Let S = { alpha : log_2(9^{2alpha-4} + 13) - | vedclass

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(B) Given $\log_2(9^{2\alpha-4} + 13) - \log_2(\frac{5}{2} \cdot 3^{2\alpha-4} + 1) = 2$. Let $y = 3^{2\alpha-4}$. Then $9^{2\alpha-4} = y^2$. The equ

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(B) Given $\log_2(9^{2\alpha-4} + 13) - \log_2(\frac{5}{2} \cdot 3^{2\alpha-4} + 1) = 2$. Let $y = 3^{2\alpha-4}$. Then $9^{2\alpha-4} = y^2$. The equation becomes $\log_2(\frac{y^2 + 13}{\frac{5}{2}y + 1}) = 2$. $\frac{y^2 + 13}{\frac{5}{2}y + 1} = 4 \implies y^2 + 13 = 10y + 4$. $y^2 - 10y + 9 = 0 \implies (y-1)(y-9) = 0$. So $y = 1$ or $y = 9$. If $3^{2\alpha-4} = 1$,then $2\alpha-4 = 0 \implies \alpha = 2$. If $3^{2\alpha-4} = 9$,then $2\alpha-4 = 2 \implies \alpha = 3$. Thus,$S = \{2, 3\}$. $\sum_{\alpha \in S} \alpha = 2 + 3 = 5$. $\sum_{\alpha \in S} (\alpha+1)^2 = (2+1)^2 + (3+1)^2 = 9 + 16 = 25$. The quadratic equation is $x^2 - 2(5)^2 x + 25\beta = 0$,which is $x^2 - 50x + 25\beta = 0$. For real roots,the discriminant $D \geq 0$. $D = (-50)^2 - 4(1)(25\beta) = 2500 - 100\beta \geq 0$. $100\beta \leq 2500 \implies \beta \leq 25$. The maximum value of $\beta$ is $25$.

List-$I$	List-$II$
$(i)$ $\alpha = \beta$	$(A)$ $(ac^2)^{1/3} + (a^2c)^{1/3} + b = 0$
$(ii)$ $\alpha = 2\beta$	$(B)$ $2b^2 = 9ac$
$(iii)$ $\alpha = 3\beta$	$(C)$ $b^2 = 6ac$
$(iv)$ $\alpha = \beta^2$	$(D)$ $3b^2 = 16ac$
	$(E)$ $b^2 = 4ac$
	$(F)$ $(ac^2)^{1/3} + (a^2c)^{1/3} = b$

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

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