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A log equation with a hidden identity collapses to (m-n)^2 + (n-1)^2 = 0 - JEE Advanced

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

Question
Let mm, nn be real numbers satisfying the relation
log⁡2(m2+n2+1)+log⁡3(1+2sin⁡2(2π7)2−cos⁡(4π7))=log⁡2n+log⁡2(2m+2−n).\log_2\left(m^2+n^2+1\right)+\log_3\left(\frac{1+2\sin^2\left(\frac{2\pi}{7}\right)}{2-\cos\left(\frac{4\pi}{7}\right)}\right) = \log_2 n+\log_2\left(2m+2-n\right).
 Identify which of the following statement(s) is/are correct?
Am+n=2m+n = 2
B∣ ∣m∣−∣n∣ ∣=4\left|\,|m|-|n|\,\right| = 4
CArea of the figure enclosed by ∣x∣+∣y∣=∣m∣+∣n∣|x|+|y| = |m|+|n| is 88 square unitscorrect
DArea of the figure enclosed by ∣x∣+∣y∣=∣m∣+∣n∣|x|+|y| = |m|+|n| is 1616 square unitscorrect
Solution
Step 1: Simplify the middle logarithm using a double-angle identity.
2sin⁡2θ=1−cos⁡2θwithθ=2π7  ⟹  2sin⁡22π7=1−cos⁡4π7.2\sin^2\theta = 1-\cos2\theta \quad\text{with}\quad \theta = \frac{2\pi}{7} \;\Longrightarrow\; 2\sin^2\frac{2\pi}{7} = 1-\cos\frac{4\pi}{7}.
 Therefore
1+2sin⁡22π7=2−cos⁡4π7,1+2\sin^2\frac{2\pi}{7} = 2-\cos\frac{4\pi}{7},
 so the fraction inside log⁡3\log_3 is exactly 11, and
log⁡3(1)=0.\log_3(1) = 0 .
 Step 2: The relation reduces to a single base-2 statement.
log⁡2(m2+n2+1)=log⁡2n+log⁡2(2m+2−n)=log⁡2[n(2m+2−n)].\log_2\left(m^2+n^2+1\right) = \log_2 n+\log_2\left(2m+2-n\right) = \log_2\left[n\left(2m+2-n\right)\right].
 Step 3: Remove the logarithms (both sides are defined, so the arguments are positive and equal).
m2+n2+1=2mn+2n−n2.m^2+n^2+1 = 2mn+2n-n^2 .
 Step 4: Bring everything to one side and look for squares.
m2+2n2−2mn−2n+1=0,m^2+2n^2-2mn-2n+1 = 0 ,
(m2−2mn+n2)+(n2−2n+1)=0,\left(m^2-2mn+n^2\right)+\left(n^2-2n+1\right) = 0 ,
(m−n)2+(n−1)2=0.\left(m-n\right)^2+\left(n-1\right)^2 = 0 .
 Step 5: A sum of two real squares vanishes only if each does.
m=nandn=1  ⟹  m=n=1.m = n \quad\text{and}\quad n = 1 \;\Longrightarrow\; m = n = 1 .
 Check the domain: n=1>0n = 1>0 and 2m+2−n=2+2−1=3>02m+2-n = 2+2-1 = 3>0 , so all three logarithms are defined. Step 6: Test options (1) and (2).
m+n=2 - true;∣ ∣m∣−∣n∣ ∣=∣1−1∣=0≠4 - false.m+n = 2 \ \textbf{- true};\qquad \left|\,|m|-|n|\,\right| = |1-1| = 0 \ne 4 \ \textbf{- false}.
 Step 7: Test options (3) and (4). ∣m∣+∣n∣=2|m|+|n| = 2, so the figure is ∣x∣+∣y∣=2|x|+|y| = 2. This is a square whose diagonals lie along the axes and have length 44 each. For a rhombus (here a square) the area is half the product of the diagonals:
Area=12×4×4=8 square units.\text{Area} = \frac12\times4\times4 = 8 \ \text{square units}.
 (Equivalently, ∣x∣+∣y∣=a|x|+|y| = a encloses area 2a2=2(2)2=82a^2 = 2(2)^2 = 8.) **(3) is correct; (4) is false.**
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