DTD Mathematics Challenge (2025-12-07)

Solution:

1. Using integration by part,

∫ ln(x) = xln(x) - x

Since the definite integral is improper,

we rewrite the result as below:

lim[a-->0] -1 - aln(a) + a

Note that:

aln(a) = ln(a) / (1/a)

By L'Hospital's Rule,

(1/a)/(-1/a^2) = - a

Now taking the limit,

we have the result -1.

Alternative solution:

The inverse of ln(x) is e^x.

Integration of e^x from -∞ to 0 = 1

Since the required area is above ln(x) from 0 to 1, we simply change the sign when reflecting the area above e^x, i.e. -1.

2. You can do it by MI,

but clearly: (x^m - x^-m)^2 ≥2

where m is any rational number > 0

e.g. (√x - 1/√x)^2 = x + 1/x - 2 ≥ 0

Now we simply take 2 cases:

Case 1： n is odd

Take n=5, the result is fucking obvious.

Case 2: n is even

On the left hand side, notice a term x^0=1

Q.E.D.

but clearly: (x^m - x^-m)^2 ≥ 0

i.e. x^(2m) + x^(-2m) ≥ 2

Footnote:

Real number is the subset of complex number, so we can define a variable z as a real number by stating that lm(z) = 0

For n to be odd, say 5,

you will see (x^(5/2) + x(-5/2))^2 = 0

Interestingly, x is any real number,

but in this case x is always non-zero,

because the domain of x^(n/2) is (0, ∞).

Believe it or not, plot x=y^2,

you will see the equivalent y=√x.

Remember that the square root of a negative number is always purely imaginary, i.e., lm(z) > 0 or lm(z) < 0, and thus non-real. Only real numbers can be compared by inequalities.

(x^(5/2) + x(-5/2))^2 ≥ 0

i.e. x^5 + 1/x^5 ≥ 2

Recall z = a+ bi

where a = Re(z) and b = Im(z)

Both a and b are real, but z not necessarily.

Take z =√(-5) = i√5 or -i√5

Since the function is multi-valued,

y is even not the function of x.

And for the above 2 values, Re(z) = 0.

In AP Precalculus,

we have horizontal line test to see if a mathematical relation satisfies the condition of a 2D bijective function y=f(x).

For every value of x, y can have only 1 value, otherwise the horizontal line will cut y at 2 distinct points and thus y is not bijective.

For example, y=x^2 is not bijective for x=/=0.

Google surjective and injective for more info.