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Question

The function f:R[12,12] defined as f(x)=x1+x2, is

A
neither injective nor surjective
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B
surjective but not injective
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C
injective but not surjective
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D
invertible
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Solution

The correct option is B surjective but not injective
f(x)=x1+x2
Let f(x1)=f(x2), x1,x2R
x11+x21=x21+x22
x11+x21x21+x22=0
(x1x2)(1x1x2)(1+x21)(1+x21)=0
x1=x2, x1=1x2
Thus, f is not injective.

Let y=x1+x2
yx2x+y=0
x=1±14y22y
xR, 14y20 & y0
y[12,12]{0}
Also, for x=0, y=0
Rf=[12,12]=Co-domain of f

Alternatively, put x=tanθ, θ(π2,π2)
Then y=12sin2θ
1sin2θ1
12y12
Thus, f is surjective.

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