2. Sample Papers
  3. JEE Main & Advanced

JEE Main & Advanced Sample Paper JEE Main Sample Paper-26

  • question_answer
    For \[\lambda \in R\], let \[f(\lambda )=\] det \[(A-\lambda I)\] where \[A\left[ \begin{matrix}    1 & 2  \\    -1 & 3  \\ \end{matrix} \right]\]and I is an identity matrix of order 2. The minimum value of \[f(\lambda )\] is equal to

    A)  1                    

    B)  2

    C)  4                                

    D)  5

    Correct Answer: A

    Solution :

    \[|x|\,=\frac{-(m+6)\,\pm \,(m+2)}{2}\,(reject+sign)\] \[|x|=-(m+4)\] For two distinct solution, m + 4 < 0 \[\Rightarrow \,m<-4\] \[\Rightarrow \] Number of integral values of m in (-10, 10] are {-9, -8, -7, -6, -5} i.e., 5 values Alternating: Since equation has two distinct solution and therefore product of the roots must be less than zero. \[\Rightarrow \,2(m+4)<0\Rightarrow \,m<-4\]. \[f(\lambda )\,=\det \,(A-\lambda I)\,=\left| \begin{matrix}    1-\lambda  & 2  \\    -1 & 3-\lambda   \\ \end{matrix} \right|\,=(1-\lambda )\,(3-\lambda )\,+2\]\[={{\lambda }^{2}}-4\,\lambda +5={{(\lambda -2)}^{2}}+1\] Clearly, \[{{f}_{\min }}\,(\lambda =2)=1\]


You need to login to perform this action.
You will be redirected in 3 sec spinner