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QJEEM23Ma943f: JEE Main 2023 Mathematics - Class 11 JEEM23Ma943f Solution: Let m and be the numbers of real roots of the quadratic…

Let m and n\mathrm{n} be the numbers of real roots of the quadratic equations x2−12x+[x]+31=0x^{2}-12 x+[x]+31=0 and x2−5∣x+2∣−4=0x^{2}-5|x+2|-4=0 respectively, where [x][x] denotes the greatest integer ≤x\leq x . Then m2+mn+n2\mathrm{m}^{2}+\mathrm{mn}+\mathrm{n}^{2} is equal to __________.

Step-by-Step Verified Solution

The givne eqn is : x2−12x+[x]+31=0x^2-12 x+[x]+31=0

amp;⇒{x}−x=x2−12x+31amp;⇒{x}=x2−11x+31\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

amp;⇒x2−11x+30≤0amp;⇒(x−5)(x−6)lt;0amp;⇒x∈(5,6)amp;∴[x]=5\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}

amp;∴x2−12x+5+31=0amp;⇒x2−12x+36=0amp;⇒(x−6)2=0⇒x=6\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∈ϕx \in \phi

(∵x∈(5,6))(\because x \in(5,6))

∴m=0\therefore m=0

Another equation is x2−5[x+2]−4=0x^2-5[x+2]-4=0

Case I : x≥−2x \geq-2

x2−5x−14=0⇒x=7,−2x^2-5 x-14=0 \Rightarrow x=7,-2

Case II : x<-2

amp;x2+5x+6=0⇒x=−3−2amp;∴x∈{−3,−2,7}\begin{aligned}& x^2+5 x+6=0 \Rightarrow x=-3-2 \\\\& \therefore x \in\{-3,-2,7\}\end{aligned}

∴n=3\therefore n=3

Hence, m2+mx+n2=0+0+9=9m^2+m x+n^2=0+0+9=9

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