The highest positive integral value of $\lambda$ so that the equation $x^2-5|x|+6-|\lambda-2|=0 \qua

Question

The highest positive integral value of $\lambda$ so that the equation $x^2-5|x|+6-|\lambda-2|=0 \quad$ has four distinct roots is $\_\_\_\_$ .

Professor PC Thomas Classes · Answer

in given equation $\boldsymbol{x}^{-}-\boldsymbol{\partial}|\boldsymbol{x}|+\mathbf{0}=|\boldsymbol{\lambda}-\boldsymbol{z}|$ will nave four distinct roots it the above equation in $|\boldsymbol{x}|$ nas two distinct positive roots. I nis implies $|\boldsymbol{\lambda}-\boldsymbol{z}|<\mathbf{0}$
$$
\Rightarrow-6<\lambda-2<6 \Rightarrow-4<\lambda<8 .
$$
∴ the highest positive integral value of $\boldsymbol{\lambda}$ is 7 .