JEE maths#17

Calculus Level 4

If $f(x)$ is a continuous function such that $f(x) > 0$ for all $x ≥ 0$ , and $(f(x))^{2014} = 1 + \displaystyle\int_{0}^{x} f(t) \, dt$ , then find the value of $(f(2014))^{2013}$ .

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The answer is 2014.

1 solution

Anirudh Sreekumar
Apr 3, 2017

$\begin{aligned}(f(x))^{2014}&=1+\int_{0}^x f(t)dt\hspace {7mm}\color{#3D99F6}\small(1)\\\\ \text{We can see that ,}&\\\\ (f(0))^{2014}&=1+\int_{0}^0 f(t)dt\\\\ (f(0))^{2014}&=1\\\\ \implies(f(0))&=1\hspace {7mm}\color{#3D99F6}\small f(0)\neq-1 \text{ ,as } f(x)>0 \text{ for } x\geq0\\\\ \text{Diffentiating }\color{#3D99F6}\small(1)\color{#333333}\normalsize &\text{ we get ,}\\\\ 2014(f(x))^{2013}f'(x)&=f(x)\\\\ 2014(f(x))^{2012}f'(x)&=1\hspace {7mm}\color{#3D99F6}\small(2)\\\\ \text{Integrating }\color{#3D99F6} \small(2)\color{#333333} \normalsize\text{ we get,}\\\\ 2014\cdot \dfrac{(f(x))^{2013}}{2013}&=x+C\hspace {7mm}\color{#3D99F6}\small\text {where,} C\text{ is constant of integration}\\\\ \dfrac{2014}{2013}\cdot(f(0))^{2013}&=C\\\\ \implies C&=\dfrac{2014}{2013}\hspace {7mm}\color{#3D99F6}\small\text{As } f(0)=1\\\\ (f(x))^{2013}&=\dfrac{2013}{2014}\left(x+\dfrac{2014}{2013}\right)=\left(\dfrac{2013}{2014}\cdot x\right)+1\\\\ (f(2014))^{2013}&=\left(\dfrac{2013}{2014}\cdot 2014\right)+1=2013+1=\color{#D61F06}\boxed{2014}\end{aligned}$

Nice solution sir (+1)... Did the same way...

Rahil Sehgal - 4 years, 2 months ago

Shit man!! I just missed the C!!... Or else it would have been correct.

Md Zuhair - 4 years, 2 months ago

@Rahil Sehgal @Md Zuhair @Anirudh Sreekumar Please help me out with the step that says 'Integrating (2)'

Is it done this way?

$2014(f(x))^{2012}f'(x) = 1$

$\Rightarrow 2014(f(x))^{2012}\dfrac{d(f(x))}{dx} = 1$

$\Rightarrow 2014(f(x))^{2012}d(f(x)) = dx$

Integrating both sides

$\Rightarrow \dfrac{2014}{2013} \cdot (f(x))^{2013} = x + c$

Ankit Kumar Jain - 4 years ago

Then what is your c?

Md Zuhair - 4 years ago

I solved it the same way...but I was not very sure of that step which I mentioned in my previous post...so I asked about it...Is it the correct method??

As for calculating c , it can be done the way Anirudh sir has done it.

Ankit Kumar Jain - 4 years ago

@Ankit Kumar Jain – Ya, sure, why were you not confident with this solution. It seems perfectly ok!

Md Zuhair - 4 years ago

@Md Zuhair – Okay...Thanks!!

I was not confident about that step as I am beginner at calculus ...still struggling with the basics...so I just wanted to confirm..

Ankit Kumar Jain - 4 years ago

@Ankit Kumar Jain – No. It doesnt seem you are struggling with basics. It seems you are good at it!

Md Zuhair - 4 years ago

@Md Zuhair – It has just been 10 days or so since I have learned calculus...A long way to go..:) :)

Ankit Kumar Jain - 4 years ago

@Ankit Kumar Jain – Yes a long way to go, But you have even come long way into calculus.

Md Zuhair - 4 years ago

@Md Zuhair – No...Just learned a bit of it.

Ankit Kumar Jain - 4 years ago

@Md Zuhair – @Ankit Kumar Jain Your way is perfectly fine and you have learnt so much in just 10 days...

Rahil Sehgal - 4 years ago

@Rahil Sehgal – No..aisa kuch nahi hai.

Ankit Kumar Jain - 4 years ago

@Ankit Kumar Jain – @Ankit Kumar Jain

Let $u = f(x)$

Then

$2014 (u)^{2012} \cdot \dfrac{d}{dx} u = 1$

$\Rightarrow 2014 (u)^{2012} \; du = dx$

On integrating you will get the desired result.

Rahil Sehgal - 4 years ago

@Rahil Sehgal – Thanks!! That is better.

Ankit Kumar Jain - 4 years ago

Oh yeah c=1, so you get it.

Md Zuhair - 4 years ago

@Rahil Sehgal @Md Zuhair tell me the technique behind the last two problems of this set...I mean is there some methods of solving such questions?

Ankit Kumar Jain - 4 years ago

hey...guys.. are you there?

Ankit Kumar Jain - 4 years ago

Even i couldnt do those! So i cant help

Md Zuhair - 4 years ago

The last one is very easy.. but the second last is a bit tricky...

u can come on slack .. i will explain u...

Rahil Sehgal - 4 years ago

@Rahil Sehgal – I am online on slack..come up..

Ankit Kumar Jain - 4 years ago

JEE maths#17

For more JEE problems try my set

The answer is 2014.

1 solution

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