JEE MAIN - Mathematics (2023 - 8th April Evening Shift - No. 16)
Let m and $$\mathrm{n}$$ be the numbers of real roots of the quadratic equations $$x^{2}-12 x+[x]+31=0$$ and $$x^{2}-5|x+2|-4=0$$ respectively, where $$[x]$$ denotes the greatest integer $$\leq x$$. Then $$\mathrm{m}^{2}+\mathrm{mn}+\mathrm{n}^{2}$$ is equal to __________.
Answer
9
Explanation
The givne eqn is : $x^2-12 x+[x]+31=0$
$$ \begin{aligned} & \Rightarrow\{x\}-x=x^2-12 x+31 \\\\ & \Rightarrow\{x\}=x^2-11 x+31 \end{aligned} $$
So, $0 \leq x^2-11 x+31<1$
$$ \begin{aligned} & \Rightarrow x^2-11 x+30 \leq 0 \\\\ & \Rightarrow(x-5)(x-6)<0 \\\\ & \Rightarrow x \in(5,6) \\\\ & \therefore[x]=5 \end{aligned} $$
$$ \begin{aligned} & \therefore x^2-12 x+5+31=0 \\\\ & \Rightarrow x^2-12 x+36=0 \\\\ & \Rightarrow(x-6)^2=0 \Rightarrow x=6 \end{aligned} $$
Hence, $x \in \phi$
$$ (\because x \in(5,6)) $$
$$ \therefore m=0 $$
Another equation is $x^2-5[x+2]-4=0$
Case I : $x \geq-2$
$$ x^2-5 x-14=0 \Rightarrow x=7,-2 $$
Case II : $x<-2$
$$ \begin{aligned} & x^2+5 x+6=0 \Rightarrow x=-3-2 \\\\ & \therefore x \in\{-3,-2,7\} \end{aligned} $$
$$ \therefore n=3 $$
Hence, $m^2+m x+n^2=0+0+9=9$
$$ \begin{aligned} & \Rightarrow\{x\}-x=x^2-12 x+31 \\\\ & \Rightarrow\{x\}=x^2-11 x+31 \end{aligned} $$
So, $0 \leq x^2-11 x+31<1$
$$ \begin{aligned} & \Rightarrow x^2-11 x+30 \leq 0 \\\\ & \Rightarrow(x-5)(x-6)<0 \\\\ & \Rightarrow x \in(5,6) \\\\ & \therefore[x]=5 \end{aligned} $$
$$ \begin{aligned} & \therefore x^2-12 x+5+31=0 \\\\ & \Rightarrow x^2-12 x+36=0 \\\\ & \Rightarrow(x-6)^2=0 \Rightarrow x=6 \end{aligned} $$
Hence, $x \in \phi$
$$ (\because x \in(5,6)) $$
$$ \therefore m=0 $$
Another equation is $x^2-5[x+2]-4=0$
Case I : $x \geq-2$
$$ x^2-5 x-14=0 \Rightarrow x=7,-2 $$
Case II : $x<-2$
$$ \begin{aligned} & x^2+5 x+6=0 \Rightarrow x=-3-2 \\\\ & \therefore x \in\{-3,-2,7\} \end{aligned} $$
$$ \therefore n=3 $$
Hence, $m^2+m x+n^2=0+0+9=9$
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