Let \(\mathrm{f}(\mathrm{x})=\tan \sqrt{\mathrm{m}} \cdot \mathrm{x}\) where \(\mathrm{m}=[\mathrm{P}]=\) greatest integer less then or equal to P and Principal period of \(\mathrm{f}(\mathrm{x})\) is \(\pi\). then (a) \(2 \leq \mathrm{P} \leq 3\) (b) \(1 \leq \mathrm{P} \leq 2\) (c) \(1 \leq \mathrm{P}<2\) (d) \(3 \leq \mathrm{P}<4\)

Short Answer

Expert verified
The correct answer is (c) \(1 \leq \mathrm{P}<2\).

Step by step solution

01

Understand the function

The given function is \( \mathrm{f}(\mathrm{x})=\tan \sqrt{\mathrm{m}} \cdot \mathrm{x} \), where \( \mathrm{m}=\left[\mathrm{P}\right] \), the greatest integer less than or equal to P.
02

Find the period for the given function

The period of this function will be \( \pi / \sqrt{m} \).
03

Apply the Principal Period condition

Given that the principal period of the function is \( \pi \), equate this to the period of the function. So, \( \pi / \sqrt{m} = \pi \). This implies \( m=1 \).
04

Find the range of P based on \( m=1 \)

Since \( m=[P] \) is the greatest integer less than or equal to P, the range of P that gives \( m=1 \) is \( 1 \leq \mathrm{P}<2 \). Hence the correct answer is (c) \(1 \leq \mathrm{P}<2\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on English Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free