ax−a+ay−ais = ?ax−a+ay−ais = ?\frac{a}{{x - a}} + \frac{a}{{y - a}}\,{\text{is = ?}}" /> ax−a+ay−ais = ?ax−a+ay−ais = ?\frac{a}{{x - a}} + \frac{a}{{y - a}}\,{\text{is = ?}}" /> ax−a+ay−ais = ?ax−a+ay−ais = ?\frac{a}{{x - a}} + \frac{a}{{y - a}}\,{\text{is = ?}}" />

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If x + y = 2a, then the value of $\frac{a}{x - a} + \frac{a}{y - a} is = ?$ $\frac{a}{{x - a}} + \frac{a}{{y - a}}\,{\text{is = ?}}$

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} Given, \\ x + y = 2 a \\ ∴ \frac{a}{x - a} + \frac{a}{y - a} \\ = \frac{a (y - a) + a (x - a)}{(x - a) (y - a)} \\ = \frac{a y - a^{2} + a x - a^{2}}{(x - a) (y - a)} \\ = \frac{a (x + y) - 2 a^{2}}{(x - a) (y - a)} \\ = \frac{a .2 a - 2 a^{2}}{(x - a) (y - a)} \\ = \frac{2 a^{2} - 2 a^{2}}{(x - a) (y - a)} \\ = 0 \end{aligned}

$\eqalign{ & {\text{Given, }} \cr & x + y = 2a \cr & \therefore \frac{a}{{x - a}} + \frac{a}{{y - a}} \cr & = \frac{{a\left( {y - a} \right) + a\left( {x - a} \right)}}{{\left( {x - a} \right)\left( {y - a} \right)}} \cr & = \frac{{ay - {a^2} + ax - {a^2}}}{{\left( {x - a} \right)\left( {y - a} \right)}} \cr & = \frac{{a\left( {x + y} \right) - 2{a^2}}}{{\left( {x - a} \right)\left( {y - a} \right)}} \cr & = \frac{{a.2a - 2{a^2}}}{{\left( {x - a} \right)\left( {y - a} \right)}} \cr & = \frac{{2{a^2} - 2{a^2}}}{{\left( {x - a} \right)\left( {y - a} \right)}} \cr & = 0 \cr}$

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", "text": "If x + y = 2a, then the value of

\frac{a}{x - a} + \frac{a}{y - a} is  =  ?

$\frac{a}{{x - a}} + \frac{a}{{y - a}}\,{\text{is = ?}}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} Given, \\ x + y = 2 a \\ ∴ \frac{a}{x - a} + \frac{a}{y - a} \\ = \frac{a (y - a) + a (x - a)}{(x - a) (y - a)} \\ = \frac{a y - a^{2} + a x - a^{2}}{(x - a) (y - a)} \\ = \frac{a (x + y) - 2 a^{2}}{(x - a) (y - a)} \\ = \frac{a .2 a - 2 a^{2}}{(x - a) (y - a)} \\ = \frac{2 a^{2} - 2 a^{2}}{(x - a) (y - a)} \\ = 0 \end{aligned}

$\eqalign{ & {\text{Given, }} \cr & x + y = 2a \cr & \therefore \frac{a}{{x - a}} + \frac{a}{{y - a}} \cr & = \frac{{a\left( {y - a} \right) + a\left( {x - a} \right)}}{{\left( {x - a} \right)\left( {y - a} \right)}} \cr & = \frac{{ay - {a^2} + ax - {a^2}}}{{\left( {x - a} \right)\left( {y - a} \right)}} \cr & = \frac{{a\left( {x + y} \right) - 2{a^2}}}{{\left( {x - a} \right)\left( {y - a} \right)}} \cr & = \frac{{a.2a - 2{a^2}}}{{\left( {x - a} \right)\left( {y - a} \right)}} \cr & = \frac{{2{a^2} - 2{a^2}}}{{\left( {x - a} \right)\left( {y - a} \right)}} \cr & = 0 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} Given, \\ x + y = 2 a \\ ∴ \frac{a}{x - a} + \frac{a}{y - a} \\ = \frac{a (y - a) + a (x - a)}{(x - a) (y - a)} \\ = \frac{a y - a^{2} + a x - a^{2}}{(x - a) (y - a)} \\ = \frac{a (x + y) - 2 a^{2}}{(x - a) (y - a)} \\ = \frac{a .2 a - 2 a^{2}}{(x - a) (y - a)} \\ = \frac{2 a^{2} - 2 a^{2}}{(x - a) (y - a)} \\ = 0 \end{aligned}

$\eqalign{ & {\text{Given, }} \cr & x + y = 2a \cr & \therefore \frac{a}{{x - a}} + \frac{a}{{y - a}} \cr & = \frac{{a\left( {y - a} \right) + a\left( {x - a} \right)}}{{\left( {x - a} \right)\left( {y - a} \right)}} \cr & = \frac{{ay - {a^2} + ax - {a^2}}}{{\left( {x - a} \right)\left( {y - a} \right)}} \cr & = \frac{{a\left( {x + y} \right) - 2{a^2}}}{{\left( {x - a} \right)\left( {y - a} \right)}} \cr & = \frac{{a.2a - 2{a^2}}}{{\left( {x - a} \right)\left( {y - a} \right)}} \cr & = \frac{{2{a^2} - 2{a^2}}}{{\left( {x - a} \right)\left( {y - a} \right)}} \cr & = 0 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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