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
544 views
in Derivatives by (29.4k points)
closed by

If f(x) = \(\frac{1}{4x^2+2x+1}\), then its maximum value is :

A. \(\frac{4}{3}\)

B. \(\frac{2}{3}\)

C. 1 

D. \(\frac{3}{4}\)

1 Answer

+1 vote
by (28.3k points)
selected by
 
Best answer

Option : (A)

f(x) = \(\frac{1}{4x^2+2x+1}\) achieves it’s maximum value when g(x) = 4x2+2x+1 achieves it’s minimum value.

Differentiating g(x) with respect to x, we get

g’(x) = 8x+2 

Differentiating g’(x) with respect to x, we get

g’’(x) = 8 

For minima at x = c, 

g’(c) = 0 and g’’(c) > 0 g’(x) = 0

⇒ x = \(-\frac{1}{4}\)

g''(\(-\frac{1}{4}\)) = 8>0

Hence,

x = \(-\frac{1}{4}\) is a point of minima for g(x) and g(\(-\frac{1}{4}\)) = \(\frac{3}{4}\)is the minimum value of g(x). 

Hence the maximum value of f(x) = \(\frac{1}{g(x)}\) = \(\frac{4}{3}\)

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

...