The function f:R→R given by f(x)=cosx for all x∈R, is :
A
surjective bu nor injective
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B
injective but nor surjective
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C
neither injective nor surjective
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D
injective
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Solution
The correct option is C neither injective nor surjective Injectivity : We know that f(0)=cos0=1 and f(2π)=cos2π=1.
So, different elements in R may have the same image. Hence, f is not an injection.
Surjectivity : Since the values of cosx lie between −1 and 1, it follows that the range of f(x) is not equal to its co-domain. So, f is not a surjection.