Definite integral - Page 3 - The Student Room

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OP

Mr M

And there you go thomas .. you weren't ignored were you?

No; now I guess my earlier threads were "ignored" probably because of the difficulty of the problems rather than me.

OP

SimonM

Pick me pick me!

I think you have missed some modulus signs taking a certain square root

\sqrt{\frac{u}{2}+2+2\sqrt{u}} = |(\frac{\sqrt{u}}{\sqrt{2}}+\sqrt{2})| \not= \frac{\sqrt{u}}{\sqrt{2}}+\sqrt{2}

Actually,

\sqrt{\frac{u}{2}+2+2\sqrt{u}} = |(\frac{\sqrt{u}}{\sqrt{2}}+\sqrt{2})| = \frac{\sqrt{u}}{\sqrt{2}}+\sqrt{2}

However,

\sqrt{\frac{u}{2}+2-\sqrt{u}} = |(\frac{\sqrt{u}}{\sqrt{2}}-\sqrt{2})| \not= \frac{\sqrt{u}}{\sqrt{2}}-\sqrt{2}

Alex, I made that point immediately after posting that.

OP

SimonM

Alex, I made that point immediately after posting that.

But I saw that only just now. You should've edited your post, then.

Anyway the matter is resolved. :smile:

:smile:

OP

SimonM

Either that or the problem setters made a mistake

I too feel that the problem setters thought B) was the correct answer. Oh well...

alexmahone

No; now I guess my earlier threads were "ignored" probably because of the difficulty of the problems rather than me.

More likely is simply that nobody was interested in the problems you posted.

Which earlier threads of yours were ignored?

OP

SimonM

Which earlier threads of yours were ignored?

"Pigeonhole principle", "Sequence of primes" (received only one helpful response, which was from you) and "Find the last 6 digits.."

PHP had already been answered in a different fora and I couldn't think of a hint which didn't give the answer away instantly.

Sequence of primes, was responded to, but you'd really solved it yourself

Find the last 6 digits was also responded to

OP

Kolya

More likely is simply that nobody was interested in the problems you posted.

As a matter of fact, nobody is interested in problems that are far beyond their ability.

alexmahone

As a matter of fact, nobody is interested in problems that are far beyond their ability.

I've read plenty of books on the Riemann hypothesis

alexmahone

As a matter of fact, nobody is interested in problems that are far beyond their ability.

While I'm not sure that is true, even if it were don't forget that

(¬X \implies ¬Y)

does not imply

X \implies Y

(!)

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