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Question

Let f:XY be a function such that f(x)=x2+4x. If f(x) is both injective as well as surjective, then set X and Y is

A
[2,4] and [2,2]
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B
[3,4] and [2,2]
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C
[2,4] and [1,2]
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D
[2,3] and [1,2]
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Solution

The correct option is B [3,4] and [2,2]
f(x)=x2+4x
For the square root to exist,
x20x24x0x4
So, Df[2,4]
Now,
f(x)=012x2124x=0(x2)=4xx=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]

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