A) 0
B) 1
C) 4
D) \[1+abcd\]
Correct Answer: A
Solution :
[a] \[\frac{1}{(1-a)(1-b)(1-c)}+\frac{1}{(1-b)(1-c)(1-d)}\] \[+\frac{1}{(1-c)(1-d)(1-d)}+\frac{1}{(1-d)(1-a)(1-b)}\] \[=\frac{1-d+1-a+1-b+1-c}{(1-a)(1-b)(1-c)(1-d)}\] \[=\frac{(4-a-b-c-d)}{(1-a)(1-b)(1-c)(1-d)}\] \[=\frac{a+b+c+d-a-b-c-d}{(1-a)(1-b)(1-c)(1-d)}\] \[[\therefore a+b+c+d=4]\] \[=\frac{0}{(1-a)(1-b)(1-c)(1-d)}=0\] |
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