tan^{-1} frac{x}{pi}

Question

(B) Given the inequality: $	an^{-1} \frac{x}{\pi} < \frac{\pi}{3}$.Since the function $f(t) = 	an(t)$ is strictly increasing for $t \in (-\frac{\pi}

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) Given the inequality: $	an^{-1} \frac{x}{\pi} Since the function $f(t) = 	an(t)$ is strictly increasing for $t \in (-\frac{\pi}{2}, \frac{\pi}{2})$,we can take the tangent of both sides:$	an(	an^{-1} \frac{x}{\pi}) $\frac{x}{\pi} $x Using the approximation $\sqrt{3} \approx 1.732$ and $\pi \approx 3.14159$:$x Since $x \in N$ (the set of natural numbers),the largest integer $x$ satisfying $x Therefore,the maximum value of $x$ is $5$.

$tan^{-1} \frac{x}{\pi} < \frac{\pi}{3}, x \in N$,then the maximum value of $x$ is :-

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