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Question

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

A
injective but not surjective.
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B
surjective but not injective.
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C
neither injective nor 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
for checking whether function is injective
f(x)=(1+x2)(1)(x)(0+2x)(1+x2)2
f(x)=x2+12x2(1+x2)2
f(x)=1x21+x2
We Know that , for a function to be one-one its derivative has to always on same side of zero i.e always positive or always negative f(x)=1x21+x2
so it is neither always +ve , nor always - ve.
f(x) is not injective.
for checking whether function is surjective.
y=x1+x2
x2y+yx=0
x=+1±14y22y
we need to make D0
14y20
(12y)(1+2y)0
(2y1)(2y+1)0
y[1212]
Range of f(x) is same as codomain, So Surjective.

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