A) Neither one-one nor onto
B) One-one
C) Onto
D) One-one onto
Correct Answer: A
Solution :
Let \[{{x}_{1}},\,{{x}_{2}}\in R,\] then \[f({{x}_{1}})=\cos {{x}_{1}}\], \[f({{x}_{2}})=\cos {{x}_{2}}\], so \[f({{x}_{1}})=f({{x}_{2}})\] Þ \[\cos {{x}_{1}}=\cos {{x}_{2}}\] Þ \[{{x}_{1}}=2n\pi \pm {{x}_{2}}\] Þ \[{{x}_{1}}\ne {{x}_{2}}\], so it is not one-one. Again the value of f-image of x lies in between ?1 to 1 Þ \[f[R]=\left\{ f(x):-1\le f(x)\le 1) \right\}\] So other numbers of co-domain (besides ?1 and 1) is not f-image. \[f[R]\in R,\] so it is also not onto. So this mapping is neither one-one nor onto.
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