JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $S = \{a : \log_2(9^{2\alpha-4} + 13) - \log_2\left(\frac{5}{2} \cdot 3^{2\alpha-4} + 1\right) = 2\}$ . Then the maximum value of $\beta$ for which the equation $x^2 - 2\left(\sum_{\alpha \in S} \alpha\right)^2 x + \sum_{\alpha \in S} (\alpha + 1)^2 \beta = 0$ has real roots, is

Choose the correct answer:

Explanation

Solution

1. Solve for $\alpha$ in set $S$

Let $3^{2\alpha-4} = t$ . Then $9^{2\alpha-4} = t^2$ . The equation becomes:

\log_2\left(\frac{t^2+13}{\frac{5}{2}t+1}\right) = 2 \implies \frac{t^2+13}{\frac{5t+2}{2}} = 2^2 = 4

t^2 + 13 = 2(5t + 2) \implies t^2 - 10t + 9 = 0

(t-9)(t-1) = 0 \implies t = 9 \text{ or } t = 1

If $3^{2\alpha-4} = 9 = 3^2 \implies 2\alpha-4 = 2 \implies \alpha = 3$
If $3^{2\alpha-4} = 1 = 3^0 \implies 2\alpha-4 = 0 \implies \alpha = 2$

So, $S = \{2, 3\}$ .

2. Calculate the Summations

$\sum_{\alpha \in S} \alpha = 2 + 3 = 5$
$\sum_{\alpha \in S} (\alpha+1)^2 = (2+1)^2 + (3+1)^2 = 3^2 + 4^2 = 9 + 16 = 25$

3. Set up the Quadratic Equation

Substitute the sums back into the original equation:

x^2 - 2(5)^2x + 25\beta = 0 \implies x^2 - 50x + 25\beta = 0

4. Find the Condition for Real Roots

For real roots, the discriminant $D$ must be $\geq 0$ :

D = (-50)^2 - 4(1)(25\beta) \geq 0

2500 - 100\beta \geq 0 \implies 100\beta \leq 2500 \implies \beta \leq 25

The maximum value of $\beta$ is 25.

Choose the correct answer:

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