Hi,

Approach of

VeritasKarishma, is brilliant. The approach is very much logical. Its like calculations in brain.

However what if i needed to solve algebrically

So here it is

we are given that u(u=v)\(\neq{0}\)

means u\(\neq{0}\), (u+v) \(\neq{0}\) and u\(\neq{-v}\)

we are also given that u>0

we are asked it \(\frac{1}{u+v}< \frac{1}{u}+ v\)

Lets try to rephrase out question

Because u>0

\(\frac{1}{u+v}< \frac{1+ uv}{u}\)

\(\frac{1}{u+v}- \frac{1+ uv}{u}\) <0

\(\frac{u- (1+uv)(u+v)}{(u+v)}\) <0

\(\frac{u- (1+uv)(u+v)}{(u+v)}\) <0

\(\frac{u- u -v -uv(u+v)}{(u+v)}\) <0

\(\frac{ -v -uv(u+v)}{(u+v)}\) <0

multiplying both sides by -1 and flipping the sign

\(\frac{ v +uv(u+v)}{(u+v)}\) >0

if you want to identify what the question is really asking simplify further (though in above step you should have identified what is being asked)

Splitting the numerator and cancelling the common terms we get

\(\frac{ v}{(u+v)}\) +uv>0

So we see above inequality holds if we are told that v>0

So question really is IS v>0

So Statement 2 answers our question

This method is lengthy and not sure you should employ on the test day but if we start thinking what is being asked in DS question we can mark the answer confidently.

Probus