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
1.4k views
in Sets, Relations and Functions by (93.6k points)
closed by
Consider the real-valued function satisfying `2f(sinx)+f(cosx)=xdot` then the domain of `f(x)i sR` domain of `f(x)i s[-1,1]` range of `f(x)` is `[-(2pi)/3,pi/3]` range of `f(x)i sR`
A. domain of `f(x)` is R
B. domain of `f(x)" is " [-1,1]`
C. range of `f(x) " is " [-(2pi)/(3),(pi)/(3)]`
D. range of `f(x)` is R

1 Answer

0 votes
by (94.6k points)
selected by
 
Best answer
Correct Answer - B::C
Given `2f(sinx)+f(cosx)+x " (1)" `
Replacing x by `(pi)/(2)-x,` we get
`2f(cosx)+f(sinx)=(pi)/(2)-x " (2)" `
Eliminating `f(cosx)` from (1) and (2), we get
`3f(sinx)=3x-(pi)/(2)`
` or f(sinx)=x-(pi)/(2)`
` or f(x) ="sin"^(-1)x-(pi)/(6)`
`f(x)` has the domain [-1, 1].
Also, `sin^(-1)x in[-(pi)/(2),(pi)/(2)] or sin^(-1)x - (pi)/(6) in [-(2pi)/(3),(pi)/(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

...