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
243 views
in Exponents by (20 points)
edited by

The solution set of inequality, \( \frac{2^{x-1}-1}{2^{x+1}+1}<2 \), contains

(A) all real numbers

(B) finite number of integers

(C) no negative number

(D) no real number

Please log in or register to answer this question.

1 Answer

0 votes
by (53.5k points)

Correct option is (A) All real numbers

\(\frac{2^{x -1} -1}{2^{x + 1} + 1} < 2\)

⇒ \(\frac{2^{x -1} -1}{2^{x + 1} + 1} -2 < 0\)

⇒ \(\frac{2^{x -1} - 1 -2.2^{x + 1} -2}{2^{x +1} + 1}< 0\)

⇒ \(2^{x-1}-2^{x +2}- 3< 0\)   (∵ \(2^{x+1}+1 > 0\) (always))

⇒ \(\frac{2^x}2 - 4.2^x - 3< 0\)

⇒ \(\frac{-7}2.2^x - 3<0\)

⇒ \(\frac 72 . 2^x > -3\)

⇒ \(2^x > \frac{-6}7\)

which is true for all real values of x.

No related questions found

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.

...