Câu 10: \(f\left(x\right)=x^4-12x^2-4\)
=>f'(x)=\(4x^3-12\cdot2x=4x^3-24x=4x\left(x^2-6\right)\)
Đặt f'(x)=0
=>\(x\left(x^2-6\right)=0\)
=>\(\left[\begin{array}{l}x=0\\ x^2-6=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x^2=6\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\left(nhận\right)\\ x=\sqrt6\left(nhận\right)\\ x=-\sqrt6\left(loại\right)\end{array}\right.\)
\(f\left(0\right)=0^4-12\cdot0^2-4=-4\)
\(f\left(\sqrt6\right)=\left(\sqrt6\right)^4-12\cdot\left(\sqrt6\right)^2-4=36-12\cdot6-4=-40\)
\(f\left(9\right)=9^4-12\cdot9^2-4=5585\)
=>Chọn B
Câu 9: \(f\left(x\right)=x^4-10x^2-4\)
=>f'(x)=\(4x^3-10\cdot2x=4x^3-20x=4x\left(x^2-5\right)\)
Đặt f'(x)=0
=>\(x\left(x^2-5\right)=0\)
=>\(\left[\begin{array}{l}x=0\\ x^2-5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x^2=5\end{array}\right.=>\left[\begin{array}{l}x=0\left(nhận\right)\\ x=\sqrt5\left(nhận\right)\\ x=-\sqrt5\left(loại\right)\end{array}\right.\)
\(f\left(0\right)=0^4-10\cdot0^2-4=-4\)
\(f\left(\sqrt5\right)=\left(\sqrt5\right)^4-10\cdot\left(\sqrt5\right)^2-4=25-10\cdot5-4=25-50-4=21-50=-29\)
\(f\left(9\right)=9^4-10\cdot9^2-4=5747\)
=>Chọn D
Câu 8: \(f\left(x\right)=x^3-21x\)
=>f'(x)=\(3x^2-21\) =3(x^2-7)
Đặt f'(x)=0
=>\(x^2-7=0\)
=>\(x^2=7\)
=>\(\left[\begin{array}{l}x=\sqrt7\left(nhận\right)\\ x=-\sqrt7\left(loại\right)\end{array}\right.\)
\(f\left(2\right)=2^3-21\cdot2=8-42=-34\)
\(f\left(19\right)=19^3-21\cdot19=6460\)
\(f\left(\sqrt7\right)^2=\left(\sqrt7\right)^3-21\cdot\sqrt7=7\sqrt7-21\sqrt7=-14\sqrt7\)
=>Chọn B
Câu 7: \(f\left(x\right)=x^3-24x\)
=>f'(x)=\(3x^2-24=3\left(x^2-8\right)\)
Đặt f'(x)=0
=>\(x^2-8=0\)
=>\(x^2=8\)
=>\(\left[\begin{array}{l}x=2\sqrt2\left(nhận\right)\\ x=-2\sqrt2\left(loại\right)\end{array}\right.\)
\(f\left(2\right)=2^3-24\cdot2=8-48=-40\)
\(f\left(2\sqrt2\right)=\left(2\sqrt2\right)^3-24\cdot2\sqrt2=8\cdot\sqrt8-48\sqrt2=16\sqrt2-48\sqrt2=-32\sqrt2\)
\(f\left(19\right)=19^3-24\cdot19=6403\)
=>Chọn C
Câu 1: C
Câu 2: B
Câu 3: A
Câu 4: B
Câu 5: C
Câu 6: C
