Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
399 views
in Continuity and Differentiability by (29.1k points)
closed by

Find the value of c prescribed by Lagrange’s mean value theorem for the function f(x) = \(\sqrt{x^2 - 4}\) defined on [2, 3].

1 Answer

+1 vote
by (28.7k points)
selected by
 
Best answer

f(x) = \(\sqrt{x^2 - 4}\)

f(x) will exist, if

x2 – 4 ≥ 0

x2 ≥ 4

\(\therefore\) x ≤ -2 or x ≥ 2

\(\because\) for each x Є [2, 3], the function f(x) has a unique definite value, f(x) is continuous on (2, 3).

Exists for all x Є (2, 3).

So, f(x) is differentiable on (2,3).

Hence, both the conditions of Lagrange’s Theorem are satisfied.

Consequently, there exists c Є (2, 3) such that,

Squaring both sides,

Hence, c = \(\sqrt{5}\) Є (2, 3) such that f'(c) = \(\frac{f(3) - f(2)}{3 - 2}\)

Hence the above explanation verifies the Lagrange’s Theorem.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...