Question

# Let f:X→Y be a function such that f(x)=√x−2+√4−x. If f(x) is both injective as well as surjective, then set X and Y is

A
[2,4] and [2,2]
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
[3,4] and [2,2]
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
[2,4] and [1,2]
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
[2,3] and [1,2]
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is B [3,4] and [√2,2] f(x)=√x−2+√4−x For the square root to exist, x−2≥0⇒x≥24−x≥0⇒x≤4 So, Df∈[2,4] Now, f′(x)=0⇒12√x−2−12√4−x=0⇒(x−2)=4−x⇒x=3 f(3)=2,f(2)=√2,f(4)=√2, The graph of the function, ∴ Range [√2,2] Now, for the function to be both injective and surjective the domain has to be restricted in either [2,3] or [3,4]. Hence, from the given options, X=[3,4], Y=[√2,2]

Suggest Corrections
0
Explore more