TS ICET 2015 Question Paper with Solutions for online practice

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Question : 90 of 200

Marks: +1, -0

If a ≠ b,

a^2 + b^2

≠ 0, then

{(a-b)(√a-√b)}/{(a+b)(√a+√b)}

1 - ${2√{ab}}/{a+b}$
1 + ${2√{ab}}/{a+b}$
1 + $2√{1/a+1/b}$
1 - $2√{1/a+1/b}$

Solution:

Given :

{(a-b)(√a-√b)}/{(a+b)(√a+√b)}

={(√{a}+√{b})(√{a}-√{b})(√{a}-√{b})}/{(a+b)(√{a}+√{b})}

[Using

x^{2}-y^{2}=(x+y)(x-y)

]

={(√{a}-√{b})^{2}}/{a+b}={a+b-2√{ab}}/{a+b}=1-2{√{ab}}/{a+b}

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