jlgdr wrote:

Any other approach for this problem over here? I did not quite get your explanation Bunuel, how can we quickly eliminate the other answer choices?

Thanks!

Cheers

J

One good thing of this question is numerator of all the expressions is 1

Now focus on denominators

More the value of denominator, less would be the sum

Least the denominator, more would be the sum value

\(\sqrt{2} = 2^{\frac{1}{2}}\)

\(2^{\frac{1}{2}} < 2^2\)

Reciprocal both sides

\(\frac{1}{2^{\frac{1}{2}}} > \frac{1}{2^2}\)

(Greater/smaller sign changes for reciprocals/negation)

Same for all other fractions

So highest addition would be of option A

Answer = A

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