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C. \(f(a + b)=f(5)=\sqrt{5}\neq{f(a) + f(b)=f(2)+f(3)=\sqrt{2}+\sqrt{3}}\)

D. \(f(a + b)=f(5)=\frac{2}{5}\neq{f(a)+f(b)=f(2)+f(3)=\frac{2}{2}+\frac{2}{3}=\frac{5}{3}}\)

E. \(f(a + b)=f(5)=-3*(5) =-15=f(a)+f(b)=f(2)+f(3)=-3*(2)-3*(3)=-15\). Correct.

It might happen that for some choices of a and b other options may be "correct" as well. If this happens just pick some other numbers and check again these "correct" options only.

Thank you Bunuel. This is really a good explaination. I did some samples based on your explaination and i am confident that I can handle these kind or problems.

C. \(f(a + b)=f(5)=\sqrt{5}\neq{f(a) + f(b)=f(2)+f(3)=\sqrt{2}+\sqrt{3}}\)

D. \(f(a + b)=f(5)=\frac{2}{5}\neq{f(a)+f(b)=f(2)+f(3)=\frac{2}{2}+\frac{2}{3}=\frac{5}{3}}\)

E. \(f(a + b)=f(5)=-3*(5) =-15=f(a)+f(b)=f(2)+f(3)=-3*(2)-3*(3)=-15\). Correct.

It might happen that for some choices of a and b other options may be "correct" as well. If this happens just pick some other numbers and check again these "correct" options only.

Hope it helps.

Thanks bunuel You have solved my problem here. i think i can handle these questions now +1
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C. \(f(a + b)=f(5)=\sqrt{5}\neq{f(a) + f(b)=f(2)+f(3)=\sqrt{2}+\sqrt{3}}\)

D. \(f(a + b)=f(5)=\frac{2}{5}\neq{f(a)+f(b)=f(2)+f(3)=\frac{2}{2}+\frac{2}{3}=\frac{5}{3}}\)

E. \(f(a + b)=f(5)=-3*(5) =-15=f(a)+f(b)=f(2)+f(3)=-3*(2)-3*(3)=-15\). Correct.

It might happen that for some choices of a and b other options may be "correct" as well. If this happens just pick some other numbers and check again these "correct" options only.

Hope it helps.

thanks a ton for the explanation. There are not many problems on functions in samples and I am glad now i know how to go about for questions ike this.

i think Karishma has explained very good point to tackle these functions problems, As per her explanation we should first try options with multiple , divide add and then subtract…. So trying option E was obvious and it fits well… saves time indeed…

C. \(f(a + b)=f(5)=\sqrt{5}\neq{f(a) + f(b)=f(2)+f(3)=\sqrt{2}+\sqrt{3}}\)

D. \(f(a + b)=f(5)=\frac{2}{5}\neq{f(a)+f(b)=f(2)+f(3)=\frac{2}{2}+\frac{2}{3}=\frac{5}{3}}\)

E. \(f(a + b)=f(5)=-3*(5) =-15=f(a)+f(b)=f(2)+f(3)=-3*(2)-3*(3)=-15\). Correct.

It might happen that for some choices of a and b other options may be "correct" as well. If this happens just pick some other numbers and check again these "correct" options only.

Hope it helps.

Thanks for really fleshing out the algebra on this problem Bunuel. The problems seem fairly easy once you understand how to work functions properly.

Well, i did not start with picking no's or keeping the no's as a and b,

But just worked in the following way :- if, f(a+b) = f(a) + f(b) Then, since our answers are in this format,

f(x+x) = f(x) + f(x) should also hold true, for all positive no's a and b (we havent been told that it has to be distinct, so this should be perfectly valid)

Hence, you need to evaluate f(2x) here, and check weather it is equal to 2f(x)

This seems to be the case only for E, hence, the answer. Quick and easy.
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PS: Like my approach? Please Help me with some Kudos.

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