Toán 10 $f^2(x)(|f(x)|-m-7)+(7m+6)|f(x)|-6m=0$ has 10 distinct roots

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3. The function $f(x)=-3x^2+6x+2$. How many integers $m$ are there, such that the equation $f^2(x)(|f(x)|-m-7)+(7m+6)|f(x)|-6m=0$ has 10 distinct roots.
4. The function $f(x)=ax^2+bx+c\, (a\neq 0)$ has the below graph. How many integers $m$ are there, such that the equation $f^2(|x|)+(m-2)f(|x|)+m-3=0$ has 6 distinct roots.

 

[7 1 2 5]

3. Let [TEX]|f(x)|=t \geq 0[/TEX].
The equation is equivalent to: [TEX]t^2(t-m-7)+(7m+6)t-6m=0 \Leftrightarrow (t-1)[t^2-(m+6)t+6m]=0 \Leftrightarrow (t-1)(t-6)(t-m)=0[/TEX]
Solve each equation [TEX]t=1[/TEX] and [TEX]t=6[/TEX] we have 6 distinct solutions.
So, the equation [TEX]t=m[/TEX] must has 4 distinct solutions, and [TEX]m \neq 6, m \neq 1[/TEX]
Considering the variable table:
[TEX] \begin{array}{c|ccccccccc} x & -\infty & & 1-\sqrt{\frac{5}{3}} & & 1 & & 1+\sqrt{\frac{5}{3}} & & +\infty \\ \hline y & +\infty & & & & 5 & & & & +\infty \\ & & \searrow & & \nearrow & & \searrow & & \nearrow & \\ & & & 0 & & & & 0 & & \end{array} [/TEX]
As we can see, when [TEX]0<m <5[/TEX] the equation [TEX]t=m[/TEX] has 4 distinct solutions. So [TEX]0<m<5, m \neq 1[/TEX]. The answer for the question is [TEX]3[/TEX].
4. Let [TEX]f(|x|)=t[/TEX]
The equation is equivalent to [TEX]t^2+(m-2)t+m-3=0 \Leftrightarrow (t-1)(t-m+3)=0 \Leftrightarrow t=1 \vee t=m-3[/TEX]
We can see [TEX]f(x)=1[/TEX] has 2 positive solutions, so [TEX]f(|x|)=1[/TEX] has 4 distinct solutions.
If the equation given has 6 distinct solutions, [TEX]f(|x|)=m-3[/TEX] must has 2 distinct solutions, which means [TEX]f(x)=m-3[/TEX] has 1 positive solution. From the function's graph, we infer that [TEX]m-3>3 \Rightarrow m>6[/TEX]
So the correct answer is D.