JEE Main 2023 — Mathematics Algebra

Given,

$${\mathrm{log}}_{2}({9}^{2\alpha -4}+13)-{\mathrm{log}}_{2}({3}^{2\alpha -4}\cdot \frac{5}{2}+1)=2$$log2​(92α−4+13)−log2​(32α−4⋅25​+1)=2

Now let ${3}^{2\alpha -4}=t$32α−4=t, so the equation becomes,

$${\mathrm{log}}_{2}({t}^{2}+13)-{\mathrm{log}}_{2}(\frac{5t}{2}+1)=2$$log2​(t2+13)−log2​(25t​+1)=2

$$\Rightarrow {\mathrm{log}}_{2}\frac{({t}^{2}+13)}{(\frac{5t}{2}+1)}=2$$⇒log2​(25t​+1)(t2+13)​=2

$$\Rightarrow \frac{({t}^{2}+13)}{(\frac{5t}{2}+1)}={2}^{2}$$⇒(25t​+1)(t2+13)​=22

$$\Rightarrow {t}^{2}+13=10t+4$$⇒t2+13=10t+4

$$\Rightarrow {t}^{2}-10t+9=0$$⇒t2−10t+9=0

$$\Rightarrow t=1\text{or}9$$⇒t=1or9

So,

$${3}^{2\alpha -4}=1\text{or}9$$32α−4=1or9

$$\Rightarrow {3}^{2\alpha -4}={3}^{0}\text{or}{3}^{2}$$⇒32α−4=30or32

$$\Rightarrow 2\alpha -4=0\mathrm{or}2$$⇒2α−4=0or2

$$\Rightarrow \alpha =2,3$$⇒α=2,3

Now,

$${x}^{2}-2{(\underset{\alpha \in s}{\sum }\alpha )}^{2}x+\underset{a\in s}{\sum }{(\alpha +1)}^{2}\beta =0$$x2−2(α∈s∑​α)2x+a∈s∑​(α+1)2β=0

$$\Rightarrow {x}^{2}-2({(2+3)}^{2}x)+({3}^{2}+{4}^{2})\beta =0$$⇒x2−2((2+3)2x)+(32+42)β=0

$$\Rightarrow {x}^{2}-50x+25\beta =0$$⇒x2−50x+25β=0

Now for real roots

$$D\geq 0$$D≥0

$$\Rightarrow {50}^{2}-4\times 25\beta \geq 0$$⇒502−4×25β≥0

$$\Rightarrow 50-2\beta \geq 0$$⇒50−2β≥0

$$\Rightarrow \beta \leq 25$$⇒β≤25

So, maximum value of $\beta$β is $25$25.