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Logarithm question

2023 · 25 Jan · Shift 1 · Q43
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  5. /2023 · 25 Jan · Shift 1 · Q43

Logarithm question

2023 · 25 Jan · Shift 1 · Q43

JEE MainMathematicsLogarithmNumerical+4 / −1
Let S={α:log⁡2(92α−4+13)−log⁡2(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 ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 25

  1. Solve for the set SSS

We are given

log⁡2(92α−4+13)−log⁡2(52 32α−4+1)=2.\log_2(9^{2\alpha-4}+13)-\log_2\left(\frac52\,3^{2\alpha-4}+1\right)=2.log2​(92α−4+13)−log2​(25​32α−4+1)=2.

Using log properties,

log⁡2(92α−4+1352 32α−4+1)=2\log_2\left(\frac{9^{2\alpha-4}+13}{\frac52\,3^{2\alpha-4}+1}\right)=2log2​(25​32α−4+192α−4+13​)=2

which implies

92α−4+1352 32α−4+1=4.\frac{9^{2\alpha-4}+13}{\frac52\,3^{2\alpha-4}+1}=4.25​32α−4+192α−4+13​=4.

Now let

t=32α−4>0.t=3^{2\alpha-4} > 0.t=32α−4>0.

Then

92α−4=(32)2α−4=34α−8=(32α−4)2=t2.9^{2\alpha-4}=(3^2)^{2\alpha-4}=3^{4\alpha-8}=(3^{2\alpha-4})^2=t^2.92α−4=(32)2α−4=34α−8=(32α−4)2=t2.

So the equation becomes

t2+1352t+1=4.\frac{t^2+13}{\frac52 t+1}=4.25​t+1t2+13​=4.

Cross-multiplying:

t2+13=4(52t+1)=10t+4.t^2+13=4\left(\frac52 t+1\right)=10t+4.t2+13=4(25​t+1)=10t+4.

Hence

t2−10t+9=0.t^2-10t+9=0.t2−10t+9=0.

Factorizing,

(t−1)(t−9)=0.(t-1)(t-9)=0.(t−1)(t−9)=0.

So

t=1ort=9.t=1 \quad \text{or} \quad t=9.t=1ort=9.

Since t=32α−4t=3^{2\alpha-4}t=32α−4,

  • if 32α−4=13^{2\alpha-4}=132α−4=1, then 2α−4=0⇒α=22\alpha-4=0 \Rightarrow \alpha=22α−4=0⇒α=2,
  • if 32α−4=9=323^{2\alpha-4}=9=3^232α−4=9=32, then 2α−4=2⇒α=32\alpha-4=2 \Rightarrow \alpha=32α−4=2⇒α=3.

Therefore,

S={2,3}.S=\{2,3\}.S={2,3}.
  1. Compute the required sums

We need

∑α∈Sα=2+3=5.\sum_{\alpha\in S}\alpha =2+3=5.α∈S∑​α=2+3=5.

So

(∑α∈Sα)2=25.\left(\sum_{\alpha\in S}\alpha\right)^2=25.(α∈S∑​α)2=25.

Also,

∑α∈S(α+1)2=(2+1)2+(3+1)2=32+42=9+16=25.\sum_{\alpha\in S}(\alpha+1)^2=(2+1)^2+(3+1)^2=3^2+4^2=9+16=25.α∈S∑​(α+1)2=(2+1)2+(3+1)2=32+42=9+16=25.
  1. Form the quadratic equation

Given equation:

x2−2(∑α∈Sα)2x+(∑α∈S(α+1)2)β=0.x^2-2\left(\sum_{\alpha\in S}\alpha\right)^2x+\left(\sum_{\alpha\in S}(\alpha+1)^2\right)\beta=0.x2−2(α∈S∑​α)2x+(α∈S∑​(α+1)2)β=0.

Substituting the sums,

x2−2(25)x+25β=0,x^2-2(25)x+25\beta=0,x2−2(25)x+25β=0,

so

x2−50x+25β=0.x^2-50x+25\beta=0.x2−50x+25β=0.
  1. Condition for real roots

For the quadratic x2−50x+25β=0x^2-50x+25\beta=0x2−50x+25β=0 to have real roots, its discriminant must be non-negative:

Δ=(−50)2−4(1)(25β)≥0.\Delta = (-50)^2-4(1)(25\beta) \ge 0.Δ=(−50)2−4(1)(25β)≥0.

So

2500−100β≥0.2500-100\beta \ge 0.2500−100β≥0.

Thus,

β≤25.\beta \le 25.β≤25.

Hence the maximum value of β\betaβ is

25.\boxed{25}.25​.
  1. Compare with stored answer

Stored correct answer: 252525

Our derived answer matches it.

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