Use app×
QUIZARD
QUIZARD
JEE MAIN 2026 Crash Course
NEET 2026 Crash Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
1.5k views
in Sets, Relations and Functions by (93.6k points)
closed by
Find the domain and range of `f(x)=sqrt(4-16x^(2))`.

1 Answer

0 votes
by (94.6k points)
selected by
 
Best answer
Correct Answer - Domain: [-1/2, 1/2], Range: [0, 2]
`f(x)=y=sqrt(4-16x^(2))`
We must have `4-16x^(2) ge 0 implies x^(2) le 1//4`
`implies x in [-1//2,1//2]`
Hence domain is `[-1//2,1//2]`
Also, minimum value of function is 0 and maximum value 2 when `16x^(2)=4`
Hence range is `[0,2]`

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

...