What after JEE Main results 2024? All you need to know about JEE Adv, CSAB, JoSAA counselling

Latest Question