A) one-one onto
B) one-one into
C) many-one onto
D) many-one into
Correct Answer: D
Solution :
Since \[f(x)\] and \[g(x)\]are mirror images of each other about the line\[y=a,\]\[f(x)\]and \[g(x)\]are at equal distances from the line \[y=a\], Let for some particular \[{{x}_{0}}\] |
\[f\left( {{x}_{0}} \right)=a+k,\operatorname{then}\,g\text{ }\left( {{x}_{0}} \right)=a-k,\]then \[h\left( {{x}_{0}} \right)=f\left( {{x}_{0}} \right)+g\left( {{x}_{0}} \right)=2a\] |
\[\therefore h\left( x \right)=2a\,\,\forall \,\,x\in R.\]So,\[h(x)\] must be a constant function, which is many -one into. |
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