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Minimum of alpha + beta for (log3 x)^2 - 6 log3 x + k = 0 | JEE Advanced

JEE Maths question with a full step-by-step solution.

Question
If α\alpha and β\beta are real roots of
(log⁡3x)2−6log⁡3x+k=0,k∈R,\left(\log_3 x\right)^{2} - 6\log_3 x + k = 0, \qquad k \in \mathbb{R},
 then the minimum integral value of (α+β)(\alpha + \beta) is
A5353
B5454correct
C5555
D5656
Solution
Step 1: Substitute to make it an ordinary quadratic. Put t=log⁡3xt = \log_3 x:
t2−6t+k=0.t^{2} - 6t + k = 0 .
 Step 2: Solve for tt.
t=6±36−4k2=3±9−k,t = \frac{6 \pm \sqrt{36 - 4k}}{2} = 3 \pm \sqrt{9-k} ,
 which is real provided 9−k≥09 - k \ge 0. Step 3: Go back to xx. Since t=log⁡3xt = \log_3 x means x=3tx = 3^{t},
α=3 3+9−k,β=3 3−9−k.\alpha = 3^{\,3+\sqrt{9-k}}, \qquad \beta = 3^{\,3-\sqrt{9-k}} .
 Step 4: Find the product, which is pleasantly free of kk.
αβ=3(3+9−k)+(3−9−k)=36=729.\alpha\beta = 3^{\left(3+\sqrt{9-k}\right)+\left(3-\sqrt{9-k}\right)} = 3^{6} = 729 .
 Step 5: Apply AM ≥\ge GM to the two positive numbers α\alpha and β\beta.
α+β2≥αβ=729=27  ⟹  α+β≥54.\frac{\alpha+\beta}{2} \ge \sqrt{\alpha\beta} = \sqrt{729} = 27 \implies \alpha + \beta \ge 54 .
 Step 6: Check that 5454 is actually reached. Equality in AM ≥\ge GM needs α=β\alpha = \beta, i.e. 9−k=0\sqrt{9-k} = 0, i.e. k=9k = 9. Then α=β=33=27\alpha = \beta = 3^{3} = 27 and
α+β=27+27=54.\alpha + \beta = 27 + 27 = 54.
 Step 7: So the smallest value of α+β\alpha+\beta is 5454, and it is already an integer. Answer: (2).
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