JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $m$ and 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 $⩽ x$ . Then m² + mn +n² is equal to _______.

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Explanation

The givne eqn is $: x^{2} - 12 x + [x] + 31 = 0$
$\Rightarrow {x} - x = x^{2} - 12 x + 31$
$\Rightarrow {x} = x^{2} - 11 x + 31$ So, 0 $\leq x^{2}$ - 11x+ 31< 1 $\Rightarrow x^{2} - 11 x + 30 \leq 0$ $\Rightarrow(x-5)\left(x-6\right)<0$ $\Rightarrow x \in (5, 6)$
$∴ [x] = 5$
$∴ x^{2} - 12 x + 5 + 31 = 0$
$\Rightarrow x^{2} - 12 x + 36 = 0$
$\Rightarrow (x - 6)^{2} = 0 \Rightarrow x = 6$
Hence, $x \in ϕ$
$∴ m = 0$
Another equation is $x^{2} - 5 [x + 2] - 4 = 0$
$Case I : x \geq - 2$
$x^{2} - 5 x - 14 = 0 \Rightarrow x = 7, - 2$
$C_{2}$ ase II $:x<-2$
$x^{2} + 5 x + 6 = 0 \Rightarrow x = - 3 - 2$
$∴ x \in {- 3, - 2, 7}$

n = 3
Hence, m² + mx + n² = 0 + 0 + 9 = 9

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