Question

If \( \cot^{-1}(n)>\frac{\pi}{6} \), \( n \in N \), then the maximum value of n is

• A. 1
• B. 5
• C. 9
• D. 3

Hint:

Remember that the inverse cotangent function \( \cot^{-1}(x) \) has a range of \( (0, \pi) \) and is a decreasing function. When applying trigonometric functions to inequalities involving inverse trigonometric functions, pay attention to whether the function is increasing or decreasing to maintain the correct direction of the inequality.

Answer 1

The Correct Option is B

Step 1: Take the cotangent of both sides.
Since the cotangent function is decreasing in the interval \( (0, \pi) \) (which is the range of \( \cot^{-1}(x) \)), taking the cotangent of both sides reverses the inequality:
$$\cot(\cot^{-1}(n))<\cot\left(\frac{\pi}{6}\right)$$ Step 2: Evaluate \( \cot(\cot^{-1}(n)) \) and \( \cot\left(\frac{\pi}{6}\right) \).
We know that \( \cot(\cot^{-1}(n)) = n \).
Also, \( \cot\left(\frac{\pi}{6}\right) = \frac{\cos(\frac{\pi}{6})}{\sin(\frac{\pi}{6})} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} \). Step 3: Form the inequality.
The inequality becomes:
$$n<\sqrt{3}$$ Step 4: Approximate the value of \( \sqrt{3} \).
We know that \( \sqrt{3} \approx 1.732 \).
Step 5: Find the maximum value of \( n \) given that \( n \in N \) (n is a natural number).
Since \( n \) must be a natural number (positive integer) and \( n<1.732 \), the only natural number that satisfies this condition is \( n = 1 \).
There seems to be a discrepancy with the provided correct answer of 5. Let's re-examine the problem statement and my steps.
Re-evaluation:
If \( \cot^{-1}(n)>\frac{\pi}{6} \), then taking cotangent on both sides gives \( n<\cot(\frac{\pi}{6}) = \sqrt{3} \approx 1.732 \). The largest natural number \( n \) satisfying this is \( n = 1 \).
It's possible there's an error in the provided correct answer. Based on my analysis, the maximum value of \( n \) should be 1.