JEE Main 2020LogarithmsLogarithmMediumQuestionLet S={α:log2(92α−4+13)−log2(52. 32α−4+1)=2}S = \left\{ {\alpha :{{\log }_2}({9^{2\alpha - 4}} + 13) - {{\log }_2}\left( {{5 \over 2}.\,{3^{2\alpha - 4}} + 1} \right) = 2} \right\}S={α:log2(92α−4+13)−log2(25.32α−4+1)=2}. Then the maximum value of β\betaβ for which the equation x2−2(∑α∈sα)2x+∑α∈s(α+1)2β=0{x^2} - 2{\left( {\sum\limits_{\alpha \in s} \alpha } \right)^2}x + \sum\limits_{\alpha \in s} {{{(\alpha + 1)}^2}\beta = 0} x2−2(α∈s∑α)2x+α∈s∑(α+1)2β=0 has real roots, is ____________.Answer: 2Hide SolutionSolutionlog2(92α−4+13)−log2(52⋅32α−4+1)=2⇒92α−4+135232α−4+1=4⇒α=2 or 3∑α∈Sα=5 and ∑α∈S(α+1)2=25⇒x2−50x+25β=0 has real roots ⇒β≤25⇒βmax=25\begin{aligned} & \log _2\left(9^{2 \alpha-4}+13\right)-\log _2\left(\frac{5}{2} \cdot 3^{2 \alpha-4}+1\right)=2 \\\\ & \Rightarrow \frac{9^{2 \alpha-4}+13}{\frac{5}{2} 3^{2 \alpha-4}+1}=4 \\\\ & \Rightarrow \alpha=2 \quad \text { or } \quad 3 \\\\ & \sum_{\alpha \in \mathrm{S}} \alpha=5 \text { and } \sum_{\alpha \in \mathrm{S}}(\alpha+1)^2=25 \\\\ & \Rightarrow x^2-50 x+25 \beta=0 \text { has real roots } \\\\ & \Rightarrow \beta \leq 25 \\\\ & \Rightarrow \beta_{\max }=25 \end{aligned}log2(92α−4+13)−log2(25⋅32α−4+1)=2⇒2532α−4+192α−4+13=4⇒α=2 or 3α∈S∑α=5 and α∈S∑(α+1)2=25⇒x2−50x+25β=0 has real roots ⇒β≤25⇒βmax=25