    Question

# Find the range off(x)=sin−1[x2+12]+cos−1[x2−12] where [⋅]denotes the greatest integer function.

A
{π}
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
{2π}
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
{3π}
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
{7π}
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is A {π}Given function is, f(x)=cos−1[x2−12]+sin−1[x2+12]For this function to be defined −1≤[x2+12]≤1 and −1≤[x2−12]≤1or −1≤x2+12<2 and −1≤x2−12<2or −32≤x2<32 and −12≤x2<52 but x2≥0Taking intersection of above, domain of f(x) is x2∈[0,32)Now taking different cases,case 1. x2∈[0,12)f(x)=cos−1[x2−12]+sin−1[x2+12]=cos−1(−1)+sin−1(0)=π−0=πcase 2. x2∈[12,32)f(x)=cos−1[x2+12]+sin−1[x2−12]=cos−1(0)+sin−1(1)=π2+π2=πHence range of f(x) is only singleton set {π}  Suggest Corrections  1      Similar questions
Join BYJU'S Learning Program
Select...  Related Videos   Differentiation under Integral Sign
MATHEMATICS
Watch in App  Explore more
Join BYJU'S Learning Program
Select...