JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $A = 0 a 1 am p; 1 am p; 0 am p; c am p; 2 am p; 3 am p; 0$ , where $a, c \in R$ . If $A^{3} = A$ and the positive value of $a$ belongs to the interval $(n - 1, n]$ , where $n \in N$ , then $n$ is equal to ________.

Choose the correct answer:

Explanation

Given $A^{2} = A \times A$ :
$A^{2} = 0 a 1 am p; 1 am p; 0 am p; c am p; 2 am p; 3 am p; 0 \times 0 a 1 am p; 1 am p; 0 am p; c am p; 2 am p; 3 am p; 0 = a + 2 3 a c am p; 2 c am p; a + 3 c am p; 1 am p; 3 am p; 2 a am p; 2 + 3 c$
And $A^{3} = A^{2} \times A = A$ :
$2 a c + 3 a^{2} + 3 a c + 2 a a + 2 + 3 c am p; a + 2 + 3 c am p; 3 + 2 a c am p; a c + 2 c + 3 c^{2} am p; 2 a + 4 + 6 c am p; 6 + 3 a + 9 c am p; 2 a c + 3 = 0 a 1 am p; 1 am p; 0 am p; c am p; 2 am p; 3 am p; 0$
On comparing the corresponding elements:
$2 a c + 3 = 0 ⟹ c = \frac{- 3}{2 a} \dots (i)$
$a + 2 + 3 c = 1 ⟹ a + 3 c = - 1 \dots (ii)$
Substitute $(i)$ in $(ii)$ :
$a + 3 (\frac{- 3}{2 a}) = - 1 ⟹ 2 a^{2} - 9 = - 2 a ⟹ 2 a^{2} + 2 a - 9 = 0$
Using quadratic formula:
$a = \frac{- 2 \pm 4 + 4 \times 2 \times 9}{2 \times 2} = \frac{- 2 \pm 2 19}{4} = \frac{\pm 19 - 1}{2}$
Positive value of $a = \frac{19 - 1}{2} \approx \frac{4.35 - 1}{2} = 1.67$ .
Since $1.67 \in (1, 2]$ , the value of $n = 2$ .