question_answer

If both the roots of the quadratic equation \[{{x}^{2}}-mx+4=0\] are real and distinct and they lie in the interval \[\left[ 1,\text{ }5 \right]\], then m lies in the interval: [JEE Main Online Paper (Held On 09-Jan-2019 Evening]

A) (4, 5)                

B)               (\[-5,\,\,-4\])

C)               (3, 4)                            

D)               (5, 6)

Correct Answer: A

Solution :

\[{{x}^{2}}-mx+4=0~\] Case-I: \[D>0\] \[{{m}^{2}}-16>0\] \[\Rightarrow \,\,\,\,m\,\,\in \text{ }\left( -\infty ,\,-4 \right),\text{ }\left( 4,\text{ }\infty  \right)\] Case-II: \[\Rightarrow \,\,\,1<\frac{-b}{2a}\,\,<\,\,5\] \[\Rightarrow \,\,\,1<\frac{m}{2}\,\,<\,\,5\,\,\,\,\Rightarrow \,\,m\in \,(2,\,\,10)\] Case-III: \[f\left( 1 \right)\,\,>\,\,0~~~~~~and\text{ }f\left( 5 \right)>0\] \[1-m+4>0\,\,\,\,\text{ }and\text{ }25-5m+4>0\] \[m<5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,and\,\,m<\frac{29}{5}\] Case-IV: Let one root is \[x=1\] \[1-m+4=0\] \[m=5\] Now equation \[{{x}^{2}}-5x+4=0\]                         \[\left( x-1 \right)\left( x-4 \right)=0\] \[x=1\text{ }i.e.\text{ }m=5\] is also included hence \[m\,\,\in \,\,\left( 4,\text{ }5 \right]\] So given option is \[\left( 4,\text{ }5 \right)\]