Mathematics Quesion Review (2025-11-05)

學術
#1 廢老終結者
05/11/25 02:06

Question 1:

Given b≥x≥a , show that:

1/(x - a)^2 + 1/(b-x)^2 ≥ 8/(b-a)^2

#2 廢老終結者
05/11/25 02:07

Solution:

Consider AM ≥ GM

LHS will be the least when x=(a+b)/2

QED

#3 廢老終結者
05/11/25 02:10

Question 2:

Consider f'(x) = tan(x)/x for π/2≥x≥0,

show that f(x) ≥ π/4.

#4 廢老終結者
05/11/25 02:13

Solution:

tan(x) ≥ x ≥ sin(x) ≥ 0

So tan(x)/x ≥ 1

Integrating both side from 0 to π/4,

we obtain the desired result.

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