Question

Beautiful calculus problem

Answer 1

Let\[f'(x) = \frac{192x^3}{2 + \sin^4 (\pi x)}\]for all \(x \in \mathbb R\) with \(f(1/2) = 0\). If \(m\le \int_{1/2}^1 f(x)dx \le M\) then the possible values of \(m\) and \(M\) are:

A: m = 13, M = 24
B: m = 1/4, M = 1/2
C: m = -11, M = 0
D: m = 1, M = 12

Answer 2

ok

Answer 3

ty

Answer 4

ok

Answer 5

@mayankdevnani @ganeshie8

Answer 6

what answer this omg

Answer 7

LOL

Answer 8

@ParthKohli you asked JEE ADVANCE 2015 Question !

Answer 9

you can refer to solutions easily !
GOOGLE IT

Answer 10

Bad solution.

Answer 11

The first equation is the first derivative of f(x), so the inequality will change to a double integral.
is it correct?

Answer 12

In a way, no.

Answer 13

\[\int\limits f \prime dx =f_{(x)} \rightarrow \int\limits f _{x} dx=\int\limits \int\limits f \prime dx^{2}\]

Answer 14

I still think you should go one at a time.

Answer 15

Waiting for reply.

Answer 16

The only thing I could think of is\[m\leq f'(x)\leq M\implies \frac{1}{2}m\leq \int_{1/2}^1 f(x) \,dx\leq \frac{1}{2}M\]

Answer 17

\[f'(x) = \frac{192x^3}{2 + \sin^4 (\pi x)} \implies f(x) = \int\limits_{1/2}^{x} \frac{192t^3}{2 + \sin^4 (\pi t)}\,dt\]

Answer 18

Cannot afford to calculate that in three minutes, can you?

Answer 19

However, I see you're up to something - setting bounds and stuff :D

Answer 20

\[\int\limits_{1/2}^1f(x)\, dx = \int\limits_{1/2}^1~\int\limits_{1/2}^{x} \frac{192t^3}{2 + \sin^4 (\pi t)}\,dt\, dx\]

Answer 21

I thought of changing the order of integration... but it definitely takes more than 3 minutes

Answer 22

Also I am a bit hopeless to start with.. idk if its gonna work or not...

Answer 23

Observe that \(f'(x)\) is increasing on the interval 1/2 to 1 and accordingly "squeeze" our function and then its integral.

Answer 24

Not so much of a lack of time but couldn't figure out how to use the f(1/2)=0 condition.

Answer 25

I think f(1/2) = 0 and f(x) increasing tell us that the curve is above x axis

Answer 26

\(-1\leq\sin^4(x)\leq1\)?

Answer 27

Should I disclose the solution then?

Answer 28

\[8 \le f'(x) \le96 \]

Answer 29

\[4\le f(x) \le 48\]

Answer 30

\[2\le \int\limits_{1/2}^1f(x)\, dx \le 24\]

?

Answer 31

is it A?

Answer 32

yeah A according to above squeezing thingy

Answer 33

Nope. The lower bound has to be smaller than 2 and the upper bound has to be bigger than 24. None of the options satisfy these two criteria. The bounds are not strong enough.

Answer 34

yeah the actual answer is not a range of values as we're given initial condition and it is a definite integral. the answer must be some real number. we get different bounds with different methods i guess

Answer 35

D?

Answer 36

scratch that
i made a terrible mistake earlier

Answer 37

\[8 \le f'(x) \le96 \]

\[8(x-\frac{1}{2})\le \int\limits_{1/2}^x f(x) \le96(x-\frac{1}{2}) \]

Answer 38

does that look correct ?

Answer 39

WELL DONE!

Answer 40

That's it, you'll get the answer now

Answer 41

Who got the answer?

Answer 42

Here i have fixed all mistakes, hopefully :)

\[8 \le f'(x) \le96 \]

\[8(x-\frac{1}{2})\le \int\limits_{1/2}^x f'(x) \le96(x-\frac{1}{2}) \]

\[ \int\limits_{1/2}^18(x-\frac{1}{2})\le \int\limits_{1/2}^1 f(x) \le \int\limits_{1/2}^196(x-\frac{1}{2}) \]

Answer 43

Well done. The beauty of this approach is that the answer exactly matches. It's not like Mayank's.

Answer 44

The rare occasion where a weaker bound is preferred lol.

Answer 45

haha, exactly what i was thinking.

Answer 46

Nice! then the integral doesn't evaluate to some number in the bounds in other options is it ?

Answer 47

Oops! sorry i have deleted your reply by mistake

Answer 48

Yeah if you evaluate the left and right sides you get the exact figures, i.e., 1 and 12.

Answer 49

I found the solution using Mayank's method then weakened the bound. Did not realise he posted a solution since Parth said it was a bad solution and I thought it was a wrong solution.

PS: Using my iPad and it isn't working very well.

Answer 50

I thought my iPad wasn't working very well since one of my reply disappeared... Then I deleted the other one as well since it was rather incoherent without the first one lol.

Answer 51

idk ganeshie will relate, but I prefer getting the exact answers. it gives me some confidence that I'm in the same boat as the guy who created the question.

Answer 52

"thomas5267 I got D using the same method as Mayank."
"thomas5267 Then I weakened the bound."

Answer 53

D is the only one that satisfies the stronger bound so you should not be too unconfident. That said, JEE penalise incorrect answer so I get where you are from.

Answer 54

Wow, are you from India?

Answer 55

Nope. I was comparing the math syllabus of A Level, IB and Chinese Gao Kao with a friend. My friend studied A-Level and I studied IB. He was interested in the Gao Kao since there was one exercise question from Gao Kao that I couldn't solve. On the contrary, his Chinese friend claimed that Gao Kao is much easier compared to AL. Somehow I managed to stumble upon a JEE past paper while finding the syllabus of Gao Kao. Found out that JEE penalise wrong answers unlike A-Level and IB.

I have posted and still haven't closed the Gao Kao exercise question yet. Still couldn't figure it out.

Answer 56

This should be the Gao Kao question link if my iPad works.

http://openstudy.com/study#/updates/56654662e4b0525915ef5098