Question

Find the maximum value of n for which $cot^{-1}\frac{n}{\pi}>\frac{\pi}{6}, n \epsilon N$.

• A. 1
• B. 3
• C. 5
• D. 9

Answer 1

Answer: (C)

5

Answer 2

The given inequality is $cot^{-1}\frac{n}{\pi}>\frac{\pi}{6}$. We need to find the maximum natural number $n$ that satisfies this inequality.

The domain of $cot^{-1}x$ is $(-\infty, \infty)$ and its range is $(0, \pi)$. Since $n \in N$, $n$ is a positive integer, so $\frac{n}{\pi}$ is positive. For positive arguments, the range of $cot^{-1}x$ is $(0, \frac{\pi}{2})$. The value $\frac{\pi}{6}$ lies within this range.

The function $f(x) = cot^{-1}x$ is a strictly decreasing function. This means that if $cot^{-1}A > cot^{-1}B$, then $A < B$.

We know the exact value of $cot(\frac{\pi}{6})$: $cot(\frac{\pi}{6}) = \sqrt{3}$ Therefore, $\frac{\pi}{6} = cot^{-1}(\sqrt{3})$.

Substitute this into the given inequality: $cot^{-1}\frac{n}{\pi}>cot^{-1}(\sqrt{3})$

Since $cot^{-1}x$ is a decreasing function, we can remove the $cot^{-1}$ from both sides and reverse the inequality sign: $\frac{n}{\pi} < \sqrt{3}$

Now, we need to solve for $n$: $n < \pi\sqrt{3}$

To find the maximum integer value of $n$, we need to approximate the value of $\pi\sqrt{3}$. Using approximate values: $\pi \approx 3.14159$ $\sqrt{3} \approx 1.73205$

Multiply these values: $\pi\sqrt{3} \approx 3.14159 \times 1.73205 \approx 5.44139$

So, the inequality becomes: $n < 5.44139$

Since $n$ must be a natural number ($n \in N$, which means $n$ is a positive integer), the possible integer values for $n$ that satisfy this condition are $1, 2, 3, 4, 5$.

The maximum value of $n$ among these integers is $5$.