a1−2aa1−2a\frac{a}{{1 - 2a}} + b1−2b" role="presentation">b1−2bb1−2b\frac{b}{{1 - 2b}} + c1−2c" role="presentation">c1−2cc1−2c\frac{c}{{1 - 2c}} = 12," role="presentation">12,12,\frac{1}{2}{\text{,}} then the value of 11−2a" role="presentation">11−2a11−2a\frac{1}{{1 - 2a}} + 11−2b" role="presentation">11−2b11−2b\frac{1}{{1 - 2b}} + 11−2c" role="presentation">11−2c11−2c\frac{1}{{1 - 2c}} is?" /> a1−2aa1−2a\frac{a}{{1 - 2a}} + b1−2b" role="presentation">b1−2bb1−2b\frac{b}{{1 - 2b}} + c1−2c" role="presentation">c1−2cc1−2c\frac{c}{{1 - 2c}} = 12," role="presentation">12,12,\frac{1}{2}{\text{,}} then the value of 11−2a" role="presentation">11−2a11−2a\frac{1}{{1 - 2a}} + 11−2b" role="presentation">11−2b11−2b\frac{1}{{1 - 2b}} + 11−2c" role="presentation">11−2c11−2c\frac{1}{{1 - 2c}} is?" /> a1−2aa1−2a\frac{a}{{1 - 2a}} + b1−2b" role="presentation">b1−2bb1−2b\frac{b}{{1 - 2b}} + c1−2c" role="presentation">c1−2cc1−2c\frac{c}{{1 - 2c}} = 12," role="presentation">12,12,\frac{1}{2}{\text{,}} then the value of 11−2a" role="presentation">11−2a11−2a\frac{1}{{1 - 2a}} + 11−2b" role="presentation">11−2b11−2b\frac{1}{{1 - 2b}} + 11−2c" role="presentation">11−2c11−2c\frac{1}{{1 - 2c}} is?" />

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If $\frac{a}{1 - 2 a}$ $\frac{a}{{1 - 2a}}$ + $\frac{b}{1 - 2 b}$ $\frac{b}{{1 - 2b}}$ + $\frac{c}{1 - 2 c}$ $\frac{c}{{1 - 2c}}$ = $\frac{1}{2},$ $\frac{1}{2}{\text{,}}$ then the value of $\frac{1}{1 - 2 a}$ $\frac{1}{{1 - 2a}}$ + $\frac{1}{1 - 2 b}$ $\frac{1}{{1 - 2b}}$ + $\frac{1}{1 - 2 c}$ $\frac{1}{{1 - 2c}}$ is?

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} \frac{a}{1 - 2 a} + \frac{b}{1 - 2 b} + \frac{c}{1 - 2 c} = \frac{1}{2} \\ Multiply by 2 both side \\ \Rightarrow \frac{2 a}{1 - 2 a} + \frac{2 b}{1 - 2 b} + \frac{2 c}{1 - 2 c} = 1 \\ Adding 3 both side \\ \Rightarrow 1 + \frac{2 a}{1 - 2 a} + 1 + \frac{2 b}{1 - 2 b} + 1 + \frac{2 c}{1 - 2 c} = 1 + 3 \\ \Rightarrow \frac{1}{1 - 2 a} + \frac{1}{1 - 2 b} + \frac{1}{1 - 2 c} = 4 \end{aligned}

$\eqalign{ & \frac{a}{{1 - 2a}} + \frac{b}{{1 - 2b}} + \frac{c}{{1 - 2c}} = \frac{1}{2} \cr & {\text{Multiply by 2 both side}} \cr & \Rightarrow \frac{{2a}}{{1 - 2a}} + \frac{{2b}}{{1 - 2b}} + \frac{{2c}}{{1 - 2c}} = 1 \cr & {\text{Adding 3 both side}} \cr & \Rightarrow 1 + \frac{{2a}}{{1 - 2a}} + 1 + \frac{{2b}}{{1 - 2b}} + 1 + \frac{{2c}}{{1 - 2c}} = 1 + 3 \cr & \Rightarrow \frac{1}{{1 - 2a}} + \frac{1}{{1 - 2b}} + \frac{1}{{1 - 2c}} = 4 \cr}$

