1(1−a)(1−b)(1−c)1(1−a)(1−b)(1−c)\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}} + 1(1−b)(1−c)(1−d)" role="presentation">1(1−b)(1−c)(1−d)1(1−b)(1−c)(1−d)\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} + 1(1−c)(1−d)(1−a)" role="presentation">1(1−c)(1−d)(1−a)1(1−c)(1−d)(1−a)\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}} + 1(1−d)(1−a)(1−b)" role="presentation">1(1−d)(1−a)(1−b)1(1−d)(1−a)(1−b)\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}} is?" /> 1(1−a)(1−b)(1−c)1(1−a)(1−b)(1−c)\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}} + 1(1−b)(1−c)(1−d)" role="presentation">1(1−b)(1−c)(1−d)1(1−b)(1−c)(1−d)\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} + 1(1−c)(1−d)(1−a)" role="presentation">1(1−c)(1−d)(1−a)1(1−c)(1−d)(1−a)\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}} + 1(1−d)(1−a)(1−b)" role="presentation">1(1−d)(1−a)(1−b)1(1−d)(1−a)(1−b)\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}} is?" /> 1(1−a)(1−b)(1−c)1(1−a)(1−b)(1−c)\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}} + 1(1−b)(1−c)(1−d)" role="presentation">1(1−b)(1−c)(1−d)1(1−b)(1−c)(1−d)\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} + 1(1−c)(1−d)(1−a)" role="presentation">1(1−c)(1−d)(1−a)1(1−c)(1−d)(1−a)\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}} + 1(1−d)(1−a)(1−b)" role="presentation">1(1−d)(1−a)(1−b)1(1−d)(1−a)(1−b)\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}} is?" />

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If a + b + c + d = 4, then the value of $\frac{1}{(1 - a) (1 - b) (1 - c)}$ $\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$ + $\frac{1}{(1 - b) (1 - c) (1 - d)}$ $\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$ + $\frac{1}{(1 - c) (1 - d) (1 - a)}$ $\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$ + $\frac{1}{(1 - d) (1 - a) (1 - b)}$ $\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$ is?

10
21
34
41 + abcd

Answer:- 1

Explanation:-

Solution:

\begin{aligned} \frac{1}{(1 - a) (1 - b) (1 - c)} + \frac{1}{(1 - b) (1 - c) (1 - d)} + \frac{1}{(1 - c) (1 - d) (1 - a)} + \frac{1}{(1 - d) (1 - a) (1 - b)} \\ = \frac{1 - d + 1 - a + 1 - b + 1 - c}{(1 - a) (1 - b) (1 - c) (1 - d)} \\ = \frac{4 - (a + b + c + d)}{(1 - a) (1 - b) (1 - c) (1 - d)} \\ = \frac{4 - 4}{(1 - a) (1 - b) (1 - c) (1 - d)} \\ = 0 \end{aligned}

$\eqalign{ & \frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}} + \frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} + \frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}} + \frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}} \cr & = \frac{{1 - d + 1 - a + 1 - b + 1 - c}}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} \cr & = \frac{{4 - \left( {a + b + c + d} \right)}}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} \cr & = \frac{{4 - 4}}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} \cr & = 0 \cr}$

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+

\frac{1}{(1 - b) (1 - c) (1 - d)}

$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$ +

\frac{1}{(1 - c) (1 - d) (1 - a)}

$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$ +

\frac{1}{(1 - d) (1 - a) (1 - b)}

$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$ is?", "text": "If a + b + c + d = 4, then the value of

\frac{1}{(1 - a) (1 - b) (1 - c)}

$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$ +

\frac{1}{(1 - b) (1 - c) (1 - d)}

$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$ +

\frac{1}{(1 - c) (1 - d) (1 - a)}

$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$ +

\frac{1}{(1 - d) (1 - a) (1 - b)}

$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$ 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{1}{(1 - a) (1 - b) (1 - c)} + \frac{1}{(1 - b) (1 - c) (1 - d)} + \frac{1}{(1 - c) (1 - d) (1 - a)} + \frac{1}{(1 - d) (1 - a) (1 - b)} \\ = \frac{1 - d + 1 - a + 1 - b + 1 - c}{(1 - a) (1 - b) (1 - c) (1 - d)} \\ = \frac{4 - (a + b + c + d)}{(1 - a) (1 - b) (1 - c) (1 - d)} \\ = \frac{4 - 4}{(1 - a) (1 - b) (1 - c) (1 - d)} \\ = 0 \end{aligned}

$\eqalign{ & \frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}} + \frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} + \frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}} + \frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}} \cr & = \frac{{1 - d + 1 - a + 1 - b + 1 - c}}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} \cr & = \frac{{4 - \left( {a + b + c + d} \right)}}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} \cr & = \frac{{4 - 4}}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} \cr & = 0 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} \frac{1}{(1 - a) (1 - b) (1 - c)} + \frac{1}{(1 - b) (1 - c) (1 - d)} + \frac{1}{(1 - c) (1 - d) (1 - a)} + \frac{1}{(1 - d) (1 - a) (1 - b)} \\ = \frac{1 - d + 1 - a + 1 - b + 1 - c}{(1 - a) (1 - b) (1 - c) (1 - d)} \\ = \frac{4 - (a + b + c + d)}{(1 - a) (1 - b) (1 - c) (1 - d)} \\ = \frac{4 - 4}{(1 - a) (1 - b) (1 - c) (1 - d)} \\ = 0 \end{aligned}

$\eqalign{ & \frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}} + \frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} + \frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}} + \frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}} \cr & = \frac{{1 - d + 1 - a + 1 - b + 1 - c}}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} \cr & = \frac{{4 - \left( {a + b + c + d} \right)}}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} \cr & = \frac{{4 - 4}}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} \cr & = 0 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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