326:
\(f\left(x\right)=x^2+\left(m+1\right)x+2m+7\)
\(\Delta=\left(m+1\right)^2-4\cdot1\cdot\left(2m+7\right)\)
\(=m^2+2m+1-8m-28\)
\(=m^2-6m-27\)
\(=m^2-9m+3m-27=\left(m-9\right)\left(m+3\right)\)
Để f(x)>0 với mọi x thì \(\left\{{}\begin{matrix}\Delta< 0\\a>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(m-9\right)\left(m+3\right)< 0\\1>0\left(đúng\right)\end{matrix}\right.\)
=>(m-9)(m+3)<0
TH1: \(\left\{{}\begin{matrix}m-9>0\\m+3< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m>9\\m< -3\end{matrix}\right.\)
=>Loại
TH2: \(\left\{{}\begin{matrix}m-9< 0\\m+3>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m< 9\\m>-3\end{matrix}\right.\)
=>-3<m<9
=>Chọn A
Câu 327:
\(f\left(x\right)=mx^2-12x-5\)
TH1: m=0
\(f\left(x\right)=0\cdot x^2-12x-5=-12x-5\)
f(x)=-12x-5 không thể luôn dương với mọi x được
=>Loại
TH2: m<>0
\(f\left(x\right)=mx^2-12x-5\)
\(\Delta=\left(-12\right)^2-4\cdot m\cdot\left(-5\right)=144+20m\)
Để f(x)>0 với mọi x thì \(\left\{{}\begin{matrix}\Delta< 0\\m>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}20m+144< 0\\m>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}20m< -144\\m>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m< -7,2\\m>0\end{matrix}\right.\Leftrightarrow m\in\varnothing\)
=>Chọn C
Câu 328:
\(f\left(x\right)=3x^2-6x-m-2\)
\(\Delta=\left(-6\right)^2-4\cdot3\cdot\left(-m-2\right)\)
\(=36+12\left(m+2\right)\)
\(=36+12m+24=12m+60\)
Để f(x) luôn dương thì \(\left\{{}\begin{matrix}\Delta< 0\\a>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3>0\left(đúng\right)\\12m+60< 0\end{matrix}\right.\)
=>12m+60<0
=>12(m+5)<0
=>m+5<0
=>m<-5
=>Chọn A
Câu 329:
\(f\left(x\right)=x^2-\left(m+2\right)x+8m+1\)
\(\Delta=\left(-m-2\right)^2-4\left(8m+1\right)\)
\(=m^2+4m+4-32m-4\)
\(=m^2-28m=m\left(m-28\right)\)
Để f(x) luôn dương thì \(\left\{{}\begin{matrix}\Delta< 0\\a>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}1>0\left(đúng\right)\\m\left(m-28\right)< 0\end{matrix}\right.\Leftrightarrow m\left(m-28\right)< 0\)
TH1: \(\left\{{}\begin{matrix}m>0\\m-28< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m>0\\m< 28\end{matrix}\right.\)
=>0<m<28
TH2: \(\left\{{}\begin{matrix}m< 0\\m-28>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m< 0\\m>28\end{matrix}\right.\)
=>Loại
=>Chọn C
Câu 330:
\(f\left(x\right)=-x^2+4\left(m+1\right)x+1-m^2\)
\(\Delta=\left[4\left(m+1\right)\right]^2-4\cdot\left(-1\right)\left(1-m^2\right)\)
\(=\left(4m+4\right)^2+4\left(1-m^2\right)\)
\(=16m^2+32m+16+4-4m^2\)
\(=12m^2+32m+20\)
\(=4\left(3m^2+8m+5\right)\)
\(=4\left(m+1\right)\left(3m+5\right)\)
Để \(f\left(x\right)< 0\) với mọi x thì \(\left\{{}\begin{matrix}\Delta< 0\\a< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4\left(m+1\right)\left(3m+5\right)< 0\\-1< 0\left(đúng\right)\end{matrix}\right.\)
=>\(4\left(m+1\right)\left(3m+5\right)< 0\)
=>\(\left(m+1\right)\left(3m+5\right)< 0\)
TH1: \(\left\{{}\begin{matrix}m+1>0\\3m+5< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m>-1\\3m< -5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m>-1\\m< -\dfrac{5}{3}\end{matrix}\right.\)
=>Loại
TH2: \(\left\{{}\begin{matrix}m+1< 0\\3m+5>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m< -1\\m>-\dfrac{5}{3}\end{matrix}\right.\)
=>\(-1>m>-\dfrac{5}{3}\)
=>Chọn A
