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Let f : R → R and g : R → R be two one - one and onto functions such that they are mirror images of each other about the line y = 0, then h (x) = f(x) + g(x) is
  • a)
    one - one and onto
  • b)
    one - one but not onto
  • c)
    not one - one but onto
  • d)
    None of these
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
Let f : R → R and g : R → R be two one - one and onto functi...
f : R → R & g : R →R be two one-one onto functions such that f & g are mirror images of each other about line                                            y = 0. It means one is –ve of the other
i.e. f(x) = – g(x)
⇒ f(x) + g(x) = 0
⇒ h(x) = 0
h(x) is not onto as well as not one-one
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Most Upvoted Answer
Let f : R → R and g : R → R be two one - one and onto functi...

Explanation:

Given:
Let f : R → R and g : R → R be two one - one and onto functions such that they are mirror images of each other about the line y = 0.

Analysis:
When two functions are mirror images of each other about the line y = 0, it implies that f(x) = -g(x) for all x in the domain.

h(x) = f(x) + g(x)
Since f(x) = -g(x), h(x) = f(x) + g(x) = -g(x) + g(x) = 0 for all x in the domain.

Therefore, h(x) is a constant function with h(x) = 0 for all x.

Conclusion:
Since h(x) is a constant function, it is not one-to-one as multiple inputs will map to the same output (0 in this case). Additionally, it is not onto as the range of h(x) is {0}, which is not the entire real number line. Therefore, the correct answer is option 'D' - None of these.
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