Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • provide solution for RD Sharma maths class 12 chapter Mean Value Theoram exercise 14.2 question 1 sub question (x)

provide solution for RD Sharma maths class 12 chapter Mean Value Theoram exercise 14.2 question 1 sub question (x)

Answers (1)

Answer:  \sqrt{\frac{4}{\pi}-1}

Hint: You must know the formula of Lagrange’s Mean Value Theorem.

Given: f(x)=\tan ^{-1} x \text { on }[0,1]

Solution:

f(x)=\tan ^{-1} x

f(x)  is a polynomial function.

It is continuous in [0, 1] 

f^{\prime}(x)=\frac{1}{\left(1+x^{2}\right)}

(Which is defined in [0, 1])

\therefore f(x) is differentiable in [0, 1]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.

Now, there exists at least one value,

\begin{aligned} &c \in[0,1] \\ &f^{\prime}(c)=\frac{f(1)-f(0)}{1-0} \\ &\frac{1}{1+c^{2}}=\frac{\tan ^{-1}(1)-\tan ^{-1}(0)}{1} \end{aligned}

\begin{aligned} &\frac{1}{1+c^{2}}=\frac{\frac{\pi}{4}-0}{1} \\ &\frac{1}{1+c^{2}}=\frac{\pi}{4} \end{aligned}

\begin{aligned} &\pi\left(1+c^{2}\right)=4 \\ &\pi+\pi c^{2}=4 \\ &\pi c^{2}=4-\pi \end{aligned}

\begin{aligned} &c^{2}=\frac{4-\pi}{\pi} \\ &c=\sqrt{\frac{4-\pi}{\pi}} \end{aligned}

c=\sqrt{\frac{4}{\pi}-1}

Posted by

infoexpert26

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

Board Exam Time Table 2025 Live: UP Board exams from February 24; CBSE, BSEB date sheet soon
anam-khan

CBSE Date Sheet 2025 Live: Class 10, 12 board exam dates soon; syllabus, sample papers
anam-khan

Extra time in CBSE exams for students with type 1 diabetes: Kerala SHRC seeks report on plea

Latest Question