Hello,

I have a question about solution of 2020 examination, question 37.

The solution looks confusing to me because when I integrate t (1 - (1-u)^{t-1}), I get:

\(\int_{0}^{u} t (1 - (1-x)^{t-1}) dx = tu - 1 + (1-u)^{t}\) What is different from expected.

Shouldn't we say:

\( t(1-u)^{t-1} \leq t \)

Then integrate both sides between 0 and u:

\(\int_{0}^{u}t (1-x)^{t-1} dx = 1 - (1-u)^{t}\leq tu\)

Then consider: \( (1 - (1-u)^{t})^2 \leq 1 - (1-u)^{t}\leq tu\)

Yes, sounds like you're right. There's a typo in the solution. It could be t(1-(1-u)^{t-1}) leq 0 or equivalently, what you said. What is written sounds like a combination of the two :) Thanks for reporting.

## Solution Exam 2020 question 37

Hello,

I have a question about solution of 2020 examination, question 37.

The solution looks confusing to me because when I integrate t (1 - (1-u)^{t-1}), I get:

\(\int_{0}^{u} t (1 - (1-x)^{t-1}) dx = tu - 1 + (1-u)^{t}\)

What is different from expected.

Shouldn't we say:

\( t(1-u)^{t-1} \leq t \)

Then integrate both sides between 0 and u:

\(\int_{0}^{u}t (1-x)^{t-1} dx = 1 - (1-u)^{t}\leq tu\)

Then consider:

\( (1 - (1-u)^{t})^2 \leq 1 - (1-u)^{t}\leq tu\)

Yes, sounds like you're right.

There's a typo in the solution. It could be t(1-(1-u)^{t-1}) leq 0 or equivalently, what you said.

What is written sounds like a combination of the two :)

Thanks for reporting.

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