I replaced x by x+h and y by x and took \(\lim_{x\rightarrow0}\) i.e
\[|\lim_{x\rightarrow 0}\dfrac{f(x+h)-f(x)}{h}|\leq0\]
and since Modulus of a real no. cannot be less than 0, we get
\[f'(x)=0 \Rightarrow f(x)=Constant=C(let)\]
And since f(2)=5 \(\Rightarrow \) f(x)=5..
Or f(4)=5.
Maybe I have done some mistake as I am lacking time at the moment.. Anyways can you find the error??

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestI replaced x by x+h and y by x and took \(\lim_{x\rightarrow0}\) i.e \[|\lim_{x\rightarrow 0}\dfrac{f(x+h)-f(x)}{h}|\leq0\] and since Modulus of a real no. cannot be less than 0, we get

\[f'(x)=0 \Rightarrow f(x)=Constant=C(let)\] And since f(2)=5 \(\Rightarrow \) f(x)=5..

Or f(4)=5.

Maybe I have done some mistake as I am lacking time at the moment.. Anyways can you find the error??

Log in to reply

Pretty much did the same thing, seems like they misprinted the answer. Thanks for solving.

Log in to reply

Why must the function be differentiable?

Log in to reply

I am getting f(4)= 5. Is it correct?

Log in to reply

Thanks a lot for solving all my doubts patiently.

Log in to reply

That's what i was getting but the answer given is 25. If you could please write your solution so i can check whether my approach is correct.

Log in to reply

@Rishabh Cool,@parv mor. Please if you could post a solution to the question.

Log in to reply