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
685 views
in Sets, relations and functions by (15 points)
edited by

f(x) = log7(log5(log3(log2(2x3 + 5x2 - 14x))))

Please log in or register to answer this question.

1 Answer

0 votes
by (41.6k points)

f(x) = log7(log5(log3(log2(2x3 + 5x2 - 14x))))

To defined function f(x), we have

log5(log3(log2(2x3 + 5x2 - 14x))) > 0

⇒ log3(log2(2x3 + 5x2 - 14x)) >5° or 1

⇒ log2(2x3 + 5x2 - 14x) > 31 or 3 ( ∵ log is increasing function)

⇒ 2x3 + 5x2 - 14x >23 or 8

⇒ 2x3 + 5x2 - 14x - 8 > 0

⇒ (x - 2) (2x2 + 9x + 4) > 0

⇒ (x - 2) (x + 4) (2x + 1) > 0

Domain of given function f is (-4, -1/2) U (2, \(\infty\))

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.

...