Question

已知函數(shù)f(x)=mx³-x的圖象上,以N(1,n)為切點的切線的傾斜角為.

(1)求m、n的值;

(2)是否存在最小的正整數(shù)k,使得不等式f(x)≤k-1 991對于x∈［-1,3］恒成立?如果存在,請求出最小的正整數(shù)k;如果不存在,請說明理由;

(3)求證:|f(Sinx)+f(coSx)|≤2f(t+)(x∈R,t＞0).

Answer 1

解:(1)f′(x)=3Mx²-1,依題意,得tan=f′(1),即1=3M-1,M=.?

∴f(x)=x³-x.把N(1,n)代入,得n=f(1)=-.?

∴M=,n=-.                                                                                                      

(2)令f′(x)=2x²-1=0,得x=±,?

當(dāng)-1＜x＜-時,f′(x)=2x²-1＞0;?

當(dāng)-＜x＜時,f′(x)=2x²-1＜0;?

當(dāng)＜x＜3時,f′(x)=2x²-1＞0.?

又f(-1)=,f(-)=,f()=-,f(3)=15,?

因此,當(dāng)x∈［-1,3］時,-≤f(x)≤15.                                                               ?

要使得不等式f(x)≤k-1 991對于x∈［-1,3］恒成立,則k≥15+1 991=2 006.  ?

∴存在最小的正整數(shù)k=2 006,使得不等式f(x)≤k-1 991對于x∈［-1,3］恒成立.?

(3)證法一:|f(sinx)+f(cosx)|?

=|(sin³x-sinx)+(cos³x-cosx)|?

=|(sin³x+cos³x)-(sinx+cosx)|?

=|(sinx+cosx)［(sin²x-sinxcosx+cos²x)-1］|?

=|sinx+cosx|·|-sinxcosx-|?

=|sinx+cosx|³?

=|2sin(x+)|³≤.                                                                                  

又∵t＞0,

∴t+≥,t²+≥1.?

∴2f(t+)=2［(t+)³-(t+)］=2(t+)［(t²+1+)-1］?

=2(t+)［(t²+)-］≥2(-)=.?

綜上,可得|f(sinx)+f(cosx)|≤2f(t+)(x∈R,t＞0).                                      ?

證法二:由(2)知函數(shù)f(x)在［-1,-］上是增函數(shù);在［-,］上是減函數(shù);在［,

1］上是增函數(shù).?

又f(-1)= ,f(-)=,f()=-,f(1)=-,?

∴當(dāng)x∈［-1,1］時,-≤f(x)≤,?

即|f(x)|≤.?

∵sinx,cosx∈［-1,1］,?

∴|f(sinx)|≤,|f(cosx)|≤.?

∴|f(sinx)+f(cosx)|≤|f(sinx)|+ |f(cosx)|≤+=.                            ?

又∵t＞0,∴T+≥＞1,且函數(shù)f(x)在［1,+∞)上是增函數(shù).?

∴2f(t+)≥2f()=2［()³-2］=.?

綜上,可得|f(sinx)+f(cosx)|≤2f(t+)(x∈R,t＞0).