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

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

Question

\log_{\tan 30°}\!\left(\dfrac{2|z|^2 + 2|z| - 3}{|z| + 1}\right) < -2

, then

|z| < \dfrac{3}{2}

|z| > \dfrac{3}{2}

|z| > 2

correct

|z| < 2

Solution

Step 1: Examine the base.

\tan 30° = \dfrac{1}{\sqrt 3} \approx 0.577

, which is between

0

and

1

. For a logarithm with base in

(0, 1)

, the function is decreasing, so when we remove the log the inequality sign reverses. Step 2: Convert

\log_b X < -2

(with

0 < b < 1

) to

X > b^{-2}

\frac{2|z|^2 + 2|z| - 3}{|z| + 1} > (\tan 30°)^{-2} = \left(\frac{1}{\sqrt 3}\right)^{-2} = 3

Step 3: Since

|z| \geq 0

, the denominator

|z| + 1 > 0

, so we may multiply across without flipping the sign:

2|z|^2 + 2|z| - 3 > 3(|z| + 1) = 3|z| + 3

2|z|^2 + 2|z| - 3 - 3|z| - 3 > 0 \implies 2|z|^2 - |z| - 6 > 0

Step 4: Factorise the quadratic in

|z|

. Solve

2t^2 - t - 6 = 0

t = \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) > 0

Step 5: Analyse the sign. Since

|z| \geq 0

, the factor

2|z| + 3 > 0

always. Therefore the product is positive exactly when:

|z| - 2 > 0 \implies |z| > 2

(Also check the domain of the original log: at

|z| > 2

the argument is positive, so the logarithm is defined.) Correct answer: (3)

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