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Logarithmic Inequality with tan30° Base for |z| | JEE

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Question
If log⁡tan⁡30° ⁣(2∣z∣2+2∣z∣−3∣z∣+1)<−2\log_{\tan 30°}\!\left(\dfrac{2|z|^2 + 2|z| - 3}{|z| + 1}\right) < -2, then
A∣z∣<32|z| < \dfrac{3}{2}
B∣z∣>32|z| > \dfrac{3}{2}
C∣z∣>2|z| > 2correct
D∣z∣<2|z| < 2
Solution
Step 1: Examine the base. tan⁡30°=13≈0.577\tan 30° = \dfrac{1}{\sqrt 3} \approx 0.577, which is between 00 and 11. For a logarithm with base in (0,1)(0, 1), the function is decreasing, so when we remove the log the inequality sign reverses. Step 2: Convert log⁡bX<−2\log_b X < -2 (with 0<b<10 < b < 1) to X>b−2X > b^{-2}:
2∣z∣2+2∣z∣−3∣z∣+1>(tan⁡30°)−2=(13)−2=3\frac{2|z|^2 + 2|z| - 3}{|z| + 1} > (\tan 30°)^{-2} = \left(\frac{1}{\sqrt 3}\right)^{-2} = 3
 Step 3: Since ∣z∣≥0|z| \geq 0, the denominator ∣z∣+1>0|z| + 1 > 0, so we may multiply across without flipping the sign:
2∣z∣2+2∣z∣−3>3(∣z∣+1)=3∣z∣+32|z|^2 + 2|z| - 3 > 3(|z| + 1) = 3|z| + 3
2∣z∣2+2∣z∣−3−3∣z∣−3>0  ⟹  2∣z∣2−∣z∣−6>02|z|^2 + 2|z| - 3 - 3|z| - 3 > 0 \implies 2|z|^2 - |z| - 6 > 0
 Step 4: Factorise the quadratic in ∣z∣|z|. Solve 2t2−t−6=02t^2 - t - 6 = 0: t=1±1+484=1±74=2 or −32t = \dfrac{1 \pm \sqrt{1 + 48}}{4} = \dfrac{1 \pm 7}{4} = 2 \text{ or } -\dfrac{3}{2}. So:
2∣z∣2−∣z∣−6=(∣z∣−2) (2∣z∣+3)>02|z|^2 - |z| - 6 = (|z| - 2)\,(2|z| + 3) > 0
 Step 5: Analyse the sign. Since ∣z∣≥0|z| \geq 0, the factor 2∣z∣+3>02|z| + 3 > 0 always. Therefore the product is positive exactly when:
∣z∣−2>0  ⟹  ∣z∣>2|z| - 2 > 0 \implies |z| > 2
 (Also check the domain of the original log: at ∣z∣>2|z| > 2 the argument is positive, so the logarithm is defined.) Correct answer: (3)
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