Cho hàm số f(x)=x^3-3x+m+2 . Có bao nhiêu số nguyên dương

Đáp án C

Đặt gx=x3−3x+2⇒max−1;3gx=20;min−1;3gx=0. Khi đó fx=m+gx.

Ta có: fa+fb>fc,∀a,b,c∈−1;3⇒m>gc−ga+gb,∀a,b,c∈−1;3⇒m>max−1;3gx−2min−1;3gx⇒m>20.

f2a+f2b>f2c,∀a,b,c∈−1;3⇒m+ga2+m+gb2>m+gc2,∀a,b,c∈−1;3⇒m2+2ga+gb−gcm+g2a+g2b−g2c>0,∀a,b,c∈−1;3⇒m+ga+gb−gc2−2gagb+2gagc+2gbgc−2g2c>0,∀a,b,c∈−1;3⇒m+ga+gb−gc2>2ga−gcgb−gc,∀a,b,c∈−1;3⇒m+ga+gb−gc2>2max−1;3gx−min−1;3gx2,∀a,b,c∈−1;3m>max−1;3gx+202−2min−1;3gx⇒m≥49.