又有行家突個龜頭出嚟

睇嚟大陸學制嘅吸引力已經超越DSE

一睇就知係2023最後一屆大陸高考理科數學試題

2024新高考連Linear Programming 都唔駛考

新不如舊

BTW, 試下用DSE程度數學做上述題目

提示:

Solution

We try to compare 2 sides:

LHS: x (e^x -1)

RHS: 1+ ln(x)(x+1) - 2x

Given x>0, then x^2 >0

Divide both sides by x^2 won't affect the sign of both LHS and RHS

Now LHS= (e^x -1)/x

RHS = (1+ ln(x)(x+1) - 2x)/x^2

Take limit on both side as x--> 0+

Then LHS= 1

Since deg(ln(x))=0, deg(RHS) <0

Now RHS = -∞ without using L'Hospital Rule

Clearly, LHS > RHS

QED

類似問題成日喺度翻炒:

Given f(x) = e^x/ax - b and g(x) = ax/ln(x) - b.

Both functions have the same minmum.

(a) Compute a.

(b) Prove that if f(x) and g(x) have 4 roots x_n,

where x_1 < x_2 < x_3 < x_4,

then x_1‧ x_4 = x_2 ‧ x_3

今晚放假, 順手寫埋答案

Solution:

(a)

f'(x) =[e^x (x-1)]/ax^2

and g'(x) = [a (ln(x)-1)]/(ln(x))^2

f'(1) = 0

and g'(e) = 0

Now f(1) = e/a - b

and g(e) = ae - b

Note that a is a leading coefficient.

Now e/a = ae

Then a = 1 or -1

By noticing the domains of both functions, and using the property of rectangular hyperbola (y=1/x),

we can deduce the convexity without differentiating f'(x) and g'(x) or discussing monotonicity within the specified domains.

Since e^x/x is always convex for x>0, i.e. f"(1) >0,

and lnx/x is always convex for x>1, i.e. g"(e) > 0,

then a >0.

If a < 0 , only local maximum, not local minimum as given will be computed. So we will reject a = -1.

QED

(b)

Now, since the domain of g(x) >1 and that of f(x) >0, geometrically the roots of f(x) are x_1 and x_3, where the roots of g(x) are x_2 and x_4, because f(x)=g(x) has exactly 1 solution:

e^x/x = x/lnx

(e^x) (lnx) = x^2

Sinec e^x > x and lnx < x,

Then 1<x<e

Now e^x_1/x_1 = x_4/ ln(x_4)

and e^x_3/x_3 = x_2/ ln(x_2)

summarized as (*)

Also, since f(x) = g(e^x), we have:

f(x_1) = g(e^(x_1)) = g(x_2)

--> x_2 = e^(x_1)

and

f(x_3) = g(e^(x_3)) = g(x_4)

--> x_4 = e^(x_3)

summarized as (**)

By putting (**) into (*), we have:

x_2/ x_1 = e^(x_1) / x_1

= x_4 / ln(x_4) = x_4 / x_3

Rearranging the 4 roots, the result is obvious.

QED

基本上DSE係模仿緊大陸高考

CE 座標幾何題一定全部畫哂圖俾考生睇，

但DSE開始成題一幅圖都無要考生自己畫

轉數稍為慢啲好易答唔切

就算答得哂，都因為有一大堆爛數要爆，

考生一唔少心犯manipulation mistakes 就死得

DSE嘅趨勢就係鳩計

Core惟一考天份嘅課題得Combinatorics

至於M1同M2, 鳩計之餘又要鬥快運算

基本上唔太需要數學思維，

因此，部分大陸高考題目就咁睇比AL PMath更難，

但實際上只是將考生打造成人肉計數機而已

RHS嘅進一步剖析：

留意 (1+lnx (x+1))/x^2

ln(0+) = -∞

For any finite a and b,

a ln(0) + b = -∞

喺RHS, a=1=b

個分母x^2永遠係正數，不影響分子嘅正負值

Also, 留意 -2x/x^2

-1/(0+) = 1/(0-) = -∞

-∞/(0+)^2

= +∞ / -(0+)^2

Notice a/(-x^2) for a>0:

Consider -1/x^2,

when x-> 0+, -∞ will be guaranteed

Then consider a=+∞,

We know that 正負得負,

So (+∞)(-∞)=-∞

Note that LHS = d(e^x)/dx = e^x

Just recall the first principle.

Then we know that e^x is always convex and increasing from x=0.

From RHS, there is a local maxima (a, b)

Just prove that b < e^a.

Consider RHS, it is a concave function for x>0.

When x=1, RHS = -1

When x=e, RHS = 2-e> -1

Since RHS is still increasing for x>1,

we argue that a > 1.

So, ln(x) < x < x^2 for x >1

We now argue (1+ ln(x)(x+1) - 2x) < x^2

such that b = RHS <1 = e^x for all x >0

Now 1+ ln(x)(x+1) - 2x

< 1+x(x+1)-2x

< x^2 - x + 1 < x^2 -1 + 1 < x^2

as x >1

Hence, RHS <1

QED