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+

\frac{b}{1 - 2 b}

$\frac{b}{{1 - 2b}}$ +

\frac{c}{1 - 2 c}

$\frac{c}{{1 - 2c}}$ =

\frac{1}{2},

$\frac{1}{2}{\text{,}}$ then the value of

\frac{1}{1 - 2 a}

$\frac{1}{{1 - 2a}}$ +

\frac{1}{1 - 2 b}

$\frac{1}{{1 - 2b}}$ +

\frac{1}{1 - 2 c}

$\frac{1}{{1 - 2c}}$ is?", "text": "If

\frac{a}{1 - 2 a}

$\frac{a}{{1 - 2a}}$ +

\frac{b}{1 - 2 b}

$\frac{b}{{1 - 2b}}$ +

\frac{c}{1 - 2 c}

$\frac{c}{{1 - 2c}}$ =

\frac{1}{2},

$\frac{1}{2}{\text{,}}$ then the value of

\frac{1}{1 - 2 a}

$\frac{1}{{1 - 2a}}$ +

\frac{1}{1 - 2 b}

$\frac{1}{{1 - 2b}}$ +

\frac{1}{1 - 2 c}

$\frac{1}{{1 - 2c}}$ is?", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} \frac{a}{1 - 2 a} + \frac{b}{1 - 2 b} + \frac{c}{1 - 2 c} = \frac{1}{2} \\ Multiply by 2 both side \\ \Rightarrow \frac{2 a}{1 - 2 a} + \frac{2 b}{1 - 2 b} + \frac{2 c}{1 - 2 c} = 1 \\ Adding 3 both side \\ \Rightarrow 1 + \frac{2 a}{1 - 2 a} + 1 + \frac{2 b}{1 - 2 b} + 1 + \frac{2 c}{1 - 2 c} = 1 + 3 \\ \Rightarrow \frac{1}{1 - 2 a} + \frac{1}{1 - 2 b} + \frac{1}{1 - 2 c} = 4 \end{aligned}

$\eqalign{ & \frac{a}{{1 - 2a}} + \frac{b}{{1 - 2b}} + \frac{c}{{1 - 2c}} = \frac{1}{2} \cr & {\text{Multiply by 2 both side}} \cr & \Rightarrow \frac{{2a}}{{1 - 2a}} + \frac{{2b}}{{1 - 2b}} + \frac{{2c}}{{1 - 2c}} = 1 \cr & {\text{Adding 3 both side}} \cr & \Rightarrow 1 + \frac{{2a}}{{1 - 2a}} + 1 + \frac{{2b}}{{1 - 2b}} + 1 + \frac{{2c}}{{1 - 2c}} = 1 + 3 \cr & \Rightarrow \frac{1}{{1 - 2a}} + \frac{1}{{1 - 2b}} + \frac{1}{{1 - 2c}} = 4 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} \frac{a}{1 - 2 a} + \frac{b}{1 - 2 b} + \frac{c}{1 - 2 c} = \frac{1}{2} \\ Multiply by 2 both side \\ \Rightarrow \frac{2 a}{1 - 2 a} + \frac{2 b}{1 - 2 b} + \frac{2 c}{1 - 2 c} = 1 \\ Adding 3 both side \\ \Rightarrow 1 + \frac{2 a}{1 - 2 a} + 1 + \frac{2 b}{1 - 2 b} + 1 + \frac{2 c}{1 - 2 c} = 1 + 3 \\ \Rightarrow \frac{1}{1 - 2 a} + \frac{1}{1 - 2 b} + \frac{1}{1 - 2 c} = 4 \end{aligned}

$\eqalign{ & \frac{a}{{1 - 2a}} + \frac{b}{{1 - 2b}} + \frac{c}{{1 - 2c}} = \frac{1}{2} \cr & {\text{Multiply by 2 both side}} \cr & \Rightarrow \frac{{2a}}{{1 - 2a}} + \frac{{2b}}{{1 - 2b}} + \frac{{2c}}{{1 - 2c}} = 1 \cr & {\text{Adding 3 both side}} \cr & \Rightarrow 1 + \frac{{2a}}{{1 - 2a}} + 1 + \frac{{2b}}{{1 - 2b}} + 1 + \frac{{2c}}{{1 - 2c}} = 1 + 3 \cr & \Rightarrow \frac{1}{{1 - 2a}} + \frac{1}{{1 - 2b}} + \frac{1}{{1 - 2c}} = 4 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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