Let $S=\left\{\alpha: \log _{2}\left(9^{2 \alpha-4}+13\right)-\log _{2}\left(\frac{5}{3} \cdot 3^{2

Question

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

Professor PC Thomas Classes · Answer

(25) : Given\ $\log _{2}\left(9^{2 \alpha-4}+13ight)-\log _{2}\left(\frac{5}{2} \cdot 3^{2 \alpha-4}+1ight)=2$\ $\Rightarrow \log _{2}\left(\frac{9^{2 \alpha-4}+13}{\frac{5}{2} \cdot 3^{2 \alpha-4}+1}ight)=2 \Rightarrow \frac{9^{2 \alpha-4}+13}{\frac{5}{2} \cdot 3^{2 \alpha-4}+1}=4$\ $\Rightarrow \quad 9^{2 \alpha-4}+13=10 \cdot 3^{2 \alpha-4}+4$\ $\Rightarrow \quad 10 \cdot 3^{2 \alpha-4}-9^{2 \alpha-4}=9$\ Put $\alpha=1,10 \cdot 3^{-2}-9^{-2}$\ $=\frac{10}{9}-\frac{1}{81}=\frac{89}{81} 
eq 9$ not satisfy\ Put $\alpha=2,10 \cdot 3^{0}-9^{0}=10-1=9$ satisfy.\ Put $\alpha=3,10 \cdot 3^{2}-9^{2}=90-81=9$ satisfy\ Hence, $\alpha=2$, or 3\ Now, $$ \begin{aligned} & \begin{aligned} \sum_{\alpha \in s} \alpha & =2+3=5 	ext { and } \sum_{\alpha \in s}(\alpha+1)^{2} \ & =\sum_{\alpha \in s} \alpha^{2}+2 \sum_{\alpha \in s} \alpha+\sum_{\alpha \in s} 1 \end{aligned} \ & \quad=4+9+2(5)+2=25 \ & 	ext { So, } x^{2}-2(5)^{2} x+25 \beta=0=x^{2}-50 x+25 \beta=0 \end{aligned} $$ For real roots, $D \geq 0$\ $\Rightarrow \quad 2500-4(25 \beta) \geq 0$\ $\Rightarrow 100 \beta \leq 2500 \Rightarrow \beta \leq 25$\ Hence, $\beta_{\max }=25$\