JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Choose the correct answer:

Explanation

Solution

1. Factorize the equation:

x(x^6 + 3x^4 - 13x^2 - 15) = 0

Let $t = x^2$ :

x(t^3 + 3t^2 - 13t - 15) = 0

2. Find the roots of the cubic in $t$ :

By inspection, $t = 3$ is a root ( $27 + 27 - 39 - 15 = 0$ ).

Dividing $(t^3 + 3t^2 - 13t - 15)$ by $(t-3)$ :

(t-3)(t^2 + 6t + 5) = 0

(t-3)(t+1)(t+5) = 0

So, $t = 3, -1, -5$ .

3. Find the roots $x$ :

From $t=3$ , $x = \pm \sqrt{3}$
From $t=-1$ , $x = \pm i$
From $t=-5$ , $x = \pm \sqrt{5}i$
From the original $x$ factor, $x = 0$

4. Arrange by magnitude ( $|\alpha_i|$ ):

$|\pm \sqrt{5}i| = \sqrt{5} \approx 2.23$
$|\pm \sqrt{3}| = \sqrt{3} \approx 1.73$
$|\pm i| = 1$
$|0| = 0$

So: $\alpha_1, \alpha_2 = \sqrt{5}i, -\sqrt{5}i$ ; $\alpha_3, \alpha_4 = \sqrt{3}, -\sqrt{3}$ ; $\alpha_5, \alpha_6 = i, -i$ ; $\alpha_7 = 0$ .

5. Calculate the expression:

\alpha_1 \alpha_2 - \alpha_3 \alpha_4 + \alpha_5 \alpha_6

(\sqrt{5}i)(-\sqrt{5}i) - (\sqrt{3})(-\sqrt{3}) + (i)(-i)

(-5i^2) - (-3) + (-i^2)

5 + 3 + 1 = 9

Final Answer:

9

Choose the correct answer:

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