Question

Let $\mathrm{f}(\mathrm{x})=\lim _{\mathrm{n} \rightarrow \infty} \frac{1}{\left(\frac{3}{\pi} \tan ^{-1} 2 \mathrm{x}\right)^{2 \mathrm{n}}+5}$. Then the set of values of \(x\) for which \(f(x)=0\), is (A) \(|2 x|>\sqrt{3}\) (B) \(|(2 x)|<\sqrt{3}\) (C) \(|2 x| \geq \sqrt{3}\) (D) \(|2 x| \leq \sqrt{3}\)

Short answer

Answer: (B) \(|(2 x)|<\sqrt{3}\).

Step-by-step solution

Analyze the given function

The given function is $f(x)=\lim _{n \rightarrow \infty} \frac{1}{\left(\frac{3}{\pi} \tan ^{-1} 2 x\right)^{2 n}+5}\(. For \)f(x) = 0\(, the denominator should tend to infinity as \)n \rightarrow \infty$. So, let's analyze the denominator.

Analyze the denominator of the function

The denominator is given by $\left(\frac{3}{\pi} \tan ^{-1} 2 x\right)^{2 n}+5$. Consider the term inside the bracket, which is \(\frac{3}{\pi} \tan ^{-1} 2 x\). Since the arctangent function has a range of \((-\pi / 2,\pi / 2)\), the maximum value of \(\tan ^{-1} 2 x\) is \(\pi / 2\). Hence, the maximum value of \(\frac{3}{\pi} \tan ^{-1} 2 x\) is 1.5. Therefore, for the denominator to tend to infinity, \(0<(\frac{3}{\pi} \tan ^{-1} 2 x)<1\).

Find the range of x from the condition on the denominator

We've found that \(0<(\frac{3}{\pi} \tan ^{-1} 2 x)<1\). We need to find the range of \(x\) that satisfies this condition. To do so, let's first multiply all parts of the inequality by \(\frac{\pi}{3}\): \(0 < \tan ^{-1} 2 x < \frac{\pi}{3}\) Now, we will take the tangent of all parts of the inequality, which gives: \(0 < 2x < \tan{\frac{\pi}{3}}\) Since \(\tan{\frac{\pi}{3}} = \sqrt{3}\), we obtain: \(0<2 x<\sqrt{3}\) Dividing by 2 gives: \(0<x<\frac{\sqrt{3}}{2}\) Since \(f(x)\) is symmetric around the y-axis, we can extend the range to both negative and positive values of x: \(-\frac{\sqrt{3}}{2}<x<\frac{\sqrt{3}}{2}\) Thus, \(|2x| < \sqrt{3}\). So, the correct answer is option (B) \(|(2 x)|<\sqrt{3}\).