/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 In which of the following functi... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 9

In which of the following functions Rolle's theorem is applicable - (A) \(f(x)= \begin{cases}x & , 0 \leq x<1 \\ 0 & , x=1\end{cases}\) (B) $f(x)= \begin{cases}\frac{\sin x}{x},-\pi \leq x<0 \\ 0, & x=0\end{cases}$ (C) \(f(x)=\frac{x^{2}-x-6}{x-1}\) on \([-2,3]\) (D) $f(x)= \begin{cases}\frac{x^{3}-2 x^{2}-5 x+6}{x-1} & \text { if } x \neq 1, x \in[-2,3] \\ -6 & \text { if } x=1\end{cases}$

Short Answer

Expert verified
Answer: (B) is the only function for which Rolle's theorem is applicable.

Step by step solution

01

Check function (A)

For function (A), we have: \(f(x)= \begin{cases}x & , 0 \leq x<1 \\\ 0 & , x=1\end{cases}\) The function is continuous and differentiable in the given domain except at x = 1, where it has a discontinuity. Condition 1 is not satisfied, so Rolle's theorem is not applicable.
02

Check function (B)

For function (B), we have: \(f(x)= \begin{cases}\frac{\sin x}{x}, -\pi \leq x<0 \\\ 0, & x=0\end{cases}\) The function is continuous and differentiable for every point in its domain, and the function value f(-蟺) = f(0) = 0. All three conditions are satisfied, so Rolle's theorem is applicable for this function.
03

Check function (C)

For function (C), we have: \(f(x)=\frac{x^{2}-x-6}{x-1}\) on \([-2,3]\) This function is undefined at x = 1, but since we're considering the function only on the interval [-2, 3], we can ignore this point. The function is continuous and differentiable in the given interval. However, f(-2) = -5 and f(3) = 2, so Condition 3 is not satisfied. Consequently, Rolle's theorem is not applicable for this function.
04

Check function (D)

For function (D), we have: $f(x)= \begin{cases}\frac{x^{3}-2 x^{2}-5 x+6}{x-1} & \text { if } x \neq 1, x \in[-2,3] \\\ -6 & \text { if } x=1\end{cases}$ The function is continuous and differentiable for each point in [-2, 3] except for x = 1, where it is defined but not differentiable. Condition 2 is not satisfied, so Rolle's theorem is not applicable for this function. In conclusion, Rolle's theorem is applicable only for function (B).

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 91影视!

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

Consider a polynomial \(\mathrm{y}=\mathrm{P}(\mathrm{x})\) of the least degree passing through \(\mathrm{A}(-1,1)\) and whose graph has two points of inflexion \(\mathrm{B}(1,2)\) and \(\mathrm{C}\) with abscissa \(0 \mathrm{at}\) which the curve is inclined to the positive axis of abscissas at an angle of $\sec ^{-1} \sqrt{2}$. The area of \(\triangle \mathrm{ABC}\) equals (A) \(\frac{1}{2}\) (B) 1 (C) \(\frac{1}{8}\) (D) \(\frac{1}{12}\)

The set of value of \(\mathrm{c}\) for which $\sin \\{\ln (\cos \mathrm{x}+\mathrm{c})\\}=1$ has at most one solution in \([0, \pi]\) is (A) \((2 \pi, \infty)\) (B) \(\left(\mathrm{e}^{2 n}, \infty\right)\) (C) \(\left(\frac{\mathrm{e}^{2 \pi}+1}{\mathrm{e}^{2 \pi}-1}, \infty\right)\) (D) null set

If $f(\theta)=\frac{\sin ^{4} \frac{\theta}{n}-\sin ^{2} \frac{\theta}{n}+1}{\sin ^{4} \frac{\theta}{n}+\sin ^{2} \frac{\theta}{n}+1}$, where \(\theta \neq \mathrm{k} \mathrm{n} \pi .\) \(\mathrm{k} \in \mathrm{I}\), then range of \(f(\theta)\) is (A) \((1 / 3,1]\) (B) \([1 / 3,1]\) (C) \([1 / 3,1)\) (D) \((1 / 3,1)\)

Let \(f(x)=a x^{5}+b x^{4}+c x^{3}+d x^{2}+e x\), where \(a, b_{1} c, d, e \in R\) and \(f(x)=0\) has a positive root \(\alpha\), then (A) \(\mathrm{f}(\mathrm{x})=0\) has root al such that \(0<\alpha_{1}<\alpha\) (B) \(\mathrm{f}^{\prime}(\mathrm{x})=0\) has at least one real root (C) \(\mathrm{f}^{\prime}(\mathrm{x})=0\) has at least two real roots (D) All of the above

Let \(f(x)=x^{3}-3(7-a) x^{2}-3\left(9-a^{2}\right) x+2\) The values of a if \(f(x)\) has points of extrema which are of opposite signs, are (A) \(\bar{\phi}\) (B) \((-3,3)\) (C) \(\left(-\infty, \frac{58}{14}\right)\) (D) None of these
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Calculus

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation
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.