Cho hàm số y = x3 + 3x2 - 2m + 1. Gọi S là tập hợp tất cả các giá trị nguyên của tham số m thuộc đoạn [–30;30] sao cho . Số phần tử của S là

A. 55

B. 56

C. 61

D. 57

Trả lời

Xét hàm số $\text{y}=\text{f}\left(\text{x}\right)={\text{x}}^3+3{\text{x}}^2-2 \text{m}+1$ trên đoạn [1;3].

Ta có: ${\text{f}}^{\text{'}}\left(\text{x}\right)=3{\text{x}}^2+6\text{x}>0,\forall \text{x}\in [1;3]$ Suy ra $\left\{\begin{array}{l}{\min }_{[1;3]}\text{f}\left(\text{x}\right)=\text{f}\left(1\right)=5-2 \text{m}\\ {\max }_{[1;3]}\text{f}\left(\text{x}\right)=\text{f}\left(3\right)=55-2 \text{m}\end{array}\right.$

Trường hợp 1: $(5-2 \text{m})(55-2 \text{m})>0\iff \left[\begin{array}{l}\text{m}<\frac{5}{2}\\ \text{m}>\frac{55}{2}\end{array}\right.$

Khi đó ${\max }_{[1;3]}\left|f\right(x\left)\right|+{\min }_{[1;3]}\left|f\right(x\left)\right|\ge 10\iff |5-2m|+|55-2 \text{m}|\ge 10$(1) .

Nếu $\text{m}<\frac{5}{2}$ ta có $\left(1\right)\iff 5-2 \text{m}+55-2 \text{m}\ge 10\iff \text{m}\le \frac{50}{4}\Rightarrow \text{m}\in \{-30;-29;\dots \dots \mathrm{..2}\}$

Nếu $\text{m}>\frac{55}{2}$ ta có $\left(1\right)\iff 2 \text{m}-5+2 \text{m}-55\ge 10\iff \text{m}\ge \frac{70}{4}\Rightarrow \text{m}\in \{28;29;30\}$

Trường hợp 2: $(5-2 \text{m})(55-2 \text{m})\le 0\iff \frac{5}{2}\le \text{m}\le \frac{55}{2}$ (*)

Khi đó ${\min }_{[1;3]}\left|f\right(x\left)\right|=0$ và ${\max }_{[1;3]}\left|f\right(x\left)\right|+{\min }_{[1;3]}\left|f\right(x\left)\right|\ge 10\iff {\max }_{[1;3]}\left|f\right(x\left)\right|\ge 10$ (2)

$\iff \left[\begin{array}{l}\left\{\begin{array}{l}|5-2m|\ge |55-2m|\\ |5-2m|\ge 10\end{array}\right.\\ \left\{\begin{array}{l}|55-2m|\ge |5-2m|\\ |55-2m|\ge 10\end{array}\right.\end{array}\right.\iff \left[\begin{array}{l}\left\{\begin{array}{l}25-20m+4m^2\ge 3025-220m+4m^2\\ m\ge \frac{15}{2}\\ m\le -\frac{5}{2}\end{array}\right.\\ \left\{\begin{array}{l}3025-220m+4m^2\ge 25-20m+4m^2\\ m\ge \frac{65}{2}\\ m\le \frac{45}{2}\end{array}\right.\end{array}\right.\iff \left[\begin{array}{l}\left\{\begin{array}{l}m\ge 15\\ m\ge \frac{15}{2}\\ m\le -\frac{5}{2}\end{array}\right.\\ \left\{\begin{array}{l}\text{m}\le 15\\ \text{m}\ge \frac{65}{2}\\ \text{m}\le \frac{45}{2}\end{array}\right.\end{array}\right.$

$\iff \left[\begin{array}{l}\text{m}\ge 15\\ \text{m}\le 15\end{array}\iff \text{m}\in ℝ\Rightarrow \text{m}\in \{3;4;\dots ;27\}\right.$. Vậy có 61 số m thỏa yêu cầu bài toán